How do you explain to a 5th grader why division by zero is meaningless?
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I wish to explain my younger brother: he is interested and curious, but he cannot grasp the concepts of limits and integration just yet. What is the best mathematical way to justify not allowing division by zero?
divisibility education
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I wish to explain my younger brother: he is interested and curious, but he cannot grasp the concepts of limits and integration just yet. What is the best mathematical way to justify not allowing division by zero?
divisibility education
63
How do you divide, say, $5$ apples between zero people? There is no meaningful way to distribute the apples.
â Alvin Lepik
Aug 15 at 10:36
12
Division by zero should not be justified at all.
â Peter
Aug 15 at 11:39
23
Just ask Siri. It explains this very well.
â DonielF
Aug 15 at 15:32
12
This would also fit well on the Mathematics Educators site.
â Chase Ryan Taylor
Aug 15 at 16:10
21
"Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends." - Siri
â jkd
Aug 15 at 19:12
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up vote
77
down vote
favorite
up vote
77
down vote
favorite
I wish to explain my younger brother: he is interested and curious, but he cannot grasp the concepts of limits and integration just yet. What is the best mathematical way to justify not allowing division by zero?
divisibility education
I wish to explain my younger brother: he is interested and curious, but he cannot grasp the concepts of limits and integration just yet. What is the best mathematical way to justify not allowing division by zero?
divisibility education
edited yesterday
SQB
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1,6241926
asked Aug 15 at 10:33
Shubh Khandelwal
537128
537128
63
How do you divide, say, $5$ apples between zero people? There is no meaningful way to distribute the apples.
â Alvin Lepik
Aug 15 at 10:36
12
Division by zero should not be justified at all.
â Peter
Aug 15 at 11:39
23
Just ask Siri. It explains this very well.
â DonielF
Aug 15 at 15:32
12
This would also fit well on the Mathematics Educators site.
â Chase Ryan Taylor
Aug 15 at 16:10
21
"Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends." - Siri
â jkd
Aug 15 at 19:12
 |Â
show 17 more comments
63
How do you divide, say, $5$ apples between zero people? There is no meaningful way to distribute the apples.
â Alvin Lepik
Aug 15 at 10:36
12
Division by zero should not be justified at all.
â Peter
Aug 15 at 11:39
23
Just ask Siri. It explains this very well.
â DonielF
Aug 15 at 15:32
12
This would also fit well on the Mathematics Educators site.
â Chase Ryan Taylor
Aug 15 at 16:10
21
"Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends." - Siri
â jkd
Aug 15 at 19:12
63
63
How do you divide, say, $5$ apples between zero people? There is no meaningful way to distribute the apples.
â Alvin Lepik
Aug 15 at 10:36
How do you divide, say, $5$ apples between zero people? There is no meaningful way to distribute the apples.
â Alvin Lepik
Aug 15 at 10:36
12
12
Division by zero should not be justified at all.
â Peter
Aug 15 at 11:39
Division by zero should not be justified at all.
â Peter
Aug 15 at 11:39
23
23
Just ask Siri. It explains this very well.
â DonielF
Aug 15 at 15:32
Just ask Siri. It explains this very well.
â DonielF
Aug 15 at 15:32
12
12
This would also fit well on the Mathematics Educators site.
â Chase Ryan Taylor
Aug 15 at 16:10
This would also fit well on the Mathematics Educators site.
â Chase Ryan Taylor
Aug 15 at 16:10
21
21
"Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends." - Siri
â jkd
Aug 15 at 19:12
"Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends." - Siri
â jkd
Aug 15 at 19:12
 |Â
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18 Answers
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âÂÂOne of the ways to look at division is as how many of the smaller number you need to make up the bigger number, right? So 20/4 means: how many groups of 4 do you need to make 20? If you want 20 apples, how many bags of 4 apples do you need to buy?
So for dividing by 0, how many bags of 0 apples would make up 20 apples in total? ItâÂÂs impossible â however many bags of 0 apples you buy, youâÂÂll never get any apples â youâÂÂll certainly never get to 20 apples! So thereâÂÂs no possible answer, when you try to divide 20 by 0.âÂÂ
62
A good enough answer till the little brother grows up and asks why it can't be $+infty$
â Peeyush Kushwaha
Aug 15 at 13:17
36
@PeeyushKushwaha, then you introduce them to a number system where 1/0 is $infty$
â ilkkachu
Aug 15 at 13:22
43
@JackM: If you look at my answer carefully, youâÂÂll see I really am describing division as the inverse of multiplication (specifically, in âÂÂ), and arguing that 0 has no multiplicative inverse. IâÂÂm just presenting it concretely, because to just about anyone short of a mathematically fairly mature undergraduate, thatâÂÂs clearer and more convincing than a formal algebraic proof. This isnâÂÂt âÂÂdumbing downâ either â the real-world understanding of division has just as much claim to being the âÂÂreal thingâÂÂ, and mathematics was being done well for centuries without modern notions of definition and proof.
â Peter LeFanu Lumsdaine
Aug 15 at 15:14
8
@ctrl-alt-delor, that's why I said to pick a system where it works: en.wikipedia.org/wiki/Riemann_sphere
â ilkkachu
Aug 15 at 15:36
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@Kimball: I wanted to conjure up the image of OP talking to their younger brother, rather than of me addressing OP.
â Peter LeFanu Lumsdaine
2 days ago
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When we first start teaching multiplication, we use successive additions. So,
3 x 4 = 3 | 3
+ 3 | 6
+ 3 | 9
+ 3 | 12
=12
Division can be taught as successive subtractions. So 12 / 3 becomes,
12 - 3 -> 9 (1)
9 - 3 -> 6 (2)
6 - 3 -> 3 (3)
3 - 3 -> 0 (4)
Now apply the second algorithm with zero as a divisor. Tell your brother to get back to you when he's done.
While this algorithmic approach is not rigorous, I think it is probably a good way of developing an intuitive understanding of the concept.
5
I should comment that in some schools of thought, teaching young students that multiplication is repeated addition can be hindering. I've certainly seen kids who only count on their fingers - they can't actually multiply, just multiple add. So there are arguments from Ring Theory to education where you may be wrong, although it is perfectly reasonable to say in the integers.
â theREALyumdub
Aug 15 at 13:49
31
I can't comment on the first part, but for the second part; We're talking about a 9 year old. One of the problems with mathematical pedagogy is we prioritize rigour over understanding. I suggest we provide the understanding first, then make it rigourous. After all, we have thousands of years of mathematical development before the hard rigour came in in the 19th century.
â Chris Cudmore
Aug 15 at 13:54
2
I agree completely, and I have hit walls with mathematics involving rigour and a lack of understanding. I was more making the point that this sounds much like a computerized argument, and I have blindly heard of education arguments against this method of approach for young children - it's more or less my best understanding of division, but it can be made more abstract and perhaps more practical for education.
â theREALyumdub
Aug 15 at 13:58
3
@theREALyumdub, the ring theorist will appreciate that every abelian group is a $mathbb Z$-module in a natural way.
â Carsten S
Aug 15 at 17:55
5
@ChrisCudmore Seeing as most college educated people never get a rigorous definition of almost any of the mathematical concepts they were taught, I don't think an over-emphasis on rigor is the problem. You seem to be confusing calculation with rigor. Frankly, I don't think the wave of formalization had almost any impact on how math was/is taught, at least pre-college. I strongly suspect that there is even more of an emphasis of intuitive understanding in grade school classrooms now than in the 1810s say.
â Derek Elkins
Aug 15 at 19:13
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New story
Suppose that we can divide numbers with $0$. So if I would divide $1$ with zero i would get some new number name it $a$. Now what can we say about this number $a$?
Remember:
If I divide say $21$ with $3$ we get $7$. Why? Because $3cdot 7 = 21$.
And similiary if I divide $36$ with $9$ we get $4$. Why? Because $9cdot 4 = 36$.
So if I divide $1$ with $0$ and we get $a$ then we have $acdot 0 =1$ which is clearly nonsense since $acdot 0 =0$.
(I hope this is better explanation?)
Older explanation:
Suppose that $1over 0$ is some number $a$. So $$1over 0 =a.$$ Remember that $$boxedbover c = diff b = ccdot d$$ So we get $$1= acdot 0=0$$ a contradiction. So $1over 0$ doesn't exist.
88
5th grader.....
â Lightness Races in Orbit
Aug 15 at 18:02
3
I see a lot of people doesn't like this answer although it is perfectly correct. I agree it is perhaps to advance for 5th grader but then again how else could I do it? There is a nice way Peter gave, but what if brother asks what is 5/4. How many bags with 4 apples do we need to get 5 apples?
â greedoid
2 days ago
3
The logic proposed in this answer seems constructive, but a plain-English explanation would probably be more helpful. Separately, it's probably best to avoid saying that $frac10$ "doesn't exist"; it'd seem a bit better to say that $frac10$ doesn't cleanly match up with a number.
â Nat
2 days ago
1
You have to be careful when declaring contradictions. If you arrive at a "contradiction", it's actually a consequence of asserting the truth of the statements used. If you arrive at $12 = 0$ as a result, it really is just asserting a mod 12 system. if you accept $frac10 = a$ as a statement, then you are in a mod 1 system as a consequence. Suppose that you assert a statement that only after millions of manipulations asserts that $0 = 2^32$. Not only is it not a "contradiction", but it leads to the most general solution. On a computer, you have to deal with this frequently.
â Rob
2 days ago
1
I really think we should be striving to get to a point where the 'older argument' is something a fifth grader can comprehend.
â Prince M
2 days ago
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An explanation that might make sense to a fifth grader is one that gets to the heart of why we have invented these operations in the first place.
Multiplication is a trick we use to add similar things to form a sum. When we say 5 x 3, what we really mean is take five things of size three each and add them all together. We invented this trick because we are frequently in the situation where we have many of a similar thing, and we wish to know their sum.
Division is the same trick but the other way. When we say 15 / 3, we are asking the question "how many times would we have to add a thing of size three starting from nothing to make a thing of size fifteen?" We'd have to add five things of size three together to make a thing of size fifteen. Again, division is just a trick we use to answer questions about sums.
Now it becomes clear why division by zero is not defined. There is no number of times you can add zero to itself to get a non-zero sum.
A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!
For non-zero divided by zero, there is no number at all of times that you can add zero to itself to get non-zero. For zero divided by zero, every number of times you add zero to itself, you get zero, so the solution is not unique. We like our mathematical questions to have unique answers where possible and so we by convention say that 0 / 0 is also not defined.
3
Actually I have never thought of division like that. 15/2 has always meant 'how big is each half when you cut it in two' to me.
â Sentinel
2 days ago
2
"how many times would we have to add a thing of size three to itself to make a thing of size fifteen?" 3+3=6; 6+3=9; 9+3=12; 12+3=15. I count four additions. Your question is therefore not worded correctly. It's not adding three to itself, it's "Starting with zero (nothing), how many 3s do you need to add to get 15?"
â Monty Harder
2 days ago
@MontyHarder: Good point; I'll change the wording.
â Eric Lippert
2 days ago
2
"A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!" Simple enough. Grab a plate and set it on the table and say "there are 0 cookies on this plate". Grab another and set it down. "there are 0 cookies on each plate, for a total of 0 cookies. If you leave the room and all you know is that there are 0 total cookies, how can you know how many empty plates I put on the table (or picked back up) when you were gone?"
â Monty Harder
2 days ago
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11
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The Wikipedia article Division by zero lists the usual arguments why there is no good choice for the result of such an operation.
I prefer the algebraic argument, that there is no multiplicative inverse of $0$,
this would need you to explain a bit about algebra.
The argument from calculus, looking at limits of $1/x$, I find also useful, but perphaps harder to explain.
1
To me this is the answer, though I would combine it with "let the 5th grader try to come up with some ideas, and help them see why they fail." Personally, I like this approach because then, when they come across the sqrt(-1), they're going to be more comfortable when we say "actually, there is a good choice for how to deal with this."
â Cort Ammon
yesterday
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7
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How many nothings do you need to add together to get 12?
4
Add all the nothings!
â Don Branson
Aug 15 at 19:54
4
@Sentinel Ok, Donnie. I did that. I found all the nothings and added them, but I still have nothing. What now?
â Sentinel
2 days ago
Not in 5th grade, maybe, but by 7th or 8th I think I might have asked "what if you added an infinite number of nothings?" Running into the same issue as some other explainations.
â Ian D. Scott
yesterday
@DonBranson Of course. Mathematicians have done for 4 centuries already... Infinitesimals.. :) Imagine integrating constant function with value 12 over $[0,1]$ or value 1 over $[0,12]$
â mathreadler
yesterday
@IanD.Scott infinitesimals and calculus. That is why you should not tell kids that/why you can't divide by 0. Playing with the idea of multiplying "almost 0" with infinity is so fruitful it is more valuable if they wrestle around a bit with it for themselves. It is not inconceivable that a kid could come up with some variant of infinitesimal calculus based on adding infinite many very small slices.
â mathreadler
yesterday
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Division by zero is meaningless because that's what we decided division means. All you can do is explain why such a convention is a useful one for ordinary arithmetic.
It might even help to demonstrate some other context (e.g. arithmetic in the projectively extended number line) where it can be useful to define division by zero, so that the student is able to compare and contrast the reasons why we might or might not like to define something.
Your question might be better placed on https://matheducators.stackexchange.com/
I believe your first sentence is circular logic
â Ruadhan2300
Aug 15 at 15:59
@Ruadhan2300 No, it's not circular. I think Hurkyl's point is that all mathematical concepts are just definitions that we decided on. You could define division differently, as $1/0 = 37$ and still develop all of modern mathematics; it would only be less convenient, not less "correct".
â 6005
yesterday
That being said, I think this meta-explanation may be a bit too difficult to grasp for a 5th grader. At that stage, most students think of definitions as immutable truth.
â 6005
yesterday
Much like most of science, all mathematics is rooted in modelling reality. The rules and concepts we produce exist entirely because we found they apply to real situations and remain consistently effective. I maintain that "because that's the definition we gave it" is circular and unhelpful.
â Ruadhan2300
yesterday
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5
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Ask Siri.
Imagine that you have zero cookies and you split them evenly among
zero friends. How many cookies does each person get? See? It doesn't
make sense. And Cookie Monster is sad that there are no cookies, and
you are sad that you have no friends.
5
If you include your imaginary friend, then the general form for dividing $X$ cookies is $-Xi$, because $(0+i) * (-Xi) = X$
â Chronocidal
Aug 15 at 22:43
That is like saying that division by two is slicing evenly in half. Most other answers here think it is about groups of two.
â Sentinel
yesterday
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You shouldn't try to do that. Instead make counter question.
