Why is the set over which we one takes the max for gcd(a, b) is a subset of the set for gcd(a, b − a)?

Question

In my book, the definition for gcd(a,b) is the following:

$$\gcd (a,b) = \max \{ d \in \Bbb{Z} : d|a \, \land \, d|b \}$$

However, I don't understand why the set for $\gcd(a,b)$ is necessarily a subset of that of $\gcd(a,b-a)$.

Answer 1

Take any $d\in\{k\in\mathbb{Z}:k|a\land k|b\}$, then $d|(b-a)$, so...

Answer 2

$\gcd(a,b)=d$

$a=dk$ and $b=dn$, for $n,k \in \mathbb{N} $

Doesn't it $\implies$ $d|d(n-k) \implies d|a-b$?

Answer 3

This follows from the fact that if $d$ divides $a$ and $d$ divides $b$, then necessarily $d$ divides $b-a$. Namely, $$(b-a)/d = b/d - a/d$$ Since the quantity on the right hand side is an integer.