The identity element of $(\mathbb{N} , \gcd)$.

Question

Let $x*y= \gcd(x,y)$ on $\mathbb N$. I have to prove that $(\mathbb{N} , *)$ is a commutative monoid.

I knew how to show the associativity, but I have a problem on the identity element.

$\gcd(x,e)=x \to x\mid e$ so $e=xk,k \in \mathbb N ^*$

And I don't know what to do next.

Answer 1

In this context, $0\in\Bbb N$. It is the identity because $0=n\times 0$ for all $n\in\Bbb N$, i.e., $n\mid 0$, hence $\gcd(n, 0)=n=\gcd(0, n)$.

The monoid is commutative because $\gcd(a, b)$ is $d$ such that $d$ divides both $a$ & $b$ and if both $c\mid a$ and $c\mid b$, then $c\mid d$; so just swap $a\leftrightarrow b$ in the definition of $\gcd$ to get $\gcd(a, b)=\gcd(b, a)$.