"What should it be, then?" and let them think about it.
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I don't have kids (my wife says one 3-year-old in the house is enough for her) and it's been a while since I was in the 5th grade (although at work sometimes...), but I'll give it a go.
I know you're too old to play with blocks, but lets start with 12 blocks.
Let's start with $12/6$ - that's $2$, right? Take $6$ at a time and there are two "sets". There are $2$ sets of $6$ in $12$.
Then $12/4$ is $3$ - $3$ sets of $4$ in $12$.
Then $12/3$ is $4$ - $4$ sets of $3$ in $12$ (commutation of the last case).
Then $12/2$ is $6$ - $2$ sets of $6$ in $12$ (commutation of first case).
Then $12/1$ is $12$ - $1$ set of $12$ in $12$ (degenerate case).
Notice the size of the result set is getting bigger as the denominator (the number on the bottom) gets smaller.
Before we go to $0$ let's try something between $1$ and $0$ - $1/2$ or $0.5$. Think of just splitting each block into two (take a hatchet to the wooden blocks blocks, or just imagine it if mom doesn't want you handling a hatchet).
$12/0.5$ is $24$ - $24$ sets of $0.5$ (half-pieces) in $12$
$12/0.25$ is $48 - 48$ sets of $0.25$ (quarter-pieces) in $12$
$12/0.125$ is $96 - 96$ sets of $0.125$ (pieces of eight**) in $12$
$12/0.0625$ is $192 - 192$ sets of $0.0625$ (pieces of 16) in $12$
The close you get to zero, the larger the set you get gets.
$12/0.000000001$ (a billionth) is $12$ billion sets of a billionth of a block (aka, sawdust)
The as you approach zero, the resulting set size is too large to represent (not enough paper in this room, not enough memory on this computer) and the size of the pieces approach zero.
A cheat for "Too large to represent" is "infinity".
** pirate reference - do 5th graders still like pirates these days?
The problem I think is a conceptual one, As you describe, it's actually talking about sets, and while you can have a single set containing everything, you can't logically have less sets than you started with. "I want you to put all of these objects into a box too small for any of them" would be a similar task. The only appropriate response is to take Exception at the impossible task.
â Ruadhan2300
Aug 15 at 14:30
This is how my grandma explained it to me when I was like 5. I got it immediately and it stuck with me ever since.
â htmlcoderexe
2 days ago
A "piece of eight" (or, in what is perhaps a more "piraty" version, "piece o' eight") was a Spanish coin that was worth eight Spanish reales. So the piece of eight was the whole, while the real was the eighth. Another term for a real was a "bit". This survives in a bit being one-eighth of a byte, and in "shave and a hair-cut, two bits [i.e. 25 cents]". So you could say "half-piece, quarter-piece, bit-piece".
â Acccumulation
2 days ago
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One would need to first explain what we mean by division. That is, what does $/$ mean in the expression $a/b,$ where $a$ and $b$ are integers?
Well, whatever it is, it is a way of combining two numbers. Now recall that every time we defined an operation (say addition), we always had a unique result as the product of the combination, so that we would like this to continue to hold. What else? We define $/$ indirectly, by looking at what we want $a/b$ to mean. Well, we want it to stand for the number $c$ which when multiplied together with $b$ recovers $a.$ (Recall how we similarly defined subtraction as the inverse operation of $+.$)
Therefore, in summary, if we let $a/b=c,$ then by definition this equality is equivalent to $cÃÂb=a.$ Also, we want $c$ to be unique for all possible integers $a$ and $b.$
Now consider the expression $a/0.$ First let us take $ane0.$ Then if we let $a/0=c,$ it follows by definition that $cÃÂ0=a.$ But with the way we defined multiplication (remind him of this), we required that $0$ must make any number vanish, so that there simply is no such $c$ as we seek. If now we let $a=0,$ then we want a unique $c$ such that $cÃÂ0=0.$ But again, by the property $rÃÂ0=0,,,forall r$ which we've previously allowed in defining $ÃÂ,$ we have infinitely many candidates for $c$ and there is no other condition we can impose to select one uniquely. We therefore do not allow ourselves to divide by $0$ in any case, in order to avoid all that mess.
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@Jack M and and @greedoid probably highlight a good point: division does not exist. It's only the inverse operation of multiplication.
You could explain your brother the complete truth: dividing 20 by 5 is about finding the only answer (if it exists) to this question: what number can be multiplied by 5 to give 20?. The unique answer is easy: 4 times 5 is 20.
And the division is only another phrasing to say the exact same thing: 20 divided by 5 is 4.
Can you always find one and only one answer? Yup, almost always...
There's only one exception...
What number, multiplied by 0, gives 20? There's none.
So "division" by 0 has no meaning, since we cannot find any number that satisfies our definition.
You could even draw his attention by mentioning that most grown-ups don't know there's no such thing as "division", and that's the first step to learn about "E-vector spaces", "rings" and other funny-named artefacts when he's in college... or before that!
Note: what if he raises a question about "0/0"?
OK, let's try: "what number, multiplied by 0, gives 0?" All of them! We cannot find one and only one answer, so, it's still impossible to divide 0 by 0!
There are a lot of buzzwords for a 5th grader, but the explanation is good.
â Alvin Lepik
2 days ago
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2
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Division is sharing:
1 / 10
:
10 boys in a class grab at a toy -- they rip the toy to tiny bits!
1 / 2
:
2 boys fight for a toy -- they rip the toy in half!
1 / 0
:
A different toy is alone -- he is a special boy!
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Because before you think about dividing something, it is more important to consider if you have someone to divide it for (he/she/it must be present, exist, etc). If you do not have anyone who can 'benefit' from the division, no point in dividing. Non rigorous, pragmatic, heuristic approach. It might pave the way for more reasoned proofs and demonstrations.
This is indeed similar with the 'division is sharing' concept in the answer of @Jason given here
â XavierStuvw
yesterday
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1
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The way I taught it, even to junior college students who were taking elementary mathematics courses, was with a calculator.
I would show them that 1/1 = 1, 1/0.1 = 10, 1/0.01 = 100, and so on. I would ask them if they saw how the numbers kept getting bigger as we divided by smaller and smaller numbers. Then I would ask them what they thought would happen when we hit zero. "We would get the biggest possible number that exists, right? But there is no biggest number. So dividing by zero gives you a number that doesn't exist. Does that make any sense? No. So we say that dividing by zero is undefined."
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To divide means to subtract many times.
It might be a duplicated answer and I apologize, in case. But, according to my experience as a teacher, this worked well.
The point, as others had observed, is what does "to divide" mean. This sometimes looked obscure to the students, whereas the concept of subtraction was more clear.
Thus, once you convey the message that "to divide" means "to subtract many times", everything becomes more clear.
How many times can we subtract $3$ from $10$? Well, usually my students got this.
How many times can we subtract $0$ from $10$? Well, how many times we want!
So there is not a precise answer, because any answer is good. This made more clear the sense of "not defined", at least to my students.
Hope it helps!
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Explain him the problems, don't enforce him as an "official view".
Explain him, what are the problems of the division by zero.
Let him to think about a possible solution.
You might also explain, that also the negative numbers don't have a suqare root, but this problem had a solution, the imaginary numbers. Let him try to think about a similar solution for the division by zero.
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Number of marbles : Number of boxes = Number of marbles in each box.
20 marbles : 4 boxes = 5 marbles per box
0 marbles : 4 boxes = 0 marbles per box
20 marbles : 0 boxes = "how many marbles in each box while no box?" ---> undefined!
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âÂÂOne of the ways to look at division is as how many of the smaller number you need to make up the bigger number, right? So 20/4 means: how many groups of 4 do you need to make 20? If you want 20 apples, how many bags of 4 apples do you need to buy?
So for dividing by 0, how many bags of 0 apples would make up 20 apples in total? ItâÂÂs impossible â however many bags of 0 apples you buy, youâÂÂll never get any apples â youâÂÂll certainly never get to 20 apples! So thereâÂÂs no possible answer, when you try to divide 20 by 0.âÂÂ
62
A good enough answer till the little brother grows up and asks why it can't be $+infty$
â Peeyush Kushwaha
Aug 15 at 13:17
36
@PeeyushKushwaha, then you introduce them to a number system where 1/0 is $infty$
â ilkkachu
Aug 15 at 13:22
43
@JackM: If you look at my answer carefully, youâÂÂll see I really am describing division as the inverse of multiplication (specifically, in âÂÂ), and arguing that 0 has no multiplicative inverse. IâÂÂm just presenting it concretely, because to just about anyone short of a mathematically fairly mature undergraduate, thatâÂÂs clearer and more convincing than a formal algebraic proof. This isnâÂÂt âÂÂdumbing downâ either â the real-world understanding of division has just as much claim to being the âÂÂreal thingâÂÂ, and mathematics was being done well for centuries without modern notions of definition and proof.
â Peter LeFanu Lumsdaine
Aug 15 at 15:14
8
@ctrl-alt-delor, that's why I said to pick a system where it works: en.wikipedia.org/wiki/Riemann_sphere
â ilkkachu
Aug 15 at 15:36
12
@Kimball: I wanted to conjure up the image of OP talking to their younger brother, rather than of me addressing OP.
â Peter LeFanu Lumsdaine
2 days ago
 |Â
show 22 more comments
up vote
206
down vote
accepted
âÂÂOne of the ways to look at division is as how many of the smaller number you need to make up the bigger number, right? So 20/4 means: how many groups of 4 do you need to make 20? If you want 20 apples, how many bags of 4 apples do you need to buy?
So for dividing by 0, how many bags of 0 apples would make up 20 apples in total? ItâÂÂs impossible â however many bags of 0 apples you buy, youâÂÂll never get any apples â youâÂÂll certainly never get to 20 apples! So thereâÂÂs no possible answer, when you try to divide 20 by 0.âÂÂ
62
A good enough answer till the little brother grows up and asks why it can't be $+infty$
â Peeyush Kushwaha
Aug 15 at 13:17
36
@PeeyushKushwaha, then you introduce them to a number system where 1/0 is $infty$
â ilkkachu
Aug 15 at 13:22
43
@JackM: If you look at my answer carefully, youâÂÂll see I really am describing division as the inverse of multiplication (specifically, in âÂÂ), and arguing that 0 has no multiplicative inverse. IâÂÂm just presenting it concretely, because to just about anyone short of a mathematically fairly mature undergraduate, thatâÂÂs clearer and more convincing than a formal algebraic proof. This isnâÂÂt âÂÂdumbing downâ either â the real-world understanding of division has just as much claim to being the âÂÂreal thingâÂÂ, and mathematics was being done well for centuries without modern notions of definition and proof.
â Peter LeFanu Lumsdaine
Aug 15 at 15:14
8
@ctrl-alt-delor, that's why I said to pick a system where it works: en.wikipedia.org/wiki/Riemann_sphere
â ilkkachu
Aug 15 at 15:36
12
@Kimball: I wanted to conjure up the image of OP talking to their younger brother, rather than of me addressing OP.
â Peter LeFanu Lumsdaine
2 days ago
 |Â
show 22 more comments
up vote
206
down vote
accepted
up vote
206
down vote
accepted
âÂÂOne of the ways to look at division is as how many of the smaller number you need to make up the bigger number, right? So 20/4 means: how many groups of 4 do you need to make 20? If you want 20 apples, how many bags of 4 apples do you need to buy?
So for dividing by 0, how many bags of 0 apples would make up 20 apples in total? ItâÂÂs impossible â however many bags of 0 apples you buy, youâÂÂll never get any apples â youâÂÂll certainly never get to 20 apples! So thereâÂÂs no possible answer, when you try to divide 20 by 0.âÂÂ
âÂÂOne of the ways to look at division is as how many of the smaller number you need to make up the bigger number, right? So 20/4 means: how many groups of 4 do you need to make 20? If you want 20 apples, how many bags of 4 apples do you need to buy?
So for dividing by 0, how many bags of 0 apples would make up 20 apples in total? ItâÂÂs impossible â however many bags of 0 apples you buy, youâÂÂll never get any apples â youâÂÂll certainly never get to 20 apples! So thereâÂÂs no possible answer, when you try to divide 20 by 0.âÂÂ
answered Aug 15 at 11:42
Peter LeFanu Lumsdaine
4,96241740
4,96241740
62
A good enough answer till the little brother grows up and asks why it can't be $+infty$
â Peeyush Kushwaha
Aug 15 at 13:17
36
@PeeyushKushwaha, then you introduce them to a number system where 1/0 is $infty$
â ilkkachu
Aug 15 at 13:22
43
@JackM: If you look at my answer carefully, youâÂÂll see I really am describing division as the inverse of multiplication (specifically, in âÂÂ), and arguing that 0 has no multiplicative inverse. IâÂÂm just presenting it concretely, because to just about anyone short of a mathematically fairly mature undergraduate, thatâÂÂs clearer and more convincing than a formal algebraic proof. This isnâÂÂt âÂÂdumbing downâ either â the real-world understanding of division has just as much claim to being the âÂÂreal thingâÂÂ, and mathematics was being done well for centuries without modern notions of definition and proof.
â Peter LeFanu Lumsdaine
Aug 15 at 15:14
8
@ctrl-alt-delor, that's why I said to pick a system where it works: en.wikipedia.org/wiki/Riemann_sphere
â ilkkachu
Aug 15 at 15:36
12
@Kimball: I wanted to conjure up the image of OP talking to their younger brother, rather than of me addressing OP.
â Peter LeFanu Lumsdaine
2 days ago
 |Â
show 22 more comments
62
A good enough answer till the little brother grows up and asks why it can't be $+infty$
â Peeyush Kushwaha
Aug 15 at 13:17
36
@PeeyushKushwaha, then you introduce them to a number system where 1/0 is $infty$
â ilkkachu
Aug 15 at 13:22
43
@JackM: If you look at my answer carefully, youâÂÂll see I really am describing division as the inverse of multiplication (specifically, in âÂÂ), and arguing that 0 has no multiplicative inverse. IâÂÂm just presenting it concretely, because to just about anyone short of a mathematically fairly mature undergraduate, thatâÂÂs clearer and more convincing than a formal algebraic proof. This isnâÂÂt âÂÂdumbing downâ either â the real-world understanding of division has just as much claim to being the âÂÂreal thingâÂÂ, and mathematics was being done well for centuries without modern notions of definition and proof.
â Peter LeFanu Lumsdaine
Aug 15 at 15:14
8
@ctrl-alt-delor, that's why I said to pick a system where it works: en.wikipedia.org/wiki/Riemann_sphere
â ilkkachu
Aug 15 at 15:36
12
@Kimball: I wanted to conjure up the image of OP talking to their younger brother, rather than of me addressing OP.
