Proof that √2 is irrational

2024-05-2810:50 Status:Complete Tags: Proofs

Suppose that $\sqrt{2}$ were rational. Then there would exist some integers $a$ and $b$ such that $\frac{a}{b}=\sqrt{2}$.
Consequently, the set $S=\{k\sqrt{2}|k$ and $k\sqrt{2}$ are positive integers$\}$ is a nonempty set of positive integers (as $a=b\sqrt{2}$ is a member of $S$). Therefore by the well-ordering property, $S$ has a smallest element, say $s=t\sqrt{2}$. $s\sqrt{2}-s=s\sqrt{2}-t\sqrt{2}$. Because $s\sqrt{2}=2t$ and $s$ are both integers, $s\sqrt{2}-s=(s-t)\sqrt{2}$ must also be an integer. Furthermore, it is positive as $s\sqrt{2}-s=s(\sqrt{2}-1)$ and $\sqrt{2}>1$. It is less than $s$ as $s=t\sqrt{2},s\sqrt{2}=2t$ and $\sqrt{2}<2$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that the converse, $\sqrt{2}$ is irrational. QED

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Let’s suppose $\sqrt{2}$ is a rational number. Then we can write it $\sqrt{2}$ = $\frac{a}{b}$ where $a,b$ are positive integers and, $b\ne 0$. We additionally assume that  $\frac{a}{b}$ is simplified to the lowest terms, since that can obviously be done with any fraction. Notice that in order for $\frac{a}{b}$ to be in the simplest terms, both of $a$ and $b$ cannot be even. One or both must be odd. Otherwise, we could simplify $\frac{a}{b}$ further. From the equality $\sqrt{2}$ = $\frac{a}{b}$ it follows that $2=\frac{a^2}{b^2}$.  So the square of $a$ is an even number since it is two times something. From this, we know that $a$ itself is also an even number. Why? Because it can’t be odd; if $a$ itself was odd, then $a\cdot a$ would be odd too. Odd number times odd number is always odd. If $a$ itself is an even number, then $a$ is 2 times some other whole number. ($a=2k$, where k is this other number. Where $k$ is just some number)

If we substitute $a=2k$ into the original equation, $2=\frac{a^2}{b^2}$ we get: $2=\frac{(2k{)}^2}{b^2}$ $2=\frac{4k^2}{b^2}$ $2b^2=4k^2$ $b^2=2k^2$

This would imply that $b^2$ is even. As it is impossible to make an even square with odd numbers, $b$ must be even as well. This contradicts, as assuming that $\frac{a}{b}$ is in its lowest terms would imply that 2 is not a factor. It follows that the converse, $\sqrt{2}$ is irrational QED