I'm studying Real Analysis and trying to prove some properties of multiplication in $\mathbb{N}$.
In my case, I'm studying $\mathbb{N}$ where $0 \notin \mathbb{N}$, that is the natural numbers starting with 1.
I'm currently trying to prove the commutative law of multiplication with induction. Can someone please check if my proof is valid, and if it's not, tell me where I am missing something or doing anything wrong?
Notation and definitions used:
- $s(n) = n + 1$
- $s(n+p) = n + s(p)$
- $n \cdot 1 = n$
- $n \cdot s(m) = n\cdot m + n$
I've already proved and because of that can use:
- Associative law of sum
- Commutative law of sum
- Cancellation law of sum
- Distributive law of multiplication
- Associative law of multiplication
Here is my proof:
Define: $$ X = \{ n \in \mathbb{N} \ \text{ | } \forall m \in \mathbb{N}, m\cdot n = n\cdot m \} $$ Checking de base case where $n = 1$: $$ \begin{align*} s(m\cdot 1) = s(m)\cdot 1&= s(m)\\ &= m+1\\ &= m\cdot 1 + 1\cdot 1\\ &= 1(m+1)\\ &= 1\cdot s(m) \end{align*} $$ That covers all natural numbers except 1, since $1 \notin \text{im}(s)$. Because of that we show: $$ \begin{align*} 1\cdot 1 &= 1\cdot 1\\ 1 &= 1 \end{align*} $$ Hence $1 \in X$.
Then I need to show that if $n \in X$, then $s(n) \in X$.
$$ \begin{align*} m\cdot s(n) &= m(n+1)\\ &=m\cdot n + m\cdot 1\\ &=n\cdot m + 1\cdot m\\ &=(n+1)\cdot m\\ &=s(n)\cdot m \end{align*} $$
Therefore if $n \in X$ then $s(n) \in X$ and since $1 \in X$ too, we can conclude that $X=\mathbb{N}$.
Any kind of help is valid. Thank you!
I would be more explicit about the fact that for the base case you are trying to show that for any $m$:
$$1\cdot m = m \cdot 1$$
and that you do this by induction on $m$
Moreover, you should start out by:
$$s(m) \cdot 1 = s(m) = m + 1...$$
and also add the all important step:
$$m \cdot 1 + 1 \cdot 1 \overset{Inductive \ Hypothesis}{=} 1 \cdot m + 1 \cdot 1 \overset{Left-Distribution}{=} 1 \cdot (m +1)$$
Also, I am wondering: Do you have both left-distribution:
$$x \cdot (y +z) = x \cdot y + x \cdot z$$
and right-distribution:
$$(y +z) \cdot x = y \cdot x + z \cdot x$$
Because you are using right-distribution when going from $n \cdot m + 1 \cdot m = (n+1) \cdot m$