I want to express $\sqrt2$ as a fraction $\frac{a}{b}$ (where $a$ and $b$ are natural numbers). I want to find the simplest fraction, so $a$ and $b$ can’t have any common factors. Prove that this cannot be done. There is an interesting history behind this problem (see Section 11.2: Pythagoras' Theorem). Show hint

Can $a$ be odd? What does this mean for $a^2$ and $b^2$?

Let’s play around algebraically first: \begin{align*} \frac{a}{b} &= \sqrt2\\ \frac{a^2}{b^2} &= 2\\ a^2 &= 2b^2. \end{align*} This means that $a^2$ is even, so $a$ must be even (if $a$ were odd then $a^2$ would be odd too). If $a$ is even then $a^2$ is divisible by $4$. So there exists an integer $k$ such that $a^2 = 4k$. \begin{align*} a^2 &= 2b^2\\ 4k &= 2b^2\\ 2k &= b^2 \end{align*} But this means that $b^2$ is even, so $b$ is even, so $a$ and $b$ both have a factor of $2$. In other words, if $\sqrt2$ is a fraction, then both the numerator and the denominator must be even. This is absurd because we should be able to find a simplified fraction. Therefore, $\sqrt2$ can’t be expressed as a fraction.