â Peter LeFanu Lumsdaine
2 days ago
62
62
A good enough answer till the little brother grows up and asks why it can't be $+infty$
â Peeyush Kushwaha
Aug 15 at 13:17
A good enough answer till the little brother grows up and asks why it can't be $+infty$
â Peeyush Kushwaha
Aug 15 at 13:17
36
36
@PeeyushKushwaha, then you introduce them to a number system where 1/0 is $infty$
â ilkkachu
Aug 15 at 13:22
@PeeyushKushwaha, then you introduce them to a number system where 1/0 is $infty$
â ilkkachu
Aug 15 at 13:22
43
43
@JackM: If you look at my answer carefully, youâÂÂll see I really am describing division as the inverse of multiplication (specifically, in âÂÂ), and arguing that 0 has no multiplicative inverse. IâÂÂm just presenting it concretely, because to just about anyone short of a mathematically fairly mature undergraduate, thatâÂÂs clearer and more convincing than a formal algebraic proof. This isnâÂÂt âÂÂdumbing downâ either â the real-world understanding of division has just as much claim to being the âÂÂreal thingâÂÂ, and mathematics was being done well for centuries without modern notions of definition and proof.
â Peter LeFanu Lumsdaine
Aug 15 at 15:14
@JackM: If you look at my answer carefully, youâÂÂll see I really am describing division as the inverse of multiplication (specifically, in âÂÂ), and arguing that 0 has no multiplicative inverse. IâÂÂm just presenting it concretely, because to just about anyone short of a mathematically fairly mature undergraduate, thatâÂÂs clearer and more convincing than a formal algebraic proof. This isnâÂÂt âÂÂdumbing downâ either â the real-world understanding of division has just as much claim to being the âÂÂreal thingâÂÂ, and mathematics was being done well for centuries without modern notions of definition and proof.
â Peter LeFanu Lumsdaine
Aug 15 at 15:14
8
8
@ctrl-alt-delor, that's why I said to pick a system where it works: en.wikipedia.org/wiki/Riemann_sphere
â ilkkachu
Aug 15 at 15:36
@ctrl-alt-delor, that's why I said to pick a system where it works: en.wikipedia.org/wiki/Riemann_sphere
â ilkkachu
Aug 15 at 15:36
12
12
@Kimball: I wanted to conjure up the image of OP talking to their younger brother, rather than of me addressing OP.
â Peter LeFanu Lumsdaine
2 days ago
@Kimball: I wanted to conjure up the image of OP talking to their younger brother, rather than of me addressing OP.
â Peter LeFanu Lumsdaine
2 days ago
 |Â
show 22 more comments
up vote
62
down vote
When we first start teaching multiplication, we use successive additions. So,
3 x 4 = 3 | 3
+ 3 | 6
+ 3 | 9
+ 3 | 12
=12
Division can be taught as successive subtractions. So 12 / 3 becomes,
12 - 3 -> 9 (1)
9 - 3 -> 6 (2)
6 - 3 -> 3 (3)
3 - 3 -> 0 (4)
Now apply the second algorithm with zero as a divisor. Tell your brother to get back to you when he's done.
While this algorithmic approach is not rigorous, I think it is probably a good way of developing an intuitive understanding of the concept.
5
I should comment that in some schools of thought, teaching young students that multiplication is repeated addition can be hindering. I've certainly seen kids who only count on their fingers - they can't actually multiply, just multiple add. So there are arguments from Ring Theory to education where you may be wrong, although it is perfectly reasonable to say in the integers.
â theREALyumdub
Aug 15 at 13:49
31
I can't comment on the first part, but for the second part; We're talking about a 9 year old. One of the problems with mathematical pedagogy is we prioritize rigour over understanding. I suggest we provide the understanding first, then make it rigourous. After all, we have thousands of years of mathematical development before the hard rigour came in in the 19th century.
â Chris Cudmore
Aug 15 at 13:54
2
I agree completely, and I have hit walls with mathematics involving rigour and a lack of understanding. I was more making the point that this sounds much like a computerized argument, and I have blindly heard of education arguments against this method of approach for young children - it's more or less my best understanding of division, but it can be made more abstract and perhaps more practical for education.
â theREALyumdub
Aug 15 at 13:58
3
@theREALyumdub, the ring theorist will appreciate that every abelian group is a $mathbb Z$-module in a natural way.
â Carsten S
Aug 15 at 17:55
5
@ChrisCudmore Seeing as most college educated people never get a rigorous definition of almost any of the mathematical concepts they were taught, I don't think an over-emphasis on rigor is the problem. You seem to be confusing calculation with rigor. Frankly, I don't think the wave of formalization had almost any impact on how math was/is taught, at least pre-college. I strongly suspect that there is even more of an emphasis of intuitive understanding in grade school classrooms now than in the 1810s say.
â Derek Elkins
Aug 15 at 19:13
 |Â
show 4 more comments
up vote
62
down vote
When we first start teaching multiplication, we use successive additions. So,
3 x 4 = 3 | 3
+ 3 | 6
+ 3 | 9
+ 3 | 12
=12
Division can be taught as successive subtractions. So 12 / 3 becomes,
12 - 3 -> 9 (1)
9 - 3 -> 6 (2)
6 - 3 -> 3 (3)
3 - 3 -> 0 (4)
Now apply the second algorithm with zero as a divisor. Tell your brother to get back to you when he's done.
While this algorithmic approach is not rigorous, I think it is probably a good way of developing an intuitive understanding of the concept.
5
I should comment that in some schools of thought, teaching young students that multiplication is repeated addition can be hindering. I've certainly seen kids who only count on their fingers - they can't actually multiply, just multiple add. So there are arguments from Ring Theory to education where you may be wrong, although it is perfectly reasonable to say in the integers.
â theREALyumdub
Aug 15 at 13:49
31
I can't comment on the first part, but for the second part; We're talking about a 9 year old. One of the problems with mathematical pedagogy is we prioritize rigour over understanding. I suggest we provide the understanding first, then make it rigourous. After all, we have thousands of years of mathematical development before the hard rigour came in in the 19th century.
â Chris Cudmore
Aug 15 at 13:54
2
I agree completely, and I have hit walls with mathematics involving rigour and a lack of understanding. I was more making the point that this sounds much like a computerized argument, and I have blindly heard of education arguments against this method of approach for young children - it's more or less my best understanding of division, but it can be made more abstract and perhaps more practical for education.
â theREALyumdub
Aug 15 at 13:58
3
@theREALyumdub, the ring theorist will appreciate that every abelian group is a $mathbb Z$-module in a natural way.
â Carsten S
Aug 15 at 17:55
5
@ChrisCudmore Seeing as most college educated people never get a rigorous definition of almost any of the mathematical concepts they were taught, I don't think an over-emphasis on rigor is the problem. You seem to be confusing calculation with rigor. Frankly, I don't think the wave of formalization had almost any impact on how math was/is taught, at least pre-college. I strongly suspect that there is even more of an emphasis of intuitive understanding in grade school classrooms now than in the 1810s say.
â Derek Elkins
Aug 15 at 19:13
 |Â
show 4 more comments
up vote
62
down vote
up vote
62
down vote
When we first start teaching multiplication, we use successive additions. So,
3 x 4 = 3 | 3
+ 3 | 6
+ 3 | 9
+ 3 | 12
=12
Division can be taught as successive subtractions. So 12 / 3 becomes,
12 - 3 -> 9 (1)
9 - 3 -> 6 (2)
6 - 3 -> 3 (3)
3 - 3 -> 0 (4)
Now apply the second algorithm with zero as a divisor. Tell your brother to get back to you when he's done.
While this algorithmic approach is not rigorous, I think it is probably a good way of developing an intuitive understanding of the concept.
When we first start teaching multiplication, we use successive additions. So,
3 x 4 = 3 | 3
+ 3 | 6
+ 3 | 9
+ 3 | 12
=12
Division can be taught as successive subtractions. So 12 / 3 becomes,
12 - 3 -> 9 (1)
9 - 3 -> 6 (2)
6 - 3 -> 3 (3)
3 - 3 -> 0 (4)
Now apply the second algorithm with zero as a divisor. Tell your brother to get back to you when he's done.
While this algorithmic approach is not rigorous, I think it is probably a good way of developing an intuitive understanding of the concept.
answered Aug 15 at 13:32
Chris Cudmore
3,4481815
3,4481815
5
I should comment that in some schools of thought, teaching young students that multiplication is repeated addition can be hindering. I've certainly seen kids who only count on their fingers - they can't actually multiply, just multiple add. So there are arguments from Ring Theory to education where you may be wrong, although it is perfectly reasonable to say in the integers.
â theREALyumdub
Aug 15 at 13:49
31
I can't comment on the first part, but for the second part; We're talking about a 9 year old. One of the problems with mathematical pedagogy is we prioritize rigour over understanding. I suggest we provide the understanding first, then make it rigourous. After all, we have thousands of years of mathematical development before the hard rigour came in in the 19th century.
â Chris Cudmore
Aug 15 at 13:54
2
I agree completely, and I have hit walls with mathematics involving rigour and a lack of understanding. I was more making the point that this sounds much like a computerized argument, and I have blindly heard of education arguments against this method of approach for young children - it's more or less my best understanding of division, but it can be made more abstract and perhaps more practical for education.
â theREALyumdub
Aug 15 at 13:58
3
@theREALyumdub, the ring theorist will appreciate that every abelian group is a $mathbb Z$-module in a natural way.
â Carsten S
Aug 15 at 17:55
5
@ChrisCudmore Seeing as most college educated people never get a rigorous definition of almost any of the mathematical concepts they were taught, I don't think an over-emphasis on rigor is the problem. You seem to be confusing calculation with rigor. Frankly, I don't think the wave of formalization had almost any impact on how math was/is taught, at least pre-college. I strongly suspect that there is even more of an emphasis of intuitive understanding in grade school classrooms now than in the 1810s say.
â Derek Elkins
Aug 15 at 19:13
 |Â
show 4 more comments
5
I should comment that in some schools of thought, teaching young students that multiplication is repeated addition can be hindering. I've certainly seen kids who only count on their fingers - they can't actually multiply, just multiple add. So there are arguments from Ring Theory to education where you may be wrong, although it is perfectly reasonable to say in the integers.
â theREALyumdub
Aug 15 at 13:49
31
I can't comment on the first part, but for the second part; We're talking about a 9 year old. One of the problems with mathematical pedagogy is we prioritize rigour over understanding. I suggest we provide the understanding first, then make it rigourous. After all, we have thousands of years of mathematical development before the hard rigour came in in the 19th century.
â Chris Cudmore
Aug 15 at 13:54
2
I agree completely, and I have hit walls with mathematics involving rigour and a lack of understanding. I was more making the point that this sounds much like a computerized argument, and I have blindly heard of education arguments against this method of approach for young children - it's more or less my best understanding of division, but it can be made more abstract and perhaps more practical for education.
â theREALyumdub
Aug 15 at 13:58
3
@theREALyumdub, the ring theorist will appreciate that every abelian group is a $mathbb Z$-module in a natural way.
â Carsten S
Aug 15 at 17:55
5
@ChrisCudmore Seeing as most college educated people never get a rigorous definition of almost any of the mathematical concepts they were taught, I don't think an over-emphasis on rigor is the problem. You seem to be confusing calculation with rigor. Frankly, I don't think the wave of formalization had almost any impact on how math was/is taught, at least pre-college. I strongly suspect that there is even more of an emphasis of intuitive understanding in grade school classrooms now than in the 1810s say.
â Derek Elkins
Aug 15 at 19:13
5
5
I should comment that in some schools of thought, teaching young students that multiplication is repeated addition can be hindering. I've certainly seen kids who only count on their fingers - they can't actually multiply, just multiple add. So there are arguments from Ring Theory to education where you may be wrong, although it is perfectly reasonable to say in the integers.
â theREALyumdub
Aug 15 at 13:49
I should comment that in some schools of thought, teaching young students that multiplication is repeated addition can be hindering. I've certainly seen kids who only count on their fingers - they can't actually multiply, just multiple add. So there are arguments from Ring Theory to education where you may be wrong, although it is perfectly reasonable to say in the integers.
â theREALyumdub
Aug 15 at 13:49
31
31
I can't comment on the first part, but for the second part; We're talking about a 9 year old. One of the problems with mathematical pedagogy is we prioritize rigour over understanding. I suggest we provide the understanding first, then make it rigourous. After all, we have thousands of years of mathematical development before the hard rigour came in in the 19th century.
â Chris Cudmore
Aug 15 at 13:54
I can't comment on the first part, but for the second part; We're talking about a 9 year old. One of the problems with mathematical pedagogy is we prioritize rigour over understanding. I suggest we provide the understanding first, then make it rigourous. After all, we have thousands of years of mathematical development before the hard rigour came in in the 19th century.
â Chris Cudmore
Aug 15 at 13:54
2
2
I agree completely, and I have hit walls with mathematics involving rigour and a lack of understanding. I was more making the point that this sounds much like a computerized argument, and I have blindly heard of education arguments against this method of approach for young children - it's more or less my best understanding of division, but it can be made more abstract and perhaps more practical for education.
â theREALyumdub
Aug 15 at 13:58
I agree completely, and I have hit walls with mathematics involving rigour and a lack of understanding. I was more making the point that this sounds much like a computerized argument, and I have blindly heard of education arguments against this method of approach for young children - it's more or less my best understanding of division, but it can be made more abstract and perhaps more practical for education.
â theREALyumdub
Aug 15 at 13:58
3
3
@theREALyumdub, the ring theorist will appreciate that every abelian group is a $mathbb Z$-module in a natural way.
â Carsten S
Aug 15 at 17:55
@theREALyumdub, the ring theorist will appreciate that every abelian group is a $mathbb Z$-module in a natural way.
â Carsten S
Aug 15 at 17:55
5
5
@ChrisCudmore Seeing as most college educated people never get a rigorous definition of almost any of the mathematical concepts they were taught, I don't think an over-emphasis on rigor is the problem. You seem to be confusing calculation with rigor. Frankly, I don't think the wave of formalization had almost any impact on how math was/is taught, at least pre-college. I strongly suspect that there is even more of an emphasis of intuitive understanding in grade school classrooms now than in the 1810s say.
â Derek Elkins
Aug 15 at 19:13
@ChrisCudmore Seeing as most college educated people never get a rigorous definition of almost any of the mathematical concepts they were taught, I don't think an over-emphasis on rigor is the problem. You seem to be confusing calculation with rigor. Frankly, I don't think the wave of formalization had almost any impact on how math was/is taught, at least pre-college. I strongly suspect that there is even more of an emphasis of intuitive understanding in grade school classrooms now than in the 1810s say.
