0
$\begingroup$

Why should $b$ groups of $a$ apples be the same as $a$ groups of $b$ apples?

We where taught this so it seems rather trivial but the more I think about it the more I feel that it is not.

I'm trying to avoid an argument that uses the fact that multiplication is commutative. Because I see that I am trying to PROVE that in $\mathbb{Z}^{+}-0$ multiplication is commutative if we define multiplication by repeated addition.

I would accept arguments using the fact that:

$a+b=b+a$ because if we define $+$ to be the operation combining to quantities then it should be rather trivial that $a$ apples and $b$ apples is the same as $b$ apples and $a$ apples.

Is it enough to draw $a$ groups of $b$ (1 by 1) squares and rotate this to show that it is the same as $b$ groups of $a$ (1 by 1) squares. It does not seem good enough for me because it uses a picture, and I was taught before that pictures in math do not prove anything.

|cite|improve this question
$\endgroup$
  • $\begingroup$ "picturs in maths do not prove anything" Why ever do you think that? $\endgroup$ – Zain Patel Jul 8 '16 at 21:17
  • $\begingroup$ So then it is the case that it is true. I just remember my math teacher telling me that. And I kind of feel that it is true because pictures are usually of one example, a proof often consist of an infinite amount of examples. How are we to draw $a$ groups of $b$? The only thing that would make sense is if we fix $a$ and fix $b$ then we could proceed would a specific example , that specific picture. $\endgroup$ – Ahmed S. Attaalla Jul 8 '16 at 21:21
  • $\begingroup$ I wasn't necessarily saying that it could be used here, merely contesting the claim that pictures in maths do not prove anything. For example, see this: mathoverflow.net/questions/8846/proofs-without-words $\endgroup$ – Zain Patel Jul 8 '16 at 21:23
  • 1
    $\begingroup$ Maybe a proof by picture wouldn't be a rigorous proof in this case, but it should be enough to at least convince you the result is true. Pick any positive integers $a$ and $b$ and draw the picture as you described. Then note that there's nothing special about those specific values of $a$ and $b$ you picked. Italicized for emphasis, because that's the part that should convince you it works in general. But again, not a rigorous proof. $\endgroup$ – tilper Jul 8 '16 at 21:36
  • $\begingroup$ I suspect that your math teacher told you that you could not just appeal to a picture to prove a geometry theorem. That has nothing to do with this problem. $\endgroup$ – user247327 Jul 9 '16 at 1:32
7
$\begingroup$

The usual way of proving it is by induction.

First, define multiplication recursively: $$a\cdot 1 = a\\a\cdot(b+1)=a\cdot b + a$$

Next, show, by induction on $b$ that $1\cdot b=b\cdot 1$. That's relatively easy to do.

Next, prove by induction on $b$ that that $(a+1)\cdot b = a\cdot b + b$ by induction on $b$.

Finally, prove by induction on $a$ that for all $b$, $a\cdot b = b\cdot a$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ "Usual"? I have to confess I've never seen this. By the time I got to abstract algebra it was assumed we'd simply define multiplication to be commutative. In elementary school when dinosaurs walked the earth I was taught with images of bags of groceries that if we choose one from each bag and put them in a new bag the result would go from n bags of m items to m bags of n items. $\endgroup$ – fleablood Jul 9 '16 at 1:47
  • $\begingroup$ Um, how do we prove that multiplication is distributive? Or why can't we just define it to be commutative? $\endgroup$ – fleablood Jul 9 '16 at 1:47
  • $\begingroup$ Where do I need the distributive law? I only need a special case of it $(a+1)b=ab+b$, which I've described how to prove (but I haven't proven anything above.) How do you "define" something to be commutative? I'm defining $a\cdot b$ recursively - you know how to compute $a\cdot 1$ in the first line, and the rest are done recursively. @fleablood $\endgroup$ – Thomas Andrews Jul 9 '16 at 2:08
  • $\begingroup$ @fleablood Some people use what I'm calling a definition as the axioms, instead, for multiplication. The "usual" place is in logic rather than algebra, where we are starting with minimal Peano axioms for number theory. See, for example, "Foundations of Analysis," by Landau, which starts with the Peano axioms and proceeds to give a rigorous foundation for the rationals and reals. $\endgroup$ – Thomas Andrews Jul 9 '16 at 2:11
  • $\begingroup$ You're right about distribution. I misread. "usual" is subjective. Your definition makes sense. $\endgroup$ – fleablood Jul 9 '16 at 2:47
3
$\begingroup$

Label each apple with a pair of numbers $(x,y)$ such that $x$ is the number of the group the apple was originally in (1 through $b$) and $y$ is the number of the apple within the original group (1 through $a$). Every apple gets a unique label this way. Now, change every label to reverse the two numbers: $(x,y) \rightarrow (y,x)$. Every apple still has a unique label. But, this new labeling scheme would also come about from grouping $a$ groups of $b$ apples. Since the number of labels is the same in both cases, $b$ groups of $a$ apples must have the same size as $a$ groups of $b$ apples.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

If $a*b$ means $a$ groups each with $b$ items we can:

Go through each of the $a$ groups and remove an apple and then group all those sample apples together into a new group. This new group will have $a$ apples; one from each group. Do this for each of the $b$ apples in the group. You will end up with $b$ groups; one for each apple in the groups.

You now have $b*a$ apples. Now unless we think apples can disappear and/or spontaneously appear by rearranging those numbers must be the same.

|cite|improve this answer
$\endgroup$
-1
$\begingroup$

Multiplication is commutative because a collection of objects that has $a$ groups of $b$ objects in each group can be partitioned also into $b$ groups of $a$ objects. This is the equivalent to repeated addition:
$a+a+...+a$ ($b$ times) is the same as
$b+b+...+b$ ($a$ times).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ This doesn't answer the question; it just changes the form in which the asserted fact is stated. $\endgroup$ – Greg Martin Jul 8 '16 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.