GCD Cases Proof Help

Question

∀a,b ∈ N, Prime(b) ⇒ gcd(a, b) ≤ 1 ∨ gcd(a, b) ≥ b

gcd(greatest common divisor)

I understand this is likely cases with b divides a, and.. b doesnt divide a but am pretty lost at the moment. Can someone walk me through this step by step?

thanks

Answer 1

If $b$ is a prime then the only divisors of $b$ are $1$ and $b$. Since for any pair of natural numbers $x_1,x_2$ the $gcd(x_1,x_2)\leq x_i$ then specially $gcd(a,b) \leq b$ and at the same time $gcd(a,b)$ is either $1$ (hence $\leq 1$) or $b$ (hence $\geq b$).