â Derek Elkins
Aug 15 at 19:13
 |Â
show 4 more comments
up vote
31
down vote
New story
Suppose that we can divide numbers with $0$. So if I would divide $1$ with zero i would get some new number name it $a$. Now what can we say about this number $a$?
Remember:
If I divide say $21$ with $3$ we get $7$. Why? Because $3cdot 7 = 21$.
And similiary if I divide $36$ with $9$ we get $4$. Why? Because $9cdot 4 = 36$.
So if I divide $1$ with $0$ and we get $a$ then we have $acdot 0 =1$ which is clearly nonsense since $acdot 0 =0$.
(I hope this is better explanation?)
Older explanation:
Suppose that $1over 0$ is some number $a$. So $$1over 0 =a.$$ Remember that $$boxedbover c = diff b = ccdot d$$ So we get $$1= acdot 0=0$$ a contradiction. So $1over 0$ doesn't exist.
88
5th grader.....
â Lightness Races in Orbit
Aug 15 at 18:02
3
I see a lot of people doesn't like this answer although it is perfectly correct. I agree it is perhaps to advance for 5th grader but then again how else could I do it? There is a nice way Peter gave, but what if brother asks what is 5/4. How many bags with 4 apples do we need to get 5 apples?
â greedoid
2 days ago
3
The logic proposed in this answer seems constructive, but a plain-English explanation would probably be more helpful. Separately, it's probably best to avoid saying that $frac10$ "doesn't exist"; it'd seem a bit better to say that $frac10$ doesn't cleanly match up with a number.
â Nat
2 days ago
1
You have to be careful when declaring contradictions. If you arrive at a "contradiction", it's actually a consequence of asserting the truth of the statements used. If you arrive at $12 = 0$ as a result, it really is just asserting a mod 12 system. if you accept $frac10 = a$ as a statement, then you are in a mod 1 system as a consequence. Suppose that you assert a statement that only after millions of manipulations asserts that $0 = 2^32$. Not only is it not a "contradiction", but it leads to the most general solution. On a computer, you have to deal with this frequently.
â Rob
2 days ago
1
I really think we should be striving to get to a point where the 'older argument' is something a fifth grader can comprehend.
â Prince M
2 days ago
 |Â
show 4 more comments
up vote
31
down vote
New story
Suppose that we can divide numbers with $0$. So if I would divide $1$ with zero i would get some new number name it $a$. Now what can we say about this number $a$?
Remember:
If I divide say $21$ with $3$ we get $7$. Why? Because $3cdot 7 = 21$.
And similiary if I divide $36$ with $9$ we get $4$. Why? Because $9cdot 4 = 36$.
So if I divide $1$ with $0$ and we get $a$ then we have $acdot 0 =1$ which is clearly nonsense since $acdot 0 =0$.
(I hope this is better explanation?)
Older explanation:
Suppose that $1over 0$ is some number $a$. So $$1over 0 =a.$$ Remember that $$boxedbover c = diff b = ccdot d$$ So we get $$1= acdot 0=0$$ a contradiction. So $1over 0$ doesn't exist.
88
5th grader.....
â Lightness Races in Orbit
Aug 15 at 18:02
3
I see a lot of people doesn't like this answer although it is perfectly correct. I agree it is perhaps to advance for 5th grader but then again how else could I do it? There is a nice way Peter gave, but what if brother asks what is 5/4. How many bags with 4 apples do we need to get 5 apples?
â greedoid
2 days ago
3
The logic proposed in this answer seems constructive, but a plain-English explanation would probably be more helpful. Separately, it's probably best to avoid saying that $frac10$ "doesn't exist"; it'd seem a bit better to say that $frac10$ doesn't cleanly match up with a number.
â Nat
2 days ago
1
You have to be careful when declaring contradictions. If you arrive at a "contradiction", it's actually a consequence of asserting the truth of the statements used. If you arrive at $12 = 0$ as a result, it really is just asserting a mod 12 system. if you accept $frac10 = a$ as a statement, then you are in a mod 1 system as a consequence. Suppose that you assert a statement that only after millions of manipulations asserts that $0 = 2^32$. Not only is it not a "contradiction", but it leads to the most general solution. On a computer, you have to deal with this frequently.
â Rob
2 days ago
1
I really think we should be striving to get to a point where the 'older argument' is something a fifth grader can comprehend.
â Prince M
2 days ago
 |Â
show 4 more comments
up vote
31
down vote
up vote
31
down vote
New story
Suppose that we can divide numbers with $0$. So if I would divide $1$ with zero i would get some new number name it $a$. Now what can we say about this number $a$?
Remember:
If I divide say $21$ with $3$ we get $7$. Why? Because $3cdot 7 = 21$.
And similiary if I divide $36$ with $9$ we get $4$. Why? Because $9cdot 4 = 36$.
So if I divide $1$ with $0$ and we get $a$ then we have $acdot 0 =1$ which is clearly nonsense since $acdot 0 =0$.
(I hope this is better explanation?)
Older explanation:
Suppose that $1over 0$ is some number $a$. So $$1over 0 =a.$$ Remember that $$boxedbover c = diff b = ccdot d$$ So we get $$1= acdot 0=0$$ a contradiction. So $1over 0$ doesn't exist.
New story
Suppose that we can divide numbers with $0$. So if I would divide $1$ with zero i would get some new number name it $a$. Now what can we say about this number $a$?
Remember:
If I divide say $21$ with $3$ we get $7$. Why? Because $3cdot 7 = 21$.
And similiary if I divide $36$ with $9$ we get $4$. Why? Because $9cdot 4 = 36$.
So if I divide $1$ with $0$ and we get $a$ then we have $acdot 0 =1$ which is clearly nonsense since $acdot 0 =0$.
(I hope this is better explanation?)
Older explanation:
Suppose that $1over 0$ is some number $a$. So $$1over 0 =a.$$ Remember that $$boxedbover c = diff b = ccdot d$$ So we get $$1= acdot 0=0$$ a contradiction. So $1over 0$ doesn't exist.
edited 2 days ago
answered Aug 15 at 10:36
greedoid
26.6k93574
26.6k93574
88
5th grader.....
â Lightness Races in Orbit
Aug 15 at 18:02
3
I see a lot of people doesn't like this answer although it is perfectly correct. I agree it is perhaps to advance for 5th grader but then again how else could I do it? There is a nice way Peter gave, but what if brother asks what is 5/4. How many bags with 4 apples do we need to get 5 apples?
â greedoid
2 days ago
3
The logic proposed in this answer seems constructive, but a plain-English explanation would probably be more helpful. Separately, it's probably best to avoid saying that $frac10$ "doesn't exist"; it'd seem a bit better to say that $frac10$ doesn't cleanly match up with a number.
â Nat
2 days ago
1
You have to be careful when declaring contradictions. If you arrive at a "contradiction", it's actually a consequence of asserting the truth of the statements used. If you arrive at $12 = 0$ as a result, it really is just asserting a mod 12 system. if you accept $frac10 = a$ as a statement, then you are in a mod 1 system as a consequence. Suppose that you assert a statement that only after millions of manipulations asserts that $0 = 2^32$. Not only is it not a "contradiction", but it leads to the most general solution. On a computer, you have to deal with this frequently.
â Rob
2 days ago
1
I really think we should be striving to get to a point where the 'older argument' is something a fifth grader can comprehend.
â Prince M
2 days ago
 |Â
show 4 more comments
88
5th grader.....
â Lightness Races in Orbit
Aug 15 at 18:02
3
I see a lot of people doesn't like this answer although it is perfectly correct. I agree it is perhaps to advance for 5th grader but then again how else could I do it? There is a nice way Peter gave, but what if brother asks what is 5/4. How many bags with 4 apples do we need to get 5 apples?
â greedoid
2 days ago
3
The logic proposed in this answer seems constructive, but a plain-English explanation would probably be more helpful. Separately, it's probably best to avoid saying that $frac10$ "doesn't exist"; it'd seem a bit better to say that $frac10$ doesn't cleanly match up with a number.
â Nat
2 days ago
1
You have to be careful when declaring contradictions. If you arrive at a "contradiction", it's actually a consequence of asserting the truth of the statements used. If you arrive at $12 = 0$ as a result, it really is just asserting a mod 12 system. if you accept $frac10 = a$ as a statement, then you are in a mod 1 system as a consequence. Suppose that you assert a statement that only after millions of manipulations asserts that $0 = 2^32$. Not only is it not a "contradiction", but it leads to the most general solution. On a computer, you have to deal with this frequently.
â Rob
2 days ago
1
I really think we should be striving to get to a point where the 'older argument' is something a fifth grader can comprehend.
â Prince M
2 days ago
88
88
5th grader.....
â Lightness Races in Orbit
Aug 15 at 18:02
5th grader.....
â Lightness Races in Orbit
Aug 15 at 18:02
3
3
I see a lot of people doesn't like this answer although it is perfectly correct. I agree it is perhaps to advance for 5th grader but then again how else could I do it? There is a nice way Peter gave, but what if brother asks what is 5/4. How many bags with 4 apples do we need to get 5 apples?
â greedoid
2 days ago
I see a lot of people doesn't like this answer although it is perfectly correct. I agree it is perhaps to advance for 5th grader but then again how else could I do it? There is a nice way Peter gave, but what if brother asks what is 5/4. How many bags with 4 apples do we need to get 5 apples?
â greedoid
2 days ago
3
3
The logic proposed in this answer seems constructive, but a plain-English explanation would probably be more helpful. Separately, it's probably best to avoid saying that $frac10$ "doesn't exist"; it'd seem a bit better to say that $frac10$ doesn't cleanly match up with a number.
â Nat
2 days ago
The logic proposed in this answer seems constructive, but a plain-English explanation would probably be more helpful. Separately, it's probably best to avoid saying that $frac10$ "doesn't exist"; it'd seem a bit better to say that $frac10$ doesn't cleanly match up with a number.
â Nat
2 days ago
1
1
You have to be careful when declaring contradictions. If you arrive at a "contradiction", it's actually a consequence of asserting the truth of the statements used. If you arrive at $12 = 0$ as a result, it really is just asserting a mod 12 system. if you accept $frac10 = a$ as a statement, then you are in a mod 1 system as a consequence. Suppose that you assert a statement that only after millions of manipulations asserts that $0 = 2^32$. Not only is it not a "contradiction", but it leads to the most general solution. On a computer, you have to deal with this frequently.
â Rob
2 days ago
You have to be careful when declaring contradictions. If you arrive at a "contradiction", it's actually a consequence of asserting the truth of the statements used. If you arrive at $12 = 0$ as a result, it really is just asserting a mod 12 system. if you accept $frac10 = a$ as a statement, then you are in a mod 1 system as a consequence. Suppose that you assert a statement that only after millions of manipulations asserts that $0 = 2^32$. Not only is it not a "contradiction", but it leads to the most general solution. On a computer, you have to deal with this frequently.
â Rob
2 days ago
1
1
I really think we should be striving to get to a point where the 'older argument' is something a fifth grader can comprehend.
â Prince M
2 days ago
I really think we should be striving to get to a point where the 'older argument' is something a fifth grader can comprehend.
â Prince M
2 days ago
 |Â
show 4 more comments
up vote
17
down vote
An explanation that might make sense to a fifth grader is one that gets to the heart of why we have invented these operations in the first place.
Multiplication is a trick we use to add similar things to form a sum. When we say 5 x 3, what we really mean is take five things of size three each and add them all together. We invented this trick because we are frequently in the situation where we have many of a similar thing, and we wish to know their sum.
Division is the same trick but the other way. When we say 15 / 3, we are asking the question "how many times would we have to add a thing of size three starting from nothing to make a thing of size fifteen?" We'd have to add five things of size three together to make a thing of size fifteen. Again, division is just a trick we use to answer questions about sums.
Now it becomes clear why division by zero is not defined. There is no number of times you can add zero to itself to get a non-zero sum.
A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!
For non-zero divided by zero, there is no number at all of times that you can add zero to itself to get non-zero. For zero divided by zero, every number of times you add zero to itself, you get zero, so the solution is not unique. We like our mathematical questions to have unique answers where possible and so we by convention say that 0 / 0 is also not defined.
3
Actually I have never thought of division like that. 15/2 has always meant 'how big is each half when you cut it in two' to me.
â Sentinel
2 days ago
2
"how many times would we have to add a thing of size three to itself to make a thing of size fifteen?" 3+3=6; 6+3=9; 9+3=12; 12+3=15. I count four additions. Your question is therefore not worded correctly. It's not adding three to itself, it's "Starting with zero (nothing), how many 3s do you need to add to get 15?"
â Monty Harder
2 days ago
@MontyHarder: Good point; I'll change the wording.
â Eric Lippert
2 days ago
2
"A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!" Simple enough. Grab a plate and set it on the table and say "there are 0 cookies on this plate". Grab another and set it down. "there are 0 cookies on each plate, for a total of 0 cookies. If you leave the room and all you know is that there are 0 total cookies, how can you know how many empty plates I put on the table (or picked back up) when you were gone?"
â Monty Harder
2 days ago
add a comment |Â
up vote
17
down vote
An explanation that might make sense to a fifth grader is one that gets to the heart of why we have invented these operations in the first place.
Multiplication is a trick we use to add similar things to form a sum. When we say 5 x 3, what we really mean is take five things of size three each and add them all together. We invented this trick because we are frequently in the situation where we have many of a similar thing, and we wish to know their sum.
Division is the same trick but the other way. When we say 15 / 3, we are asking the question "how many times would we have to add a thing of size three starting from nothing to make a thing of size fifteen?" We'd have to add five things of size three together to make a thing of size fifteen. Again, division is just a trick we use to answer questions about sums.
Now it becomes clear why division by zero is not defined. There is no number of times you can add zero to itself to get a non-zero sum.
A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!
For non-zero divided by zero, there is no number at all of times that you can add zero to itself to get non-zero. For zero divided by zero, every number of times you add zero to itself, you get zero, so the solution is not unique. We like our mathematical questions to have unique answers where possible and so we by convention say that 0 / 0 is also not defined.
3
Actually I have never thought of division like that. 15/2 has always meant 'how big is each half when you cut it in two' to me.
â Sentinel
2 days ago
2
"how many times would we have to add a thing of size three to itself to make a thing of size fifteen?" 3+3=6; 6+3=9; 9+3=12; 12+3=15. I count four additions. Your question is therefore not worded correctly. It's not adding three to itself, it's "Starting with zero (nothing), how many 3s do you need to add to get 15?"
â Monty Harder
2 days ago
@MontyHarder: Good point; I'll change the wording.
