# Commutativity of multiplication in $\mathbb{N}$

I'm trying to prove that $a\cdot b=b\cdot a$ when $a$ and $b$ are two natural numbers.

In the rest of this question I'm using $a'$ for the successor of $a$.

• $a+0=a$
• $a+b'=(a+b)'$

Multiplication is defined as:

• $a\cdot 0=0$
• $a\cdot b'=a+ab$

I already proved commutativity and associativity for addition. I also proved that $a\cdot 1=1\cdot a=a$.

I tried with induction on $b$. I can easily show that $a\cdot 0=0\cdot a$. Then I suppose $a\cdot b=b\cdot a$ and try to show that $a\cdot b'=b'\cdot a$.

Here I can no longer go on. The main problem is I can't use distributivity laws since I haven't proved them yet. I hope to do that immediately after this problem is fixed. Also, $b'\cdot a$ is problematic because $b'$ is at the left.

Any hints?

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Proof distributive law first. In a sense, the distributive law is more "basic" because we will often want it to be true when we don't have commutativity. So First show that $a\cdot(b+0)=(a\cdot b) + (a\cdot 0)$ then proof the general case. Not sure if you'll also need to show right-distributivity: $(a+b)\cdot c=(a\cdot c) + (b\cdot c)$ –  Thomas Andrews Apr 20 '11 at 19:56

This is how I would go about it, in three steps.

1. Prove $0=m\cdot 0=0\cdot m$ for all $m\in\omega$. As you said, you can easily show this.

2. Prove $m'n=mn+n$ for all $m,n\in\omega$. We can do this by induction. Let $$K=\{n\in\omega\ |\ m'n=mn+n\}$$ By definition, $m'\cdot 0=0$, and $m\cdot 0+0=0+0=0$, so $0\in K$. Suppose $n\in K$. Then $$m'n'=m'+m'n=m'+mn+n=m+mn+n'=mn'+n'$$ where I have used the second facts you listed for addition and multiplication, and I assume you know $a'+b=(a+b)'=a+b'$, which is usually used in proving the commutativity of addition. So $n'\in K$.

3. We can now prove $mn=nm$ for all $m,n\in\omega$. Let $S=\{m\in\omega\ |\forall_{n\in\omega}\ mn=nm\}$. By Step 1, $0\in S$. Let $m\in S$. Then $$m'n=mn+n=nm+n=n+nm=nm'$$ so $m'\in S$, so $S$ is inductive, and $S=\omega$.

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Thank you yunone! –  Francesco Turco Apr 20 '11 at 20:02
Glad I could help. –  yunone Apr 20 '11 at 20:06

Assume commutativity of the products $ab$, $ab'$ and $a'b$. Then you can rewrite $b'a'$ and $a'b'$ in terms of $a$, $b$ and $ab$ using commutativity and associativity of addition and the induction hypothesis to show that they are equal. I'll be happy to provide more details, but you asked for hints.

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If I assume $ab=ba$, $ab'=b'a$ and $a'b=ba'$ I can prove that $a'b'=b'a'$ the way you said, but I don't get how I can apply this fact to my initial problem. Can you provide some more details, please? –  Francesco Turco Apr 20 '11 at 19:10
@Francesco: I assume that question is resolved by yunone's answer? –  joriki Apr 20 '11 at 20:06
yunone completely answered my question. I tried to understand your proof but I failed, probably because I'm a little too tired now. I'll try again tomorrow. Anyway I would be happy if you could elaborate your answer a little bit. And yes, I know, you're right. I only asked for hints. But since I'm a newbie, I should have spoken differently :) –  Francesco Turco Apr 20 '11 at 20:37
Yes, I already proved $1\cdot a=a\cdot 1=a$ for any $a$. But unfortunately I cannot see how this can help me... –  Francesco Turco Apr 20 '11 at 19:30