# Validate Peano Arithmetic Approach

## Homework Statement

Show that the natural numbers satisfy commutativity of multiplication and distributivity of multiplication over addition.

## The Attempt at a Solution

I'm wondering if there is any potential circularity in this reasoning. I proved distributivity over addition first by say, fixing n, m and showing that M = {p in naturals | (n+m)p = np + mp} is equal to the set of natural numbers.

Then I used distributivity in the final step of my commutativity proof where the inductive hypothesis was mk = km and the inductive step m(k+1) = mk + m = km + m = k(m+1) (literally the last equality follows from distributivity).

This seems valid, but I know that distributivity is usually stated with both (n+m)p = np + mp and p(n+m) = pn + pm. Typically this follows from commutativity, but would this still follow even if I used the distributive law in my commutativity proof?

CompuChip
Homework Helper
I suppose it would be fine to have a sequence like:
- Show left distributivity: (n+m)p = np + mp
- Show commutativity, only using left distributivity
- Right distributivity follows

However, it looks like you are using right distributivity in the proof of commutativity. If you had in the final step: mk + m = (k + 1)m it would be fine, as you have just proven that to be true (assuming you know that 1k = k, and fun stuff like that).

By the way, note that
m(k + 1) = k(m + 1)
is not true (m = 1, k = 2 is a counter example) -- you meant: km + m = (k + 1)m ?

On closer inspection, I think you need both right and left distributivity, if you set it up this way. However, my guess is that right distributivity is an almost-copy of the proof of left distributivity [/edit]

Sorry that was a huge typo. I mean m(k+1) = mk + m = km + m = (k+1)m, which should be permissible since I proved that 1 is the multiplicative identity for the base case of the same commutativity proof. I'll try to set up right distributivity and then prove commutativity I guess.

*EDIT* I guess the smartest route to take would be left distributivity -> multiplicative identity -> right distributivity -> commutativity (use multiplicative identity for base case).
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