â Eric Lippert
2 days ago
2
"A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!" Simple enough. Grab a plate and set it on the table and say "there are 0 cookies on this plate". Grab another and set it down. "there are 0 cookies on each plate, for a total of 0 cookies. If you leave the room and all you know is that there are 0 total cookies, how can you know how many empty plates I put on the table (or picked back up) when you were gone?"
â Monty Harder
2 days ago
add a comment |Â
up vote
17
down vote
up vote
17
down vote
An explanation that might make sense to a fifth grader is one that gets to the heart of why we have invented these operations in the first place.
Multiplication is a trick we use to add similar things to form a sum. When we say 5 x 3, what we really mean is take five things of size three each and add them all together. We invented this trick because we are frequently in the situation where we have many of a similar thing, and we wish to know their sum.
Division is the same trick but the other way. When we say 15 / 3, we are asking the question "how many times would we have to add a thing of size three starting from nothing to make a thing of size fifteen?" We'd have to add five things of size three together to make a thing of size fifteen. Again, division is just a trick we use to answer questions about sums.
Now it becomes clear why division by zero is not defined. There is no number of times you can add zero to itself to get a non-zero sum.
A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!
For non-zero divided by zero, there is no number at all of times that you can add zero to itself to get non-zero. For zero divided by zero, every number of times you add zero to itself, you get zero, so the solution is not unique. We like our mathematical questions to have unique answers where possible and so we by convention say that 0 / 0 is also not defined.
An explanation that might make sense to a fifth grader is one that gets to the heart of why we have invented these operations in the first place.
Multiplication is a trick we use to add similar things to form a sum. When we say 5 x 3, what we really mean is take five things of size three each and add them all together. We invented this trick because we are frequently in the situation where we have many of a similar thing, and we wish to know their sum.
Division is the same trick but the other way. When we say 15 / 3, we are asking the question "how many times would we have to add a thing of size three starting from nothing to make a thing of size fifteen?" We'd have to add five things of size three together to make a thing of size fifteen. Again, division is just a trick we use to answer questions about sums.
Now it becomes clear why division by zero is not defined. There is no number of times you can add zero to itself to get a non-zero sum.
A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!
For non-zero divided by zero, there is no number at all of times that you can add zero to itself to get non-zero. For zero divided by zero, every number of times you add zero to itself, you get zero, so the solution is not unique. We like our mathematical questions to have unique answers where possible and so we by convention say that 0 / 0 is also not defined.
edited 2 days ago
answered Aug 15 at 13:20
Eric Lippert
3,0861318
3,0861318
3
Actually I have never thought of division like that. 15/2 has always meant 'how big is each half when you cut it in two' to me.
â Sentinel
2 days ago
2
"how many times would we have to add a thing of size three to itself to make a thing of size fifteen?" 3+3=6; 6+3=9; 9+3=12; 12+3=15. I count four additions. Your question is therefore not worded correctly. It's not adding three to itself, it's "Starting with zero (nothing), how many 3s do you need to add to get 15?"
â Monty Harder
2 days ago
@MontyHarder: Good point; I'll change the wording.
â Eric Lippert
2 days ago
2
"A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!" Simple enough. Grab a plate and set it on the table and say "there are 0 cookies on this plate". Grab another and set it down. "there are 0 cookies on each plate, for a total of 0 cookies. If you leave the room and all you know is that there are 0 total cookies, how can you know how many empty plates I put on the table (or picked back up) when you were gone?"
â Monty Harder
2 days ago
add a comment |Â
3
Actually I have never thought of division like that. 15/2 has always meant 'how big is each half when you cut it in two' to me.
â Sentinel
2 days ago
2
"how many times would we have to add a thing of size three to itself to make a thing of size fifteen?" 3+3=6; 6+3=9; 9+3=12; 12+3=15. I count four additions. Your question is therefore not worded correctly. It's not adding three to itself, it's "Starting with zero (nothing), how many 3s do you need to add to get 15?"
â Monty Harder
2 days ago
@MontyHarder: Good point; I'll change the wording.
â Eric Lippert
2 days ago
2
"A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!" Simple enough. Grab a plate and set it on the table and say "there are 0 cookies on this plate". Grab another and set it down. "there are 0 cookies on each plate, for a total of 0 cookies. If you leave the room and all you know is that there are 0 total cookies, how can you know how many empty plates I put on the table (or picked back up) when you were gone?"
â Monty Harder
2 days ago
3
3
Actually I have never thought of division like that. 15/2 has always meant 'how big is each half when you cut it in two' to me.
â Sentinel
2 days ago
Actually I have never thought of division like that. 15/2 has always meant 'how big is each half when you cut it in two' to me.
â Sentinel
2 days ago
2
2
"how many times would we have to add a thing of size three to itself to make a thing of size fifteen?" 3+3=6; 6+3=9; 9+3=12; 12+3=15. I count four additions. Your question is therefore not worded correctly. It's not adding three to itself, it's "Starting with zero (nothing), how many 3s do you need to add to get 15?"
â Monty Harder
2 days ago
"how many times would we have to add a thing of size three to itself to make a thing of size fifteen?" 3+3=6; 6+3=9; 9+3=12; 12+3=15. I count four additions. Your question is therefore not worded correctly. It's not adding three to itself, it's "Starting with zero (nothing), how many 3s do you need to add to get 15?"
â Monty Harder
2 days ago
@MontyHarder: Good point; I'll change the wording.
â Eric Lippert
2 days ago
@MontyHarder: Good point; I'll change the wording.
â Eric Lippert
2 days ago
2
2
"A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!" Simple enough. Grab a plate and set it on the table and say "there are 0 cookies on this plate". Grab another and set it down. "there are 0 cookies on each plate, for a total of 0 cookies. If you leave the room and all you know is that there are 0 total cookies, how can you know how many empty plates I put on the table (or picked back up) when you were gone?"
â Monty Harder
2 days ago
"A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!" Simple enough. Grab a plate and set it on the table and say "there are 0 cookies on this plate". Grab another and set it down. "there are 0 cookies on each plate, for a total of 0 cookies. If you leave the room and all you know is that there are 0 total cookies, how can you know how many empty plates I put on the table (or picked back up) when you were gone?"
â Monty Harder
2 days ago
add a comment |Â
up vote
11
down vote
The Wikipedia article Division by zero lists the usual arguments why there is no good choice for the result of such an operation.
I prefer the algebraic argument, that there is no multiplicative inverse of $0$,
this would need you to explain a bit about algebra.
The argument from calculus, looking at limits of $1/x$, I find also useful, but perphaps harder to explain.
1
To me this is the answer, though I would combine it with "let the 5th grader try to come up with some ideas, and help them see why they fail." Personally, I like this approach because then, when they come across the sqrt(-1), they're going to be more comfortable when we say "actually, there is a good choice for how to deal with this."
â Cort Ammon
yesterday
add a comment |Â
up vote
11
down vote
The Wikipedia article Division by zero lists the usual arguments why there is no good choice for the result of such an operation.
I prefer the algebraic argument, that there is no multiplicative inverse of $0$,
this would need you to explain a bit about algebra.
The argument from calculus, looking at limits of $1/x$, I find also useful, but perphaps harder to explain.
1
To me this is the answer, though I would combine it with "let the 5th grader try to come up with some ideas, and help them see why they fail." Personally, I like this approach because then, when they come across the sqrt(-1), they're going to be more comfortable when we say "actually, there is a good choice for how to deal with this."
â Cort Ammon
yesterday
add a comment |Â
up vote
11
down vote
up vote
11
down vote
The Wikipedia article Division by zero lists the usual arguments why there is no good choice for the result of such an operation.
I prefer the algebraic argument, that there is no multiplicative inverse of $0$,
this would need you to explain a bit about algebra.
The argument from calculus, looking at limits of $1/x$, I find also useful, but perphaps harder to explain.
The Wikipedia article Division by zero lists the usual arguments why there is no good choice for the result of such an operation.
I prefer the algebraic argument, that there is no multiplicative inverse of $0$,
this would need you to explain a bit about algebra.
The argument from calculus, looking at limits of $1/x$, I find also useful, but perphaps harder to explain.
answered Aug 15 at 10:43
mvw
30.6k22251
30.6k22251
1
To me this is the answer, though I would combine it with "let the 5th grader try to come up with some ideas, and help them see why they fail." Personally, I like this approach because then, when they come across the sqrt(-1), they're going to be more comfortable when we say "actually, there is a good choice for how to deal with this."
â Cort Ammon
yesterday
add a comment |Â
1
To me this is the answer, though I would combine it with "let the 5th grader try to come up with some ideas, and help them see why they fail." Personally, I like this approach because then, when they come across the sqrt(-1), they're going to be more comfortable when we say "actually, there is a good choice for how to deal with this."
â Cort Ammon
yesterday
1
1
To me this is the answer, though I would combine it with "let the 5th grader try to come up with some ideas, and help them see why they fail." Personally, I like this approach because then, when they come across the sqrt(-1), they're going to be more comfortable when we say "actually, there is a good choice for how to deal with this."
â Cort Ammon
yesterday
To me this is the answer, though I would combine it with "let the 5th grader try to come up with some ideas, and help them see why they fail." Personally, I like this approach because then, when they come across the sqrt(-1), they're going to be more comfortable when we say "actually, there is a good choice for how to deal with this."
â Cort Ammon
yesterday
add a comment |Â
up vote
7
down vote
How many nothings do you need to add together to get 12?
4
Add all the nothings!
â Don Branson
Aug 15 at 19:54
4
@Sentinel Ok, Donnie. I did that. I found all the nothings and added them, but I still have nothing. What now?
â Sentinel
2 days ago
Not in 5th grade, maybe, but by 7th or 8th I think I might have asked "what if you added an infinite number of nothings?" Running into the same issue as some other explainations.
â Ian D. Scott
yesterday
@DonBranson Of course. Mathematicians have done for 4 centuries already... Infinitesimals.. :) Imagine integrating constant function with value 12 over $[0,1]$ or value 1 over $[0,12]$
â mathreadler
yesterday
@IanD.Scott infinitesimals and calculus. That is why you should not tell kids that/why you can't divide by 0. Playing with the idea of multiplying "almost 0" with infinity is so fruitful it is more valuable if they wrestle around a bit with it for themselves. It is not inconceivable that a kid could come up with some variant of infinitesimal calculus based on adding infinite many very small slices.
â mathreadler
yesterday
add a comment |Â
up vote
7
down vote
How many nothings do you need to add together to get 12?
4
Add all the nothings!
â Don Branson
Aug 15 at 19:54
4
@Sentinel Ok, Donnie. I did that. I found all the nothings and added them, but I still have nothing. What now?
â Sentinel
2 days ago
Not in 5th grade, maybe, but by 7th or 8th I think I might have asked "what if you added an infinite number of nothings?" Running into the same issue as some other explainations.
â Ian D. Scott
yesterday
@DonBranson Of course. Mathematicians have done for 4 centuries already... Infinitesimals.. :) Imagine integrating constant function with value 12 over $[0,1]$ or value 1 over $[0,12]$
â mathreadler
yesterday
@IanD.Scott infinitesimals and calculus. That is why you should not tell kids that/why you can't divide by 0. Playing with the idea of multiplying "almost 0" with infinity is so fruitful it is more valuable if they wrestle around a bit with it for themselves. It is not inconceivable that a kid could come up with some variant of infinitesimal calculus based on adding infinite many very small slices.
â mathreadler
yesterday
add a comment |Â
up vote
7
down vote
up vote
7
down vote
How many nothings do you need to add together to get 12?
How many nothings do you need to add together to get 12?
answered Aug 15 at 19:54
Sentinel
2904
2904
4
Add all the nothings!
â Don Branson
Aug 15 at 19:54
4
@Sentinel Ok, Donnie. I did that. I found all the nothings and added them, but I still have nothing. What now?
â Sentinel
2 days ago
Not in 5th grade, maybe, but by 7th or 8th I think I might have asked "what if you added an infinite number of nothings?" Running into the same issue as some other explainations.
â Ian D. Scott
yesterday
@DonBranson Of course. Mathematicians have done for 4 centuries already... Infinitesimals.. :) Imagine integrating constant function with value 12 over $[0,1]$ or value 1 over $[0,12]$
â mathreadler
yesterday
@IanD.Scott infinitesimals and calculus. That is why you should not tell kids that/why you can't divide by 0. Playing with the idea of multiplying "almost 0" with infinity is so fruitful it is more valuable if they wrestle around a bit with it for themselves. It is not inconceivable that a kid could come up with some variant of infinitesimal calculus based on adding infinite many very small slices.
â mathreadler
yesterday
add a comment |Â
4
Add all the nothings!
â Don Branson
Aug 15 at 19:54
4
@Sentinel Ok, Donnie. I did that. I found all the nothings and added them, but I still have nothing. What now?
â Sentinel
2 days ago
Not in 5th grade, maybe, but by 7th or 8th I think I might have asked "what if you added an infinite number of nothings?" Running into the same issue as some other explainations.
â Ian D. Scott
yesterday
@DonBranson Of course. Mathematicians have done for 4 centuries already... Infinitesimals.. :) Imagine integrating constant function with value 12 over $[0,1]$ or value 1 over $[0,12]$
â mathreadler
yesterday
@IanD.Scott infinitesimals and calculus. That is why you should not tell kids that/why you can't divide by 0. Playing with the idea of multiplying "almost 0" with infinity is so fruitful it is more valuable if they wrestle around a bit with it for themselves. It is not inconceivable that a kid could come up with some variant of infinitesimal calculus based on adding infinite many very small slices.
â mathreadler
yesterday
4
4
Add all the nothings!
â Don Branson
Aug 15 at 19:54
Add all the nothings!
â Don Branson
Aug 15 at 19:54
4
4
@Sentinel Ok, Donnie. I did that. I found all the nothings and added them, but I still have nothing. What now?
â Sentinel
2 days ago
@Sentinel Ok, Donnie. I did that. I found all the nothings and added them, but I still have nothing. What now?
â Sentinel
2 days ago
Not in 5th grade, maybe, but by 7th or 8th I think I might have asked "what if you added an infinite number of nothings?" Running into the same issue as some other explainations.
â Ian D. Scott
yesterday
Not in 5th grade, maybe, but by 7th or 8th I think I might have asked "what if you added an infinite number of nothings?" Running into the same issue as some other explainations.
