Proof that the square root of $5$ is not a natural number I am a student and I want to know if this proof is correct.
Proof. If $\sqrt5$ is a natural number, it should be even or odd.
If it would be even, we could express it as $2k$ for some integer $k$. 
If it would be odd, we could express it as $2j+1$ for some integer $j$. 
But those integers do not exist what is a contradiction. 
Then, $\sqrt5$ is not a natural number. Q.E.D.
Using Joofan's idea, I have a second proof. Please give me your opinion!
Proof. If $\sqrt5$ is a natural number, then there exists a natural number $n^2=5$, but $2^2<5<3^2$, and there is no natural number between $2$ and $3$. Therefore, we can conclude that there does not exist a natural number $n^2=5$, and that is a contradiction. Q.E.D.
 A: If the square root of 5 is a natural number, call it $n,$ then $n^2=5,$ by definition of the square root function. Let $n=p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ be the unique factorization of $n$ into prime powers then $5=n^2=p_1^{2 e_1} p_2^{2 e_2} \cdots p_k^{2 e_k}$ must hold, i.e., all the prime factors of $5$ must appear in its factorization an even number of times. This is a contradiction since $5=5^1$.
A: Let's assume, to the contrary, that $\sqrt{5}$ is a natural number, $n \in \mathbb{N}.$
Then, since $f(x)=\sqrt{x}$ is an increasing function, we know that $\sqrt{4}<\sqrt{5}<\sqrt{9}.$  Thus, $$\sqrt{4}<n<\sqrt{9}\implies 2<n < 3 $$
Producing a contradiction.  So, it must be the case that $n \notin \mathbb{N}$.
A: OK, let's think about the proof as presented:
If square root of 5 is a natural number, it should be even or odd. 
If it would be even, we could express it as 2k for some integer k. 
If it would be odd, we could express it as 2j+1 for some integer j. 
But those integers do not exist what is a contradiction. 
Then, square root of 5 is not a natural number. 

So the question is, why have you specified $k$ and $j$ and what did you do or check to ensure that they do not exist? You separated into even and odd cases: what was the purpose of this separation? 

An alternative:
For $n,k>0$, observe that $(n+k)^2 = n^2+2kn+k^2 > n^2$.
So $m>n \implies m^2>n^2$
Note that $3^2 = 9 > 5$, so for all $n>3, n^2>3^2>5$
Therefore the square root of $5$ is less than $3$. Only $1$ and $2$ are natural numbers less than $3$, and $1^2 = 1 \neq 5$ and also  $2^2 = 4 \neq 5$
Therefore there is no natural number whose square is $5$.
A: You can prove that 5 is not rational. Suppose, without lost of generality that there exists natural numbers $p, q$ such that:
$$\sqrt5 =p/q$$
$$5q^2 = p^2$$
Then, $p$ has to be multiple of 5, so $p=5 k$. Therefore:
$$q^2=5k^2$$
So $p, q$ cannot be relatively prime. $\sqrt 5$ cannot be expressed as a ratio of integers, so it is irrational.
