Prove that multiplication is commutative

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Homework Statement


n * m = m * n where m, n are natural numbers.


Homework Equations


I am working from Terrence Tao's class notes and he includes 0 in the natural numbers. m++ stands for m+1. He calls it incrementation and uses it to explain the rules of addition.


The Attempt at a Solution


Proof. We will induct on m holding n constant. For the base case, let m = 0. Then we have n * 0 = 0 * n. Both sides of the equation equal to 0. Now we assume inductively that n * m = m * n. For the inductive step we need to show that n * (m++) = (m++) * n. Since m is a natural number, we know from a previous proposition that m++ is also a natural number. Thus we can choose q to be the successor of m or in other words q = m++. Then we have n * q = q * n. According to the inductive hypothesis, this is true. This closes our induction.

Can someone take a quick look at my proof and tell me if there is anything wrong with it? Thanks!
 
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It sounds to me like you are using "Peano's axioms" which take induction,
"If a set X contains 0 and, whenever it contains a number a, it also contains a++, then X is the set of all non-negative integers", as a defining property of the non-negative integers. In particular, a+ b is defined by "b+ 1= b++" and if c is not 0, then there exist a such that c= a++ in which case b+ c= (b+a)++.

However, the difficulty appears to be that you do not understand what induction says! You say, first, "Now we assume inductively that n * m = m * n." Later you say "Then we have n * q = q * n. According to the inductive hypothesis, this is true." No, it isn't. The "induction hypothesis" is that n*m= m*n for those particular values of m and n. q is NOT m so n*q= m*q does NOT follow from n*m= m*n. You are essentially asserting that your "induction hypothesis" is just what you are trying to prove.

I also note that you don't say anything about how multiplication is defined here. You can't very well prove something about "multiplication" without using the definition of "multiplication"! I recommend that you talk to your teacher about this.
 
HallsofIvy said:
However, the difficulty appears to be that you do not understand what induction says! You say, first, "Now we assume inductively that n * m = m * n." Later you say "Then we have n * q = q * n. According to the inductive hypothesis, this is true." No, it isn't. The "induction hypothesis" is that n*m= m*n for those particular values of m and n. q is NOT m so n*q= m*q does NOT follow from n*m= m*n. You are essentially asserting that your "induction hypothesis" is just what you are trying to prove.

That's exactly my problem! Thanks. With that clarified, I need to go back and fix a few other proofs I did before this one.

I'm trying to teach myself by using the class notes that Professor Tao at UCLA uses to teach his honors analysis course so I don't have a teacher to talk to.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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