â Ian D. Scott
yesterday
@DonBranson Of course. Mathematicians have done for 4 centuries already... Infinitesimals.. :) Imagine integrating constant function with value 12 over $[0,1]$ or value 1 over $[0,12]$
â mathreadler
yesterday
@DonBranson Of course. Mathematicians have done for 4 centuries already... Infinitesimals.. :) Imagine integrating constant function with value 12 over $[0,1]$ or value 1 over $[0,12]$
â mathreadler
yesterday
@IanD.Scott infinitesimals and calculus. That is why you should not tell kids that/why you can't divide by 0. Playing with the idea of multiplying "almost 0" with infinity is so fruitful it is more valuable if they wrestle around a bit with it for themselves. It is not inconceivable that a kid could come up with some variant of infinitesimal calculus based on adding infinite many very small slices.
â mathreadler
yesterday
@IanD.Scott infinitesimals and calculus. That is why you should not tell kids that/why you can't divide by 0. Playing with the idea of multiplying "almost 0" with infinity is so fruitful it is more valuable if they wrestle around a bit with it for themselves. It is not inconceivable that a kid could come up with some variant of infinitesimal calculus based on adding infinite many very small slices.
â mathreadler
yesterday
add a comment |Â
up vote
6
down vote
Division by zero is meaningless because that's what we decided division means. All you can do is explain why such a convention is a useful one for ordinary arithmetic.
It might even help to demonstrate some other context (e.g. arithmetic in the projectively extended number line) where it can be useful to define division by zero, so that the student is able to compare and contrast the reasons why we might or might not like to define something.
Your question might be better placed on https://matheducators.stackexchange.com/
I believe your first sentence is circular logic
â Ruadhan2300
Aug 15 at 15:59
@Ruadhan2300 No, it's not circular. I think Hurkyl's point is that all mathematical concepts are just definitions that we decided on. You could define division differently, as $1/0 = 37$ and still develop all of modern mathematics; it would only be less convenient, not less "correct".
â 6005
yesterday
That being said, I think this meta-explanation may be a bit too difficult to grasp for a 5th grader. At that stage, most students think of definitions as immutable truth.
â 6005
yesterday
Much like most of science, all mathematics is rooted in modelling reality. The rules and concepts we produce exist entirely because we found they apply to real situations and remain consistently effective. I maintain that "because that's the definition we gave it" is circular and unhelpful.
â Ruadhan2300
yesterday
add a comment |Â
up vote
6
down vote
Division by zero is meaningless because that's what we decided division means. All you can do is explain why such a convention is a useful one for ordinary arithmetic.
It might even help to demonstrate some other context (e.g. arithmetic in the projectively extended number line) where it can be useful to define division by zero, so that the student is able to compare and contrast the reasons why we might or might not like to define something.
Your question might be better placed on https://matheducators.stackexchange.com/
I believe your first sentence is circular logic
â Ruadhan2300
Aug 15 at 15:59
@Ruadhan2300 No, it's not circular. I think Hurkyl's point is that all mathematical concepts are just definitions that we decided on. You could define division differently, as $1/0 = 37$ and still develop all of modern mathematics; it would only be less convenient, not less "correct".
â 6005
yesterday
That being said, I think this meta-explanation may be a bit too difficult to grasp for a 5th grader. At that stage, most students think of definitions as immutable truth.
â 6005
yesterday
Much like most of science, all mathematics is rooted in modelling reality. The rules and concepts we produce exist entirely because we found they apply to real situations and remain consistently effective. I maintain that "because that's the definition we gave it" is circular and unhelpful.
â Ruadhan2300
yesterday
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Division by zero is meaningless because that's what we decided division means. All you can do is explain why such a convention is a useful one for ordinary arithmetic.
It might even help to demonstrate some other context (e.g. arithmetic in the projectively extended number line) where it can be useful to define division by zero, so that the student is able to compare and contrast the reasons why we might or might not like to define something.
Your question might be better placed on https://matheducators.stackexchange.com/
Division by zero is meaningless because that's what we decided division means. All you can do is explain why such a convention is a useful one for ordinary arithmetic.
It might even help to demonstrate some other context (e.g. arithmetic in the projectively extended number line) where it can be useful to define division by zero, so that the student is able to compare and contrast the reasons why we might or might not like to define something.
Your question might be better placed on https://matheducators.stackexchange.com/
answered Aug 15 at 10:58
Hurkyl
108k9112253
108k9112253
I believe your first sentence is circular logic
â Ruadhan2300
Aug 15 at 15:59
@Ruadhan2300 No, it's not circular. I think Hurkyl's point is that all mathematical concepts are just definitions that we decided on. You could define division differently, as $1/0 = 37$ and still develop all of modern mathematics; it would only be less convenient, not less "correct".
â 6005
yesterday
That being said, I think this meta-explanation may be a bit too difficult to grasp for a 5th grader. At that stage, most students think of definitions as immutable truth.
â 6005
yesterday
Much like most of science, all mathematics is rooted in modelling reality. The rules and concepts we produce exist entirely because we found they apply to real situations and remain consistently effective. I maintain that "because that's the definition we gave it" is circular and unhelpful.
â Ruadhan2300
yesterday
add a comment |Â
I believe your first sentence is circular logic
â Ruadhan2300
Aug 15 at 15:59
@Ruadhan2300 No, it's not circular. I think Hurkyl's point is that all mathematical concepts are just definitions that we decided on. You could define division differently, as $1/0 = 37$ and still develop all of modern mathematics; it would only be less convenient, not less "correct".
â 6005
yesterday
That being said, I think this meta-explanation may be a bit too difficult to grasp for a 5th grader. At that stage, most students think of definitions as immutable truth.
â 6005
yesterday
Much like most of science, all mathematics is rooted in modelling reality. The rules and concepts we produce exist entirely because we found they apply to real situations and remain consistently effective. I maintain that "because that's the definition we gave it" is circular and unhelpful.
â Ruadhan2300
yesterday
I believe your first sentence is circular logic
â Ruadhan2300
Aug 15 at 15:59
I believe your first sentence is circular logic
â Ruadhan2300
Aug 15 at 15:59
@Ruadhan2300 No, it's not circular. I think Hurkyl's point is that all mathematical concepts are just definitions that we decided on. You could define division differently, as $1/0 = 37$ and still develop all of modern mathematics; it would only be less convenient, not less "correct".
â 6005
yesterday
@Ruadhan2300 No, it's not circular. I think Hurkyl's point is that all mathematical concepts are just definitions that we decided on. You could define division differently, as $1/0 = 37$ and still develop all of modern mathematics; it would only be less convenient, not less "correct".
â 6005
yesterday
That being said, I think this meta-explanation may be a bit too difficult to grasp for a 5th grader. At that stage, most students think of definitions as immutable truth.
â 6005
yesterday
That being said, I think this meta-explanation may be a bit too difficult to grasp for a 5th grader. At that stage, most students think of definitions as immutable truth.
â 6005
yesterday
Much like most of science, all mathematics is rooted in modelling reality. The rules and concepts we produce exist entirely because we found they apply to real situations and remain consistently effective. I maintain that "because that's the definition we gave it" is circular and unhelpful.
â Ruadhan2300
yesterday
Much like most of science, all mathematics is rooted in modelling reality. The rules and concepts we produce exist entirely because we found they apply to real situations and remain consistently effective. I maintain that "because that's the definition we gave it" is circular and unhelpful.
â Ruadhan2300
yesterday
add a comment |Â
up vote
5
down vote
Ask Siri.
Imagine that you have zero cookies and you split them evenly among
zero friends. How many cookies does each person get? See? It doesn't
make sense. And Cookie Monster is sad that there are no cookies, and
you are sad that you have no friends.
5
If you include your imaginary friend, then the general form for dividing $X$ cookies is $-Xi$, because $(0+i) * (-Xi) = X$
â Chronocidal
Aug 15 at 22:43
That is like saying that division by two is slicing evenly in half. Most other answers here think it is about groups of two.
â Sentinel
yesterday
add a comment |Â
up vote
5
down vote
Ask Siri.
Imagine that you have zero cookies and you split them evenly among
zero friends. How many cookies does each person get? See? It doesn't
make sense. And Cookie Monster is sad that there are no cookies, and
you are sad that you have no friends.
5
If you include your imaginary friend, then the general form for dividing $X$ cookies is $-Xi$, because $(0+i) * (-Xi) = X$
â Chronocidal
Aug 15 at 22:43
That is like saying that division by two is slicing evenly in half. Most other answers here think it is about groups of two.
â Sentinel
yesterday
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Ask Siri.
Imagine that you have zero cookies and you split them evenly among
zero friends. How many cookies does each person get? See? It doesn't
make sense. And Cookie Monster is sad that there are no cookies, and
you are sad that you have no friends.
Ask Siri.
Imagine that you have zero cookies and you split them evenly among
zero friends. How many cookies does each person get? See? It doesn't
make sense. And Cookie Monster is sad that there are no cookies, and
you are sad that you have no friends.
answered Aug 15 at 21:09
Evorlor
218111
218111
5
If you include your imaginary friend, then the general form for dividing $X$ cookies is $-Xi$, because $(0+i) * (-Xi) = X$
â Chronocidal
Aug 15 at 22:43
That is like saying that division by two is slicing evenly in half. Most other answers here think it is about groups of two.
â Sentinel
yesterday
add a comment |Â
5
If you include your imaginary friend, then the general form for dividing $X$ cookies is $-Xi$, because $(0+i) * (-Xi) = X$
â Chronocidal
Aug 15 at 22:43
That is like saying that division by two is slicing evenly in half. Most other answers here think it is about groups of two.
â Sentinel
yesterday
5
5
If you include your imaginary friend, then the general form for dividing $X$ cookies is $-Xi$, because $(0+i) * (-Xi) = X$
â Chronocidal
Aug 15 at 22:43
If you include your imaginary friend, then the general form for dividing $X$ cookies is $-Xi$, because $(0+i) * (-Xi) = X$
â Chronocidal
Aug 15 at 22:43
That is like saying that division by two is slicing evenly in half. Most other answers here think it is about groups of two.
â Sentinel
yesterday
That is like saying that division by two is slicing evenly in half. Most other answers here think it is about groups of two.
â Sentinel
yesterday
add a comment |Â
up vote
5
down vote
You shouldn't try to do that. Instead make counter question.
"What should it be, then?" and let them think about it.
add a comment |Â
up vote
5
down vote
You shouldn't try to do that. Instead make counter question.
"What should it be, then?" and let them think about it.
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You shouldn't try to do that. Instead make counter question.
"What should it be, then?" and let them think about it.
You shouldn't try to do that. Instead make counter question.
"What should it be, then?" and let them think about it.
answered 2 days ago
mathreadler
13.6k71857
13.6k71857
add a comment |Â
add a comment |Â
up vote
4
down vote
I don't have kids (my wife says one 3-year-old in the house is enough for her) and it's been a while since I was in the 5th grade (although at work sometimes...), but I'll give it a go.
I know you're too old to play with blocks, but lets start with 12 blocks.
Let's start with $12/6$ - that's $2$, right? Take $6$ at a time and there are two "sets". There are $2$ sets of $6$ in $12$.
Then $12/4$ is $3$ - $3$ sets of $4$ in $12$.
Then $12/3$ is $4$ - $4$ sets of $3$ in $12$ (commutation of the last case).
Then $12/2$ is $6$ - $2$ sets of $6$ in $12$ (commutation of first case).
Then $12/1$ is $12$ - $1$ set of $12$ in $12$ (degenerate case).
Notice the size of the result set is getting bigger as the denominator (the number on the bottom) gets smaller.
Before we go to $0$ let's try something between $1$ and $0$ - $1/2$ or $0.5$. Think of just splitting each block into two (take a hatchet to the wooden blocks blocks, or just imagine it if mom doesn't want you handling a hatchet).
$12/0.5$ is $24$ - $24$ sets of $0.5$ (half-pieces) in $12$
$12/0.25$ is $48 - 48$ sets of $0.25$ (quarter-pieces) in $12$
$12/0.125$ is $96 - 96$ sets of $0.125$ (pieces of eight**) in $12$
$12/0.0625$ is $192 - 192$ sets of $0.0625$ (pieces of 16) in $12$
The close you get to zero, the larger the set you get gets.
$12/0.000000001$ (a billionth) is $12$ billion sets of a billionth of a block (aka, sawdust)
The as you approach zero, the resulting set size is too large to represent (not enough paper in this room, not enough memory on this computer) and the size of the pieces approach zero.
A cheat for "Too large to represent" is "infinity".
** pirate reference - do 5th graders still like pirates these days?
The problem I think is a conceptual one, As you describe, it's actually talking about sets, and while you can have a single set containing everything, you can't logically have less sets than you started with. "I want you to put all of these objects into a box too small for any of them" would be a similar task. The only appropriate response is to take Exception at the impossible task.
â Ruadhan2300
Aug 15 at 14:30
This is how my grandma explained it to me when I was like 5. I got it immediately and it stuck with me ever since.
â htmlcoderexe
2 days ago
A "piece of eight" (or, in what is perhaps a more "piraty" version, "piece o' eight") was a Spanish coin that was worth eight Spanish reales. So the piece of eight was the whole, while the real was the eighth. Another term for a real was a "bit". This survives in a bit being one-eighth of a byte, and in "shave and a hair-cut, two bits [i.e. 25 cents]". So you could say "half-piece, quarter-piece, bit-piece".
â Acccumulation
2 days ago
add a comment |Â
up vote
4
down vote
I don't have kids (my wife says one 3-year-old in the house is enough for her) and it's been a while since I was in the 5th grade (although at work sometimes...), but I'll give it a go.
I know you're too old to play with blocks, but lets start with 12 blocks.
Let's start with $12/6$ - that's $2$, right? Take $6$ at a time and there are two "sets". There are $2$ sets of $6$ in $12$.
Then $12/4$ is $3$ - $3$ sets of $4$ in $12$.
Then $12/3$ is $4$ - $4$ sets of $3$ in $12$ (commutation of the last case).
Then $12/2$ is $6$ - $2$ sets of $6$ in $12$ (commutation of first case).
Then $12/1$ is $12$ - $1$ set of $12$ in $12$ (degenerate case).
Notice the size of the result set is getting bigger as the denominator (the number on the bottom) gets smaller.
Before we go to $0$ let's try something between $1$ and $0$ - $1/2$ or $0.5$. Think of just splitting each block into two (take a hatchet to the wooden blocks blocks, or just imagine it if mom doesn't want you handling a hatchet).
$12/0.5$ is $24$ - $24$ sets of $0.5$ (half-pieces) in $12$
$12/0.25$ is $48 - 48$ sets of $0.25$ (quarter-pieces) in $12$
$12/0.125$ is $96 - 96$ sets of $0.125$ (pieces of eight**) in $12$
$12/0.0625$ is $192 - 192$ sets of $0.0625$ (pieces of 16) in $12$
The close you get to zero, the larger the set you get gets.
