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. 

You are almost there, suppose that $a$ and $b$ are relatively prime,
write $a^2=b^2(4m+2)=2b^2(2m+1)$, you deduce that $2$ divides $a^2$, and $a$, write $a=2a_1$, you have $2b^2(2m+1)=4a_1^2$, you deduce that $b^2(2m+1)=2a_1^2$. This implies that $2$ divides $b$, contradiction since $a$ and $b$ are relatively prime.
