Proof by contradiction to show irrationality of $\sqrt{2}$ logically

Question

I am trying to learn more about basic logic in order to make my proofs and reasoning more precise or even "mechanical". Just to make sure that my proof really shows what I wanted. (Any literature recommendations about the basics are welcome as well!)

For example, here is my attempt to prove that $\sqrt{2} $ is irrational. The idea is to use proof by contradiction (as usual).

Proof:

Denote $p=$"$\sqrt{2}$ is irrational" and $q=$"$\sqrt{2}=a/b $ s.t. integers $a$ and $b\ne 0$ have no common factors".

Suppose that $\sqrt{2}$ is rational, i.e. $\neg p $. Then $\sqrt{2}=a/b$ for some integers $a $ and $b \ne 0$. We can assume that $a$ and $b$ have no common factors, because if they had, they can be cancelled away. Thus, we have shown that the statement (or what would be the correct term?) $ \neg p \rightarrow q $ has truth value 1 (true).

Using mathematics, from $\sqrt{2}=a/b$ we can derive that both $a$ and $b$ are even. Thus, they have a common factor 2, and we have shown that logical statement $q \rightarrow \neg q $ has truth value 1.

Since both $ \neg p \rightarrow q $ and $q \rightarrow \neg q $ have truth values 1, also $ (\neg p \rightarrow q)  \wedge (q \rightarrow \neg q) $ has truth value 1. Using tautology $ (a\rightarrow b )\wedge (b \rightarrow c) \Longrightarrow a \rightarrow c $, we conclude that $ \neg p \rightarrow \neg q  $ has truth value 1.

Since $\neg p \rightarrow q $ and $\neg p \rightarrow \neg q $ have truth values 1, statement $(\neg p \rightarrow q)\wedge (\neg p \rightarrow \neg q) $ has value 1. Using tautology $ (a\rightarrow b) \wedge (a\rightarrow c) \Longleftrightarrow a\rightarrow (b \wedge c) $, we conclude that $\neg p \rightarrow (q \wedge \neg q) $ has truth value 1.

Using tautology $ \neg a \rightarrow (b \wedge \neg b) \Longleftrightarrow a $, we conclude that $p$ has truth value 1.

Thus, p is true, i.e., $\sqrt{2}$ is irrational. $ \square $

Answer 1

The biggest problem is that the core of irrationality of square-root of two is actually in the claim that every rational number is equal to $a/b$ for some integers $a,b$ such that not both of $a,b$ are even, but you clearly handwaved that part by claiming without proof: "We can assume that a and b have no common factors, because if they had, they can be cancelled away.".

The next biggest problem is that everything you wrote about $p,q$ and "truth values" is completely pointless bloat, as Mauro already said in a comment. If you truly want to understand a logically precise proof of irrationality of square-root of two, you must first learn basic FOL (first-order logic), including a practical deductive system for it, and become familiar enough with it before moving on to learning to prove things within PA (Peano Arithmetic; axioms in the linked post), which would include "$∀k,m{∈}ℕ\ ( \ k·k = m·m·2 ⇒ k = 0 \ )$". This statement is the true core behind irrationality of square-root of two.

Take note that if you truly desire logical precision, you also cannot write "$\sqrt{2}$" until you have defined "$\sqrt{\vphantom{x^2}\quad}$", and this is much harder than you might think, so forget about this now and deal with the basics first.