Prove that if $\gcd(a, b) = 1$, $x \mid a$ and $y \mid b$. then $\gcd(x, y) = 1$

Question

If $\gcd(a, b) = 1$, $x \mid a$ and $y \mid b$. I want to prove that $\gcd(x, y) = 1$.

From what I understand, I need to prove $a = xb$, for any positive integer $b$ # 0 (1) $b = ya$, for any positive integer $a$ # 0 (2), then I can prove that $\gcd(x, y) = 1$.

I think I missed something to prove the formula of GCD before I can prove $\gcd(x, y) = 1$.

Answer 1

You can prove this by contradiction. If $$\gcd(x,y)=d\ne 1$$then $$d|x|a\\d|y|b$$therefore $$d|\gcd(a,b)=1$$which is a contradiction and the proof is complete.

Answer 2

Your proposition can be stated as follows:

Let $x$ be a divisor of $a$ and $y$ a divisor of $b$. If $\gcd(a,b)=1$ then $\gcd(x,y)=1$.

I suggest we prove the contrapositive, which is to say we prove the logically equivalent proposition:

Let $x$ be a divisor of $a$ and $y$ a divisor of $b$. If $\gcd(x,y)\neq 1$ then $\gcd(a,b)\neq 1$.

So suppose $x$ and $y$ have a common divisor $d>1$. Use transitivity of divisibility.

Answer 3

From the given,

$$a=mx=mp\gcd(x,y),\\b=ny=nq\gcd(x,y).$$

Hence

$$\gcd(x,y)|\gcd(a,b).$$