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. 
 A: 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.
A: The general result is that $\sqrt x$ is Rational iff $x$ is a perfect square. But, if we have $\sqrt {2y}$ , in order for $2y$ to be a perfect square, we must have an even number of $2$s. But this cannot happen when $y$ is odd. So $2y$ is not a perfect square and so $\sqrt {2y}$ is irrational.
A: Suppose rational, then $\sqrt{2k} = a$. We can suppose $a \in \mathbb{Z}$. We have: $2k = a^{2}$ If a is odd, $a^{2}$ too, contradction. If $a$ is even, then $a = 2m$, thus $2k = 4m^{2}$, like  this $ k = 2m^{2}$, but $k$ is odd.
A: Proof by contradiction: suppose that $\sqrt{2k}=\frac{n}m$ where $k,n,m\in\Bbb N_{>0}$ and $\gcd(m,n)=1$. Then taking squares both sides we get
$$2k m^2=n^2\tag1$$
Now observe that $2km^2$ is even regardless the parity of $m$, thus $n^2$ must be even if $(1)$ holds, what imply that $n$ is even and $m$ must be odd, because $\gcd(m,n)=1$. 
Because $n$ is even we knows that $n^2=4j$ for some $j\in\Bbb N_{>0}$, thus we can rewrite $(1)$ as
$$km^2=2j\tag2$$
but $km^2$ is odd and $2j$ even.$\Box$
A: Hint:  If $k$ is odd then $k^2$ is odd.  (You can prove that).
If $k$ even then $k^2$ is not only even but $4|k^2$.  (You can prove that).
And if $k$ is odd, then $2k$ is even but $4\not \mid 2k$.  (You can prove that--- and you and use $2k = 4m + 2$ to prove it.)
So you have $b^2*(4m + 2) = a^2$.   If $b$ is odd is $a$ even or odd?  Is $a^2$ divisible by $4$?  If $b$ is even is $a$ even or odd? Is it divisible by $4$?  Are any of these possible?  Is it possible for $a$ and $b$ to both be even?
A: Hint: Use fundamental theorem of arithmetic. 
The thing inside square root has an odd power of two hence it cannot be a perfect square. 
