Prove that the set of positive rational values that are less than $ \sqrt{2}$ has no maximum value Its been a while since I've written a proof and would appreciate some feedback on this one.
Question:
Given the set of rational positive values,  $\{q | q \in \mathbb{Q} \wedge 0 \lt q \lt \sqrt{2}\}$, show that there is no maximum value for $q \lt \sqrt{2}$
Response
Given $x=\sqrt{2}$, suppose that $x$ can be defined as a positive rational number such that $x$ is composed of positive numbers $p, q$ such that $x=\frac{p}{q}$ where $q \neq 0 $ and that $p,q$ are simplified to the lowest possible terms.
It follows that $2=\frac{p^{2}}{q{^2}}$ or $p^{2} = 2 \cdot q{^2}$.  Therefore, $p^{2}$ must be an even number as it is the product of some $n$ and an even number.  As a result, $p$ is an even number because otherwiese, $p^{2}$ would be odd.
If $p$ is an even number, then $p=2n$ for some number $n$.
Substituting $p=2n$ into the original equation:
$$2= \frac{(2n)^{2}}{q^{2}}$$
$$2= \frac{(4n^{2}}{q^{2}}$$
$$2q^{2} = 4n^{2}$$
$$q^{2} = 2n^{2}$$
Therefore, $q^{2}$ is an even number, which makes $q$ even as well.  This is a contradiction as $p, q$ are defined to be simplified to the lowest possible terms, which would not be possible if $p, q$ were even.  Therefore, $\sqrt{2}$ must be an irrational number.
An irrational number is defined as a number with no terminal or repeating decimals.  Because $\sqrt{2}$ repeats to infinite decimal places, there is no exact value that can be defined as a maximal value $< \sqrt{2}$
Therefore, there is no positive rational value that can be defined to be less than or equal to the $\sqrt{2}$
My strategy for this proof is to prove that $\sqrt{2}$ is an irrational number and then explain why there is no possible maximal value for some number $\lt \sqrt{2}$.  I think I've satisfied the first section, but feel there is rigor lacking in the second section.  How can I redefine the proof that there is no maximal value to a number that is $\lt \sqrt{2}$?

To show that no maximum is attained you should show that given any positive rational number $0<\frac{a}{b} < \sqrt{2}$ that there is another rational number $\frac{c}{d}$ such that $\frac{a}{b}<\frac{c}{d}$. 
This shows that no matter what you might try to use as a maximum, it misses some rationals. Thus no max can exist.
Also, one note: irrational numbers are defined to be real numbers that are not rational. Taking the definition to be reals with non-terminating non-repeating decimal expansions is equivalent but not natural and quite messy to use (formally).
A possible patch to your proof (spoiler):

 Suppose $\frac{a}{b}<\sqrt{2}$. Then let $\epsilon=\sqrt{2}-\frac{a}{b}>0$. Since $1<2<\cdots<n<\cdots \rightarrow \infty$, $0 \leftarrow \cdots<\frac{1}{n}<\cdots<\frac{1}{2}<1$. So given any positive real numbers (such as $\epsilon$) we can find a positive integer $N$ such that $\frac{1}{N}<\epsilon$. Then $\frac{a}{b}<\frac{a}{b}+\frac{1}{N}<\frac{a}{b}+\epsilon=\sqrt{2}$. Thus $\frac{a}{b}+\frac{1}{N}$ is a rational number bigger than $\frac{a}{b}$ but still in your set.

