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.
Best Answer
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 $5=5^1$.