- #1

evinda

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For each pair of natural numbers $m \in \omega, n \in \omega$ we define the multiplication between $m,n$ (as a function $\cdot: \omega \times \omega \to \omega $) like that:

$$m \cdot 0=0$$

$$m \cdot n'=m \cdot n+m$$

I want to show that for each $m \in \omega, n \in \omega, k \in \omega$ the following properties are satisfied:

- $m \cdot n=n \cdot m$
- $(m \cdot n) \cdot k=m \cdot (n \cdot k)$
- $(m+n) \cdot k=m\cdot k+n \cdot k$
- $1 \cdot n=n$
- If $k \neq 0$ and $n \cdot k=m \cdot k$ then $n=m$.

For $m=0: 0 \cdot 0=0 \checkmark$

We suppose a $m$ such that $0 \cdot m=0$.

We want to show that $0 \cdot m'=0$.

$0 \cdot m'=0 \cdot m+0 \overset{\text{induction hypothesis}}{=}0$.

Therefore, $0 \cdot m=0$, for any $m \in \omega$.Then we want to show that $n' \cdot m=m \cdot n'$.

For $m=0$: $n' \cdot 0 \overset{\text{definition}}{=}0 \overset{\text{previous sentence}}{=}0 \cdot n' \checkmark$We suppose a $m$ such that $n' \cdot m=m \cdot n'$ for all $n \in \omega$.We want to show that $n' \cdot m'=m' \cdot n'$.

$n' \cdot m'=n' \cdot m+n'=m \cdot n'+n'$How could we continue? (Thinking)