elementary number theory - Prove $ sqrt{2k}$ is irrational where $ k$ is an odd integer.

Question:

Prove $ \sqrt{2k}$ is irrational where $ k$ is an odd integer.

My attempt:

Proof by contradiction:

Now, assume $ \sqrt{2k}$ is rational. Then, $ \sqrt{2k} = \frac{a}{b}$ where $a,b \in Z$, $b$ not equal $0$ and $a,b$ have no common factors.

$ \sqrt{2k} = \frac{a}{b} \implies 2k = \frac{a^{2}}{b^{2}} \implies (b^{2})(2k) = a^{2} \implies 2|a^{2} \implies 2|a, \ since \ 2 \ is\ prime \implies \exists c \in Z$ such that $ a = 2c$

Then, $ (2k)(b^{2}) = a^{2} \implies (2k)(b^{2}) = 4c^{2} \implies kb^{2} = 2c^{2} \implies 2|kb^{2} \implies 2 | b^{2} \implies 2|b$, since $2$ is prime.

So, $ 2|a$ and $ 2|b$ , a contradiction.