We will work with $1$-based natural numbers, that is $0\notin\Bbb{N}$. Let $s:\Bbb{N}\rightarrow\Bbb{N}$ be the function axiomatically given in peano axioms, we define addition and multiplication as follows:
Definition: $+:\Bbb{N}^2\rightarrow \Bbb{N}$
$$\forall a\in \Bbb{N}:a+1=s(a)\tag{i}$$
$$\forall a,b\in \Bbb{N}:a+s(b)=s(a+b)\tag{ii}$$
For addition, i managed to prove the associative law, commutative law.
Definition: $\cdot:\Bbb{N}^2\rightarrow \Bbb{N}$
$$\forall a\in\Bbb{N}:a\cdot 1=a\tag{iii}$$
$$\forall a,b\in \Bbb{N}:a\cdot s(b)=a+(a\cdot b)\tag{iv}$$
Here, i managed to prove the distributive law and associative law, but I'm struggling to prove the commutative law. Many proofs Wikipedia offers add the following definition in (iii). They say, furthermore that $a\cdot 1=1\cdot a=a$. I tried to prove this lemma alone by induction, but I am struggling to do this and I'm running in circles. If i managed to prove $a\cdot1=1\cdot a$ then I'm done with the commutative law. I was wondering, if i HAVE TO add this to the definition, or someone can hand in a proof for that or provide a proof for the commutative law NOT using this extension?
Best Answer
You can prove be induction that$$(\forall a\in\mathbb{N}):1\cdot a=a.$$This is trivial if $a=1$. On the other hand, if $a\in\mathbb N$ is such that $1\cdot a=a$, then\begin{align}1\cdot s(a)&=1+1\cdot a\\&=1+a\\&=a+1\\&=s(a).\end{align}Of course, I am assuming here that $+$ is commutative.