[[Proof]] Exercise from [[Understanding Analysis]] $ \huge \sqrt{ 3 } \notin \Q $ [[Universal Quantifier|Any]] [[Rational Numbers|Rational Number]] can be expressed as the ratio of two [[Integer|Integers]] $p,q \in \Z$ with $q \neq 0$ where $p$ and $q$ are [[Coprime]]. $\large \begin{align} \frac{p}{q} &= \sqrt{ 3 } \\ \left( \frac{p}{q} \right)^{2} &= 3 \\ \frac{p^{2}}{q^{2}} &= 3 \\ p^{2} &= 3q^{2} \\ \\ \end{align} $ Therefore $p$ must be a multiple of $3$. $\large \begin{align} p &= 3r \\ 3^{2}r^{2} &= 3q^{2} \\ 3 r^{2} &= q^{2} \end{align} $ This now implies that $q$ must be a multiple of $3$. This contradicts the previous statement as $p\mid