If $a|b$, then why the gcd of $a$ and $b$ is $a$?

Question

If $a$ divides $b$, then why is $(a,b)=a?$

I am trying to justify this for myself and this is what I have done. I was wondering if it is correct?

Claim: Let $b\in\mathbb{Z}$ and $a\in\mathbb{N}$. If $a|b$ then $(a,b)=a$.

Justification: Since $a|b$, there must exist an integer $m$ such that $b=am$. Now

$$ \begin{align} (a,b)&=(a,am)\\ &=a(1,m)\\ &=a\times 1\\ &=a \end{align} $$

Answer 1

If $a\mid b$ then, since $a\mid a$, $a$ is a common divisor of $a$ and $b$. And obviously no natural number greater than $a$ divides $a$. So, $a$ is the greatest common divisor of $a$ and $b$.

Answer 2

For all natural numbers $a>0$, $a|a$ (because $a \div a =1$)

For all natural numbers $a>0$ and $b>0$, $a|b$ implies that $b$ is divisible by $a$ and $b\ge a$.

With $b \ge a$, $a$ is divisible by $b$ if and only if $a=b$.

Because of that, $a$ is divisible by $a$ and $b$ is also divisible by $a$, but $a$ can't be divisible by any number larger than itself, implies $(a,b)=a.$

Answer 3

due to the uniqueness of prime decomposition, the division structure $(\mathbb{N^+,|)}$ is isomorphic to the natural lattice structure on the free abelian monoid $M = (\oplus_{j \in \omega}\mathbb{N}_j,+)$ induced by the identical linear orderings on each component $\mathbb{N}_j \cong (\mathbb{N},<)$.

if $a=2^{n_0}3^{n_1}\dots$ (where only a finite number of the $n_j$ are non-zero), then the sublattice $a\mathbb{N^+}$ corresponds to the sublattice $M_a=(\oplus_{j \in \omega}(\mathbb{N}_j+n_j),+)$.

this correspondence makes it easy to see that, (now viewing the lattice as a category, with the obvious morphisms) then the translation f$_a$: $M \to M_a$ is a faithful functor between $M$ and its full subcategory $M_a$, and f$_a$ is naturally equivalent to id$_M$