$12/0.000000001$ (a billionth) is $12$ billion sets of a billionth of a block (aka, sawdust)
The as you approach zero, the resulting set size is too large to represent (not enough paper in this room, not enough memory on this computer) and the size of the pieces approach zero.
A cheat for "Too large to represent" is "infinity".
** pirate reference - do 5th graders still like pirates these days?
The problem I think is a conceptual one, As you describe, it's actually talking about sets, and while you can have a single set containing everything, you can't logically have less sets than you started with. "I want you to put all of these objects into a box too small for any of them" would be a similar task. The only appropriate response is to take Exception at the impossible task.
â Ruadhan2300
Aug 15 at 14:30
This is how my grandma explained it to me when I was like 5. I got it immediately and it stuck with me ever since.
â htmlcoderexe
2 days ago
A "piece of eight" (or, in what is perhaps a more "piraty" version, "piece o' eight") was a Spanish coin that was worth eight Spanish reales. So the piece of eight was the whole, while the real was the eighth. Another term for a real was a "bit". This survives in a bit being one-eighth of a byte, and in "shave and a hair-cut, two bits [i.e. 25 cents]". So you could say "half-piece, quarter-piece, bit-piece".
â Acccumulation
2 days ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I don't have kids (my wife says one 3-year-old in the house is enough for her) and it's been a while since I was in the 5th grade (although at work sometimes...), but I'll give it a go.
I know you're too old to play with blocks, but lets start with 12 blocks.
Let's start with $12/6$ - that's $2$, right? Take $6$ at a time and there are two "sets". There are $2$ sets of $6$ in $12$.
Then $12/4$ is $3$ - $3$ sets of $4$ in $12$.
Then $12/3$ is $4$ - $4$ sets of $3$ in $12$ (commutation of the last case).
Then $12/2$ is $6$ - $2$ sets of $6$ in $12$ (commutation of first case).
Then $12/1$ is $12$ - $1$ set of $12$ in $12$ (degenerate case).
Notice the size of the result set is getting bigger as the denominator (the number on the bottom) gets smaller.
Before we go to $0$ let's try something between $1$ and $0$ - $1/2$ or $0.5$. Think of just splitting each block into two (take a hatchet to the wooden blocks blocks, or just imagine it if mom doesn't want you handling a hatchet).
$12/0.5$ is $24$ - $24$ sets of $0.5$ (half-pieces) in $12$
$12/0.25$ is $48 - 48$ sets of $0.25$ (quarter-pieces) in $12$
$12/0.125$ is $96 - 96$ sets of $0.125$ (pieces of eight**) in $12$
$12/0.0625$ is $192 - 192$ sets of $0.0625$ (pieces of 16) in $12$
The close you get to zero, the larger the set you get gets.
$12/0.000000001$ (a billionth) is $12$ billion sets of a billionth of a block (aka, sawdust)
The as you approach zero, the resulting set size is too large to represent (not enough paper in this room, not enough memory on this computer) and the size of the pieces approach zero.
A cheat for "Too large to represent" is "infinity".
** pirate reference - do 5th graders still like pirates these days?
I don't have kids (my wife says one 3-year-old in the house is enough for her) and it's been a while since I was in the 5th grade (although at work sometimes...), but I'll give it a go.
I know you're too old to play with blocks, but lets start with 12 blocks.
Let's start with $12/6$ - that's $2$, right? Take $6$ at a time and there are two "sets". There are $2$ sets of $6$ in $12$.
Then $12/4$ is $3$ - $3$ sets of $4$ in $12$.
Then $12/3$ is $4$ - $4$ sets of $3$ in $12$ (commutation of the last case).
Then $12/2$ is $6$ - $2$ sets of $6$ in $12$ (commutation of first case).
Then $12/1$ is $12$ - $1$ set of $12$ in $12$ (degenerate case).
Notice the size of the result set is getting bigger as the denominator (the number on the bottom) gets smaller.
Before we go to $0$ let's try something between $1$ and $0$ - $1/2$ or $0.5$. Think of just splitting each block into two (take a hatchet to the wooden blocks blocks, or just imagine it if mom doesn't want you handling a hatchet).
$12/0.5$ is $24$ - $24$ sets of $0.5$ (half-pieces) in $12$
$12/0.25$ is $48 - 48$ sets of $0.25$ (quarter-pieces) in $12$
$12/0.125$ is $96 - 96$ sets of $0.125$ (pieces of eight**) in $12$
$12/0.0625$ is $192 - 192$ sets of $0.0625$ (pieces of 16) in $12$
The close you get to zero, the larger the set you get gets.
$12/0.000000001$ (a billionth) is $12$ billion sets of a billionth of a block (aka, sawdust)
The as you approach zero, the resulting set size is too large to represent (not enough paper in this room, not enough memory on this computer) and the size of the pieces approach zero.
A cheat for "Too large to represent" is "infinity".
** pirate reference - do 5th graders still like pirates these days?
edited Aug 15 at 14:05
JayTuma
978116
978116
answered Aug 15 at 13:41
Nufosmatic
491
491
The problem I think is a conceptual one, As you describe, it's actually talking about sets, and while you can have a single set containing everything, you can't logically have less sets than you started with. "I want you to put all of these objects into a box too small for any of them" would be a similar task. The only appropriate response is to take Exception at the impossible task.
â Ruadhan2300
Aug 15 at 14:30
This is how my grandma explained it to me when I was like 5. I got it immediately and it stuck with me ever since.
â htmlcoderexe
2 days ago
A "piece of eight" (or, in what is perhaps a more "piraty" version, "piece o' eight") was a Spanish coin that was worth eight Spanish reales. So the piece of eight was the whole, while the real was the eighth. Another term for a real was a "bit". This survives in a bit being one-eighth of a byte, and in "shave and a hair-cut, two bits [i.e. 25 cents]". So you could say "half-piece, quarter-piece, bit-piece".
â Acccumulation
2 days ago
add a comment |Â
The problem I think is a conceptual one, As you describe, it's actually talking about sets, and while you can have a single set containing everything, you can't logically have less sets than you started with. "I want you to put all of these objects into a box too small for any of them" would be a similar task. The only appropriate response is to take Exception at the impossible task.
â Ruadhan2300
Aug 15 at 14:30
This is how my grandma explained it to me when I was like 5. I got it immediately and it stuck with me ever since.
â htmlcoderexe
2 days ago
A "piece of eight" (or, in what is perhaps a more "piraty" version, "piece o' eight") was a Spanish coin that was worth eight Spanish reales. So the piece of eight was the whole, while the real was the eighth. Another term for a real was a "bit". This survives in a bit being one-eighth of a byte, and in "shave and a hair-cut, two bits [i.e. 25 cents]". So you could say "half-piece, quarter-piece, bit-piece".
â Acccumulation
2 days ago
The problem I think is a conceptual one, As you describe, it's actually talking about sets, and while you can have a single set containing everything, you can't logically have less sets than you started with. "I want you to put all of these objects into a box too small for any of them" would be a similar task. The only appropriate response is to take Exception at the impossible task.
â Ruadhan2300
Aug 15 at 14:30
The problem I think is a conceptual one, As you describe, it's actually talking about sets, and while you can have a single set containing everything, you can't logically have less sets than you started with. "I want you to put all of these objects into a box too small for any of them" would be a similar task. The only appropriate response is to take Exception at the impossible task.
â Ruadhan2300
Aug 15 at 14:30
This is how my grandma explained it to me when I was like 5. I got it immediately and it stuck with me ever since.
â htmlcoderexe
2 days ago
This is how my grandma explained it to me when I was like 5. I got it immediately and it stuck with me ever since.
â htmlcoderexe
2 days ago
A "piece of eight" (or, in what is perhaps a more "piraty" version, "piece o' eight") was a Spanish coin that was worth eight Spanish reales. So the piece of eight was the whole, while the real was the eighth. Another term for a real was a "bit". This survives in a bit being one-eighth of a byte, and in "shave and a hair-cut, two bits [i.e. 25 cents]". So you could say "half-piece, quarter-piece, bit-piece".
â Acccumulation
2 days ago
A "piece of eight" (or, in what is perhaps a more "piraty" version, "piece o' eight") was a Spanish coin that was worth eight Spanish reales. So the piece of eight was the whole, while the real was the eighth. Another term for a real was a "bit". This survives in a bit being one-eighth of a byte, and in "shave and a hair-cut, two bits [i.e. 25 cents]". So you could say "half-piece, quarter-piece, bit-piece".
â Acccumulation
2 days ago
add a comment |Â
up vote
4
down vote
One would need to first explain what we mean by division. That is, what does $/$ mean in the expression $a/b,$ where $a$ and $b$ are integers?
Well, whatever it is, it is a way of combining two numbers. Now recall that every time we defined an operation (say addition), we always had a unique result as the product of the combination, so that we would like this to continue to hold. What else? We define $/$ indirectly, by looking at what we want $a/b$ to mean. Well, we want it to stand for the number $c$ which when multiplied together with $b$ recovers $a.$ (Recall how we similarly defined subtraction as the inverse operation of $+.$)
Therefore, in summary, if we let $a/b=c,$ then by definition this equality is equivalent to $cÃÂb=a.$ Also, we want $c$ to be unique for all possible integers $a$ and $b.$
Now consider the expression $a/0.$ First let us take $ane0.$ Then if we let $a/0=c,$ it follows by definition that $cÃÂ0=a.$ But with the way we defined multiplication (remind him of this), we required that $0$ must make any number vanish, so that there simply is no such $c$ as we seek. If now we let $a=0,$ then we want a unique $c$ such that $cÃÂ0=0.$ But again, by the property $rÃÂ0=0,,,forall r$ which we've previously allowed in defining $ÃÂ,$ we have infinitely many candidates for $c$ and there is no other condition we can impose to select one uniquely. We therefore do not allow ourselves to divide by $0$ in any case, in order to avoid all that mess.
add a comment |Â
up vote
4
down vote
One would need to first explain what we mean by division. That is, what does $/$ mean in the expression $a/b,$ where $a$ and $b$ are integers?
Well, whatever it is, it is a way of combining two numbers. Now recall that every time we defined an operation (say addition), we always had a unique result as the product of the combination, so that we would like this to continue to hold. What else? We define $/$ indirectly, by looking at what we want $a/b$ to mean. Well, we want it to stand for the number $c$ which when multiplied together with $b$ recovers $a.$ (Recall how we similarly defined subtraction as the inverse operation of $+.$)
Therefore, in summary, if we let $a/b=c,$ then by definition this equality is equivalent to $cÃÂb=a.$ Also, we want $c$ to be unique for all possible integers $a$ and $b.$
Now consider the expression $a/0.$ First let us take $ane0.$ Then if we let $a/0=c,$ it follows by definition that $cÃÂ0=a.$ But with the way we defined multiplication (remind him of this), we required that $0$ must make any number vanish, so that there simply is no such $c$ as we seek. If now we let $a=0,$ then we want a unique $c$ such that $cÃÂ0=0.$ But again, by the property $rÃÂ0=0,,,forall r$ which we've previously allowed in defining $ÃÂ,$ we have infinitely many candidates for $c$ and there is no other condition we can impose to select one uniquely. We therefore do not allow ourselves to divide by $0$ in any case, in order to avoid all that mess.
add a comment |Â
up vote
4
down vote
up vote
4
down vote
One would need to first explain what we mean by division. That is, what does $/$ mean in the expression $a/b,$ where $a$ and $b$ are integers?
Well, whatever it is, it is a way of combining two numbers. Now recall that every time we defined an operation (say addition), we always had a unique result as the product of the combination, so that we would like this to continue to hold. What else? We define $/$ indirectly, by looking at what we want $a/b$ to mean. Well, we want it to stand for the number $c$ which when multiplied together with $b$ recovers $a.$ (Recall how we similarly defined subtraction as the inverse operation of $+.$)
Therefore, in summary, if we let $a/b=c,$ then by definition this equality is equivalent to $cÃÂb=a.$ Also, we want $c$ to be unique for all possible integers $a$ and $b.$
Now consider the expression $a/0.$ First let us take $ane0.$ Then if we let $a/0=c,$ it follows by definition that $cÃÂ0=a.$ But with the way we defined multiplication (remind him of this), we required that $0$ must make any number vanish, so that there simply is no such $c$ as we seek. If now we let $a=0,$ then we want a unique $c$ such that $cÃÂ0=0.$ But again, by the property $rÃÂ0=0,,,forall r$ which we've previously allowed in defining $ÃÂ,$ we have infinitely many candidates for $c$ and there is no other condition we can impose to select one uniquely. We therefore do not allow ourselves to divide by $0$ in any case, in order to avoid all that mess.
One would need to first explain what we mean by division. That is, what does $/$ mean in the expression $a/b,$ where $a$ and $b$ are integers?
Well, whatever it is, it is a way of combining two numbers. Now recall that every time we defined an operation (say addition), we always had a unique result as the product of the combination, so that we would like this to continue to hold. What else? We define $/$ indirectly, by looking at what we want $a/b$ to mean. Well, we want it to stand for the number $c$ which when multiplied together with $b$ recovers $a.$ (Recall how we similarly defined subtraction as the inverse operation of $+.$)
Therefore, in summary, if we let $a/b=c,$ then by definition this equality is equivalent to $cÃÂb=a.$ Also, we want $c$ to be unique for all possible integers $a$ and $b.$
Now consider the expression $a/0.$ First let us take $ane0.$ Then if we let $a/0=c,$ it follows by definition that $cÃÂ0=a.$ But with the way we defined multiplication (remind him of this), we required that $0$ must make any number vanish, so that there simply is no such $c$ as we seek. If now we let $a=0,$ then we want a unique $c$ such that $cÃÂ0=0.$ But again, by the property $rÃÂ0=0,,,forall r$ which we've previously allowed in defining $ÃÂ,$ we have infinitely many candidates for $c$ and there is no other condition we can impose to select one uniquely. We therefore do not allow ourselves to divide by $0$ in any case, in order to avoid all that mess.
answered Aug 15 at 17:53
Allawonder
1,647414
1,647414
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4
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@Jack M and and @greedoid probably highlight a good point: division does not exist. It's only the inverse operation of multiplication.
You could explain your brother the complete truth: dividing 20 by 5 is about finding the only answer (if it exists) to this question: what number can be multiplied by 5 to give 20?. The unique answer is easy: 4 times 5 is 20.
And the division is only another phrasing to say the exact same thing: 20 divided by 5 is 4.
Can you always find one and only one answer? Yup, almost always...
There's only one exception...
What number, multiplied by 0, gives 20? There's none.
So "division" by 0 has no meaning, since we cannot find any number that satisfies our definition.
