# 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

• If a number divides both n/d and s/d, then that number will divide any multiple of them and their sum, so it will divide 1. – anon Mar 4 '13 at 3:51

## 3 Answers

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.

• If $c|a$, then $c = ax$ for some integer $x$. Similarly for $b$, that is $c = by$, How does that imply $c | ua + vb$? Because I could have $c | ax + by$ instead no? – hosun Mar 4 '13 at 3:50
• I made my comment part of the answer. See above. – gnometorule Mar 4 '13 at 3:55

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 $\gcd(a,b)=1$.

$\rm\: m A + n B = 1,\:$ 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.