Proving there's no $n \in \mathbb{Z}$ such that $n^2 = 2$ I feel that my attempt is not what is expected in a Real Analysis course and that there's something I should incorporate from the Algebraic properties of Real numbers. My hunch is that there's better proof with induction, but was not able to think of any and could be wrong.
My attempt:
$$n^2 = 2 $$
$$n^2 = 1 + 1 $$
$$n^2+(-1)=1+1+(-1) $$
$$n^2-1=1+(1+(-1)) $$
$$n^2-1=1$$
Case $n=0$: then $-1=1$
Case $n \neq 0$: then $n^2 \geq 1$ and $n^2-1 \neq 1$
 A: Your proof is correct. Well done!
Minor pet peeve: You are clearly trying to prove by contradiction. It would be nice to write $$"\text{suppose that there exists a positive integer } n \text{ such that }n^2=2"$$
Then conclude that you get a contradiction.

You could have also proved by checking just one case. Clearly the square root of $2$ is less than $2$. Thus, if there did exist a positive integer less than $2$ whose square equals $2$, the only possibility is $1$. But since $1^2=1$, we conclude that there does not exist any such integer.
A: Alternative approach:
Since $~n^2 = (-n)^2,~$ and $~0^2 \neq 2,~$ it is sufficient to prove the assertion for $n \in \Bbb{Z^+}$.
Further, for $x \in \Bbb{R^+},$ if $~f(x) = x^2,~$ then $f'(x) = 2x > 0.$
Therefore, $~f(x)~$ is a strictly increasing function, for $x \in \Bbb{R^+}.$
Therefore, for $~x_1,x_2 \in \Bbb{R^+},~: ~x_1 \neq x_2,~$ you have that
$$x_1 < x_2 \iff x_1^2 < x_2^2.$$
Based on the above analysis, the problem is immediately resolved by noting that $1^2 < 2 < 2^2.$
A: Using the algebraic properties of the integers,  you can adapt the standard rational argument in this simpler case as well
Since $2$ is prime, we have $2|2=n^2$,  thus $2|n$.   So dividing by 2 gets us $n \cdot \frac n 2=1$, the product of two integers is 1 means they are units.  The only units in the integers are $1$ and $-1$, which both square to 1, so we're done
A: $(-n)^2=n^2$ so we can restrict ourselves to $n\ge 0$
For $n\le 1$ then multiplying by $n$ we get $n^2\le n\le 1$
For $n\ge 2$ then multiplying by $n$ we get $n^2\ge 2n\ge 4$
Therefore neither $n^2=2$ nor $n^2=3$ are reachable.
