Relatively Prime Suppose $$d = un + vs$$ where $d$ is the $\gcd(n,s)$
Dividing $d$ both sides
$$1 = u(n/d) + v(s/d)$$
So $(n/d)$ and $(s/d)$ are integers that are relatively prime. Why does this show that for any integer dividing both of them must also divide $1$?
This is a part of a proof I am reading from Abstract Algebra. I like to add that I have limited knowledge in Number Theory
 A: This is (sometimes called) Bezout: $\gcd(a,b)=1$ if and only if there exist $u,v$ such that $au+bv=1$.
You are asking about $\Leftarrow$.
If $n$ divides $a$ and $b$, then $n$ divides $au$ and $bv$, so $n$ divides $au+bv$.
Thus $n$ divides $1=au+bv$.
The only divisors of $1$ are $1$ and $-1$, so $|n|=1$.
Thus $\gcd(a,b)=1$.
A: If 
$$1 = u(n/d) + v(s/d) := ua + vb,$$
and for some integer $c$ you have $c|a$ and $c|b$, then obviously 
$$c|ua+vb = 1.$$
Edit (in more detail per a former comment): If $c|a$, then $a = cx$ (not $c = ax$ - example: $2|6$, and $6 = 3 \cdot 2$, not $2 = 3 \cdot 6$). Similarly $b = cy$. Hence, 
$$ua + vb = ucx + vcy = c (ux + vy),$$
 so $c$ divides this. And as only $1|1$, you are done.
A: 
$\rm\: m A + n B = 1,\:$ so $\rm\:A\:$ and $\rm\:B\:$ are integers that are relatively prime. Why does this show that for any integer $\rm\,d\,$ dividing both $\rm\,A,B\,$ must also divide $1$?

Hint $\ $ $\rm\:d\mid A,B\:\Rightarrow\: A = ad,\ B\, =\, bd,\ $ so $\rm\ mA \!+ nB  \,=\, (ma\!+\!nb)\,d \,=\, 1\:\Rightarrow\:d\mid 1$
Remark $\ $ Since you've tagged it "abstract algebra", a slight abstraction may prove instructive.
More conceptually: if $\rm\ A,B\:$ are multiples of $\rm\:d,\:$ then so too is $\rm\:m A \!+ n B\ \,(=\, 1),\:$ since multiples are closed under  both addition and integer scalings $\rm\:x\to nx,\ n\in\Bbb Z,\:$ i.e. they form an ideal in $\,\Bbb Z.\:$ Generally, in any ring, the common multiples of some elements forms an ideal (historically one of the prototypical examples of an ideal). In a principal ideal domain (PID) such as $\rm\,\Bbb Z,\,$ the set of common multiples $\rm\,a,b,\ldots$ is the ideal $\rm\:lcm(a,b,\ldots)\,\Bbb Z.$  
Since common multiple sets are ideals, your statement is a special case of the fact that an ideal $\rm\ I = (1)\ $ when it contains two comaximal elements: $\rm\ I\supset (a),(b)\!\iff\! I \supset (a)+(b)\,\ (= 1).\ $ For principal ideals, contains = divides, i.e. $\rm\ (c)\supseteq (a)\!\iff\! c\mid a.\:$ Thus, in a PID, the above specializes to $\rm\:c\mid a,b\!\iff\!c\mid gcd(a,b)\,\ (= 1),\:$ the universal property/definition of the gcd. 
