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 $\lt \sqrt{2}$?