Let \(a\) and \(b\) be positive integers and let \(k\) be any ineteger. We need to prove:
$$ (a, b) = 1 \implies (a, ka + b) = 1 $$
The contrapositive of this is:
$$ (a, ka + b) \ne 1 \implies (a,b) \ne 1 $$
Let's say \((a, ka + b) = d \ne 1\). This means \(d|a\) and \(d|ka+b\). Therefore:
$$ dh = ka+b $$
Dividing both sides by \(d\):
$$ h = \frac{ka}{d}+\frac{b}{d} $$
Since \(d|a\):
$$ h = \text{some integer } +\frac{b}{d} $$
Since \(h\) is an integer, and the first operand is an integer, then \(b/d\) also has to be an integer. Therefore \(d|b\).
If \(d|a\) and \(d|b\), then \((a,b) \ne 1\). Thus, we have proven that is \((a, ka + b) = d \ne 1\), then \((a,b) \ne 1\). This also means if \((a,b)=1\), then \((a, ka + b) = 1\).