Proving, by contradiction, that there are infinitely $n \in \mathbb{Z^+}$ such that $\sqrt{n}$ is irrational. I would like to know whether I have proved this statement correctly using contradiction and if not get some tips or pointers on how to improve it/make it correct.
We are asked to prove  that there are infinitely $n \in \mathbb{Z^+}$ such that  $\sqrt{n}$ is irrational.
Proof: $$\text{We assume, to the contrary, that there are infinitely many } n \text{ } (n \in Z^+) \text{ such that } \sqrt{n} \text{ is rational.} \\\text{Then, } \sqrt{n}=\frac{a}{b} \text{,  } n=\frac{a^2}{b^2}  \text{ where } a \text{ and } b \text{ are integers and } \frac{a}{b} \text{ has been reduced to lowest terms.} \\\text{Hence, } a^2=nb^2 \text{ and } b^2=\frac{1}{n}a^2\\\text{We know if } n\mid a^2 \text{ then } n\mid a \text{. It follows that } a=nk, k\in\mathbb{Z} \\ b^2=\frac{1}{n}(nk)^2=nk^2 \text{ and thus } n\mid b^2 \text{ and thus } n \mid b \\
\text{Hence, } n = \frac{nb^2}{nk^2} \\
\text{This is a contradiction given that we have assumed }a \text{ and } b \text{ to be reduced to lowest terms. }\blacksquare$$  ps. This isn't homework. Answers are in the back of my book but I want to improve.
 A: What is the statement you are trying to prove, saying? It is saying that the number of elements of the set $\{n \in \mathbb N : \sqrt n \ \mbox{is irrational}\}$, is infinite.
What is the opposite of this statement? Naturally, that the number of elements of that set is not infinite, in which case it is finite. So, the contradictory statement is this :

The set $\{n \in \mathbb N : \sqrt n \ \mbox{ is irrational}\}$ is finite. Or, there are only finitely many natural numbers $n$ such that $\sqrt n$ is irrational.

Now, you have to assume this is true, and derive a contradiction.

To derive a contradiction to this statement, you have to realize, that every non-empty finite set of natural numbers has a maximum element. We know, from a classical proof by contradiction, that $\sqrt 2$ is irrational, so $2$ is an element of our subject set.
So, if the set of natural numbers $n$ such that $\sqrt n$ is irrational has a maximum, then let $L$ be that maximum number. So $\sqrt L$ is irrational.
But, as Peter mentioned in his comment, $2 \sqrt L$ is also irrational, and this is equal to $\sqrt{4L}$, so $4L$ also belongs to the set of natural numbers whose square roots are irrational... can you take it from here?

Note that Peter's comment may also be used to produce infinitely many members of the set, so the two arguments , contradiction and direct proof, are not very different from one another.
A: An idea for a different kind of proof:
It is known that $\sqrt{n}$ rational if and only if $n$ is a perfect square.
Indeed, assume that  $\sqrt{n} = \frac{a}{b}$ with $\gcd(a,b) = 1$. Hence there exist $\alpha, \beta \in \mathbb{Z}$ such that $\alpha a - \beta b = 1$.
Now $$0 = (a-b\sqrt{n})(\beta + \alpha \sqrt{n}) = \beta a - \alpha bn + (\alpha a - \beta b) \sqrt{n} = \beta a - \alpha bn + \sqrt{n}$$
And hence $\sqrt{n} \in \mathbb{Z}$ so $n$ is a perfect square. The converse is obvious.
Now it just remains to show that the set $\{n \in \mathbb{N} : n \text{ is not a perfect square}\}$ is infinite.
For this notice that numbers of the form $3k^2+2$ for $k \in \mathbb{N}$ are never perfect squares.
