So I was thinking about how to prove the $\sqrt{2}$ is irrational while on holiday. (Bear in mind that at this point, I only knew that it was indeed irrational, not how to prove it.)
Anyway, I came up with my own proof with no knowledge of what "contradiction" was, this was just the easiest way.
My proof begins familiarly:
$\sqrt{2} = \frac{a}{b}$ ==> Suppose $\sqrt{2}$ is indeed rational, it can be expressed as the ratio of $a$ to $b$ in lowest form. $$ \begin{split} 2 &= a^2/b^2 \\ 2b^2 &= a^2 \end{split} $$ At this point, my proof deviates from the usual inductive process of finding $a$ and $b$, finding that they do indeed share a common factor greater than $1$ and so the original statement is false.
I reasoned that because $1^2 = 1$ and $2^2 = 4$, $\sqrt{2}$ must be between $1$ and $2$. Therefore, $a > b$ because $1 < \frac{a}{b} < 2$. As they are both positive integers, we can rewrite the equation: $$ 2(a-x)^2 = a^2 $$ (Where $x$ is the integer difference between $a$ and $b$. In addition, $x, a,$ and $b$ share no common factors.) It will also become important to recognize that $x < b$, and so $x is not a multiple of b$. This can be proven by reasoning:
if $x \ge b$ then we could substitute in a value for $x$ into $\frac{a}{b} > 1$ but $\frac{a}{b} < 2$ (which we showed to be true earlier.)
Therefore, if $x \ge b$, $a$ could be rewritten as $(b + x)$ to give $(b + x)/b$. But if $x$ could be $b$ we can substitute $x$ as $b$ to give $2b/b = 2$. But $\sqrt{2} \ne 2$ because $2 = \left(\sqrt{2}\right)^2 \ne 2 = 4$.
Therefore, if $x \ge b$, then we get $\sqrt{2} \ge 2$, which is incorrect.
So we have shown that our equation holds based on our original assumption ($\sqrt{2} = a/b$).
If we expand our equation and simplify:
$$2(a-x)^2 = a^2$$
$$2(a^2 -2ax +x^2) = a^2$$
$$2a^2 -4ax + 2x^2 = a^2$$
$$a^2 -4ax + 2x^2 = 0$$
If we can prove this has solutions $a, x \in \Bbb Z^+$, then our original assumption ($\sqrt{2} = \frac{a}{b}$) must be invalid, as $a$ and $b$ must be integers.
I thought about how to prove this, and was stuck until I remembered completing the square. So that is what we shall do.
$$(a - 2x)^2 - 4x^2 + 2x^2 = 0$$
$$(a - 2x)^2 = 2x^2$$
$$a - 2x = \sqrt{2}(x)$$
$$a = \sqrt{2}(x) + 2x$$
We can assume that because $x$ is an integer $2x$ is also an integer.
But wait. We already know $x$ is not a factor or multiple of $b$, and $\sqrt{2}$ is not an integer. And since $a$ and $b$ are in simplest form, the smallest number we can times $\sqrt{2}$ by to produce an integer is $b$ to produce $a$. (Obviously assuming $\sqrt{2} is rational.)
But $x < b$ and not a multiple of $b$. Therefore $\sqrt{2}(x)$ does not produce integer values of $a$ or $x$.
Meaning $\sqrt{2}(x) + 2x \not \in \Bbb Z^+$. Therefore, $a \notin \Bbb Z^+$.
Meaning that there is no way of expressing $\sqrt{2}$ as a ratio of integers $\frac{a}{b}$.
Therefore $\sqrt{2} ≠ \frac{a}{b}$.
$\sqrt{2}$ is irrational.
That was quite long, but I hope everyone understands my proof. I was just wondering if anyone had seen this before and this is not original. So if you have seen it already please tell me. Also, more importantly, does it have any holes or mistakes? I understand that this may not be the simplest or best solution, but I prefer it to the usual one as I think it relies less on induction and more on algebra to produce a result.
Thank you.