You could even draw his attention by mentioning that most grown-ups don't know there's no such thing as "division", and that's the first step to learn about "E-vector spaces", "rings" and other funny-named artefacts when he's in college... or before that!
Note: what if he raises a question about "0/0"?
OK, let's try: "what number, multiplied by 0, gives 0?" All of them! We cannot find one and only one answer, so, it's still impossible to divide 0 by 0!
There are a lot of buzzwords for a 5th grader, but the explanation is good.
â Alvin Lepik
2 days ago
add a comment |Â
up vote
4
down vote
@Jack M and and @greedoid probably highlight a good point: division does not exist. It's only the inverse operation of multiplication.
You could explain your brother the complete truth: dividing 20 by 5 is about finding the only answer (if it exists) to this question: what number can be multiplied by 5 to give 20?. The unique answer is easy: 4 times 5 is 20.
And the division is only another phrasing to say the exact same thing: 20 divided by 5 is 4.
Can you always find one and only one answer? Yup, almost always...
There's only one exception...
What number, multiplied by 0, gives 20? There's none.
So "division" by 0 has no meaning, since we cannot find any number that satisfies our definition.
You could even draw his attention by mentioning that most grown-ups don't know there's no such thing as "division", and that's the first step to learn about "E-vector spaces", "rings" and other funny-named artefacts when he's in college... or before that!
Note: what if he raises a question about "0/0"?
OK, let's try: "what number, multiplied by 0, gives 0?" All of them! We cannot find one and only one answer, so, it's still impossible to divide 0 by 0!
There are a lot of buzzwords for a 5th grader, but the explanation is good.
â Alvin Lepik
2 days ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
@Jack M and and @greedoid probably highlight a good point: division does not exist. It's only the inverse operation of multiplication.
You could explain your brother the complete truth: dividing 20 by 5 is about finding the only answer (if it exists) to this question: what number can be multiplied by 5 to give 20?. The unique answer is easy: 4 times 5 is 20.
And the division is only another phrasing to say the exact same thing: 20 divided by 5 is 4.
Can you always find one and only one answer? Yup, almost always...
There's only one exception...
What number, multiplied by 0, gives 20? There's none.
So "division" by 0 has no meaning, since we cannot find any number that satisfies our definition.
You could even draw his attention by mentioning that most grown-ups don't know there's no such thing as "division", and that's the first step to learn about "E-vector spaces", "rings" and other funny-named artefacts when he's in college... or before that!
Note: what if he raises a question about "0/0"?
OK, let's try: "what number, multiplied by 0, gives 0?" All of them! We cannot find one and only one answer, so, it's still impossible to divide 0 by 0!
@Jack M and and @greedoid probably highlight a good point: division does not exist. It's only the inverse operation of multiplication.
You could explain your brother the complete truth: dividing 20 by 5 is about finding the only answer (if it exists) to this question: what number can be multiplied by 5 to give 20?. The unique answer is easy: 4 times 5 is 20.
And the division is only another phrasing to say the exact same thing: 20 divided by 5 is 4.
Can you always find one and only one answer? Yup, almost always...
There's only one exception...
What number, multiplied by 0, gives 20? There's none.
So "division" by 0 has no meaning, since we cannot find any number that satisfies our definition.
You could even draw his attention by mentioning that most grown-ups don't know there's no such thing as "division", and that's the first step to learn about "E-vector spaces", "rings" and other funny-named artefacts when he's in college... or before that!
Note: what if he raises a question about "0/0"?
OK, let's try: "what number, multiplied by 0, gives 0?" All of them! We cannot find one and only one answer, so, it's still impossible to divide 0 by 0!
edited 2 days ago
answered Aug 15 at 17:56
Evariste
1493
1493
There are a lot of buzzwords for a 5th grader, but the explanation is good.
â Alvin Lepik
2 days ago
add a comment |Â
There are a lot of buzzwords for a 5th grader, but the explanation is good.
â Alvin Lepik
2 days ago
There are a lot of buzzwords for a 5th grader, but the explanation is good.
â Alvin Lepik
2 days ago
There are a lot of buzzwords for a 5th grader, but the explanation is good.
â Alvin Lepik
2 days ago
add a comment |Â
up vote
2
down vote
Division is sharing:
1 / 10
:
10 boys in a class grab at a toy -- they rip the toy to tiny bits!
1 / 2
:
2 boys fight for a toy -- they rip the toy in half!
1 / 0
:
A different toy is alone -- he is a special boy!
add a comment |Â
up vote
2
down vote
Division is sharing:
1 / 10
:
10 boys in a class grab at a toy -- they rip the toy to tiny bits!
1 / 2
:
2 boys fight for a toy -- they rip the toy in half!
1 / 0
:
A different toy is alone -- he is a special boy!
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Division is sharing:
1 / 10
:
10 boys in a class grab at a toy -- they rip the toy to tiny bits!
1 / 2
:
2 boys fight for a toy -- they rip the toy in half!
1 / 0
:
A different toy is alone -- he is a special boy!
Division is sharing:
1 / 10
:
10 boys in a class grab at a toy -- they rip the toy to tiny bits!
1 / 2
:
2 boys fight for a toy -- they rip the toy in half!
1 / 0
:
A different toy is alone -- he is a special boy!
answered 2 days ago
Jason
1626
1626
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up vote
1
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Because before you think about dividing something, it is more important to consider if you have someone to divide it for (he/she/it must be present, exist, etc). If you do not have anyone who can 'benefit' from the division, no point in dividing. Non rigorous, pragmatic, heuristic approach. It might pave the way for more reasoned proofs and demonstrations.
This is indeed similar with the 'division is sharing' concept in the answer of @Jason given here
â XavierStuvw
yesterday
add a comment |Â
up vote
1
down vote
Because before you think about dividing something, it is more important to consider if you have someone to divide it for (he/she/it must be present, exist, etc). If you do not have anyone who can 'benefit' from the division, no point in dividing. Non rigorous, pragmatic, heuristic approach. It might pave the way for more reasoned proofs and demonstrations.
This is indeed similar with the 'division is sharing' concept in the answer of @Jason given here
â XavierStuvw
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Because before you think about dividing something, it is more important to consider if you have someone to divide it for (he/she/it must be present, exist, etc). If you do not have anyone who can 'benefit' from the division, no point in dividing. Non rigorous, pragmatic, heuristic approach. It might pave the way for more reasoned proofs and demonstrations.
Because before you think about dividing something, it is more important to consider if you have someone to divide it for (he/she/it must be present, exist, etc). If you do not have anyone who can 'benefit' from the division, no point in dividing. Non rigorous, pragmatic, heuristic approach. It might pave the way for more reasoned proofs and demonstrations.
answered 2 days ago
XavierStuvw
1496
1496
This is indeed similar with the 'division is sharing' concept in the answer of @Jason given here
â XavierStuvw
yesterday
add a comment |Â
This is indeed similar with the 'division is sharing' concept in the answer of @Jason given here
â XavierStuvw
yesterday
This is indeed similar with the 'division is sharing' concept in the answer of @Jason given here
â XavierStuvw
yesterday
This is indeed similar with the 'division is sharing' concept in the answer of @Jason given here
â XavierStuvw
yesterday
add a comment |Â
up vote
1
down vote
The way I taught it, even to junior college students who were taking elementary mathematics courses, was with a calculator.
I would show them that 1/1 = 1, 1/0.1 = 10, 1/0.01 = 100, and so on. I would ask them if they saw how the numbers kept getting bigger as we divided by smaller and smaller numbers. Then I would ask them what they thought would happen when we hit zero. "We would get the biggest possible number that exists, right? But there is no biggest number. So dividing by zero gives you a number that doesn't exist. Does that make any sense? No. So we say that dividing by zero is undefined."
add a comment |Â
up vote
1
down vote
The way I taught it, even to junior college students who were taking elementary mathematics courses, was with a calculator.
I would show them that 1/1 = 1, 1/0.1 = 10, 1/0.01 = 100, and so on. I would ask them if they saw how the numbers kept getting bigger as we divided by smaller and smaller numbers. Then I would ask them what they thought would happen when we hit zero. "We would get the biggest possible number that exists, right? But there is no biggest number. So dividing by zero gives you a number that doesn't exist. Does that make any sense? No. So we say that dividing by zero is undefined."
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The way I taught it, even to junior college students who were taking elementary mathematics courses, was with a calculator.
I would show them that 1/1 = 1, 1/0.1 = 10, 1/0.01 = 100, and so on. I would ask them if they saw how the numbers kept getting bigger as we divided by smaller and smaller numbers. Then I would ask them what they thought would happen when we hit zero. "We would get the biggest possible number that exists, right? But there is no biggest number. So dividing by zero gives you a number that doesn't exist. Does that make any sense? No. So we say that dividing by zero is undefined."
The way I taught it, even to junior college students who were taking elementary mathematics courses, was with a calculator.
I would show them that 1/1 = 1, 1/0.1 = 10, 1/0.01 = 100, and so on. I would ask them if they saw how the numbers kept getting bigger as we divided by smaller and smaller numbers. Then I would ask them what they thought would happen when we hit zero. "We would get the biggest possible number that exists, right? But there is no biggest number. So dividing by zero gives you a number that doesn't exist. Does that make any sense? No. So we say that dividing by zero is undefined."
answered yesterday
user2303321
1234
1234
add a comment |Â
add a comment |Â
up vote
0
down vote
To divide means to subtract many times.
It might be a duplicated answer and I apologize, in case. But, according to my experience as a teacher, this worked well.
The point, as others had observed, is what does "to divide" mean. This sometimes looked obscure to the students, whereas the concept of subtraction was more clear.
Thus, once you convey the message that "to divide" means "to subtract many times", everything becomes more clear.
How many times can we subtract $3$ from $10$? Well, usually my students got this.
How many times can we subtract $0$ from $10$? Well, how many times we want!
So there is not a precise answer, because any answer is good. This made more clear the sense of "not defined", at least to my students.
Hope it helps!
add a comment |Â
up vote
0
down vote
To divide means to subtract many times.
It might be a duplicated answer and I apologize, in case. But, according to my experience as a teacher, this worked well.
The point, as others had observed, is what does "to divide" mean. This sometimes looked obscure to the students, whereas the concept of subtraction was more clear.
Thus, once you convey the message that "to divide" means "to subtract many times", everything becomes more clear.
How many times can we subtract $3$ from $10$? Well, usually my students got this.
How many times can we subtract $0$ from $10$? Well, how many times we want!
So there is not a precise answer, because any answer is good. This made more clear the sense of "not defined", at least to my students.
Hope it helps!
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To divide means to subtract many times.
It might be a duplicated answer and I apologize, in case. But, according to my experience as a teacher, this worked well.
The point, as others had observed, is what does "to divide" mean. This sometimes looked obscure to the students, whereas the concept of subtraction was more clear.
Thus, once you convey the message that "to divide" means "to subtract many times", everything becomes more clear.
How many times can we subtract $3$ from $10$? Well, usually my students got this.
How many times can we subtract $0$ from $10$? Well, how many times we want!
So there is not a precise answer, because any answer is good. This made more clear the sense of "not defined", at least to my students.
Hope it helps!
To divide means to subtract many times.
It might be a duplicated answer and I apologize, in case. But, according to my experience as a teacher, this worked well.
The point, as others had observed, is what does "to divide" mean. This sometimes looked obscure to the students, whereas the concept of subtraction was more clear.
Thus, once you convey the message that "to divide" means "to subtract many times", everything becomes more clear.
How many times can we subtract $3$ from $10$? Well, usually my students got this.
How many times can we subtract $0$ from $10$? Well, how many times we want!
So there is not a precise answer, because any answer is good. This made more clear the sense of "not defined", at least to my students.
Hope it helps!
answered yesterday
Andrea Prunotto
712316
712316
add a comment |Â
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up vote
0
down vote
Explain him the problems, don't enforce him as an "official view".
Explain him, what are the problems of the division by zero.
Let him to think about a possible solution.
You might also explain, that also the negative numbers don't have a suqare root, but this problem had a solution, the imaginary numbers. Let him try to think about a similar solution for the division by zero.
add a comment |Â
up vote
0
down vote
Explain him the problems, don't enforce him as an "official view".
Explain him, what are the problems of the division by zero.
Let him to think about a possible solution.
You might also explain, that also the negative numbers don't have a suqare root, but this problem had a solution, the imaginary numbers. Let him try to think about a similar solution for the division by zero.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Explain him the problems, don't enforce him as an "official view".
Explain him, what are the problems of the division by zero.
Let him to think about a possible solution.
You might also explain, that also the negative numbers don't have a suqare root, but this problem had a solution, the imaginary numbers. Let him try to think about a similar solution for the division by zero.
Explain him the problems, don't enforce him as an "official view".
Explain him, what are the problems of the division by zero.
Let him to think about a possible solution.
You might also explain, that also the negative numbers don't have a suqare root, but this problem had a solution, the imaginary numbers. Let him try to think about a similar solution for the division by zero.
answered yesterday
peterh
2,14531631
2,14531631
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up vote
0
down vote
Number of marbles : Number of boxes = Number of marbles in each box.
20 marbles : 4 boxes = 5 marbles per box
0 marbles : 4 boxes = 0 marbles per box
20 marbles : 0 boxes = "how many marbles in each box while no box?" ---> undefined!
add a comment |Â
up vote
0
down vote
Number of marbles : Number of boxes = Number of marbles in each box.
20 marbles : 4 boxes = 5 marbles per box
0 marbles : 4 boxes = 0 marbles per box
20 marbles : 0 boxes = "how many marbles in each box while no box?" ---> undefined!
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Number of marbles : Number of boxes = Number of marbles in each box.
20 marbles : 4 boxes = 5 marbles per box
0 marbles : 4 boxes = 0 marbles per box
20 marbles : 0 boxes = "how many marbles in each box while no box?" ---> undefined!
Number of marbles : Number of boxes = Number of marbles in each box.
20 marbles : 4 boxes = 5 marbles per box
0 marbles : 4 boxes = 0 marbles per box
20 marbles : 0 boxes = "how many marbles in each box while no box?" ---> undefined!
answered 9 hours ago
Friendly Ghost
16018
16018
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protected by Daniel Fischer⦠Aug 15 at 18:03
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63
How do you divide, say, $5$ apples between zero people? There is no meaningful way to distribute the apples.
â Alvin Lepik
Aug 15 at 10:36
12
Division by zero should not be justified at all.
â Peter
Aug 15 at 11:39
23
Just ask Siri. It explains this very well.
â DonielF
Aug 15 at 15:32
12
This would also fit well on the Mathematics Educators site.
â Chase Ryan Taylor
Aug 15 at 16:10
21
"Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends." - Siri
â jkd
Aug 15 at 19:12