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I am working with formal construction of natural numbers, in terms of the empty set and successor function, a la Peano. Here is the definition of addition, defined inductively;

Definition : Function +: $ N \times N \to N$, denoted $a + b$ for $a,b \in N$ is defined so that

$$ m + 0 = m$$ $$ m + (n + 1) = (m + n) + 1$$

Now, using this definition, I showed the usual properties of natural number addition; so that $m + 0 = 0 + m = m$, $m + n = n + m$, $m + (n + l) = (m + n) + l$. In particular, I don't know about inverses or subtraction because I did not introduce $Z$ yet.

Then I defined multiplication as follows:
Definition : Function x: $ N \times N \to N$, denoted $a \times b$ or $a \cdot b$ for $a,b \in N$ is defined so that

$$ m \cdot 0 = 0$$ $$ m \cdot (n + 1) = m \cdot n + m$$

I showed the associativity, that $n \cdot 0 = 0 \cdot n = 0$ and also $m \cdot 1 = 1 \cdot m = m$. Right now, I am trying to prove commutativity by induction.

Theorem: $m \cdot n = n \cdot m$

Proof Attempt: By induction on n;

Base Case: $m \cdot 0 = 0 \cdot m $ is already shown.

Inductive Case: Suppose $m \cdot n = n \cdot m$

$$m \cdot (n + 1) = m \cdot n + m$$ [by definition of multiplication] $$= n \cdot m + m$$ [by inductive hypothesis] $$= n \cdot m + n - n + m + 1 - 1$$

I think I can complete things from here but I don't want to use subtraction in my proofs because I haven't defined such a notion yet. Can the commutative property be proven only by the things I showed above? I think it should be possible because natural number multiplication has this property. On the other hand, I know subtraction can be defined without defining full integers. Still, I am looking for a proof that uses only above properties.

José Carlos Santos
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meguli
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1 Answers1

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You wrote that $m.0=0.m$ is already shown. This is not true. By definition, $m.0=0$, but you have to prove that $0.m=0$. This is easy by induction: $0.0=0$ and if $0.m=0$, then$$0.(m+1)=0.m+0=0+0=0.$$

You can prove the equality $n.m+m=(n+1).m$ by induction on $m$. For $m=0$, this is just $n.0+0=(n+1).0$, but both of them are $0$. On the other hand, if $n.m+m=(n+1).m$, then\begin{align*}n.(m+1)+(m+1)&=n.m+n+m+1=n.m+m+n+1\\&=(n+1).m+(n+1)\\&=(n+1).(m+1).\end{align*}

José Carlos Santos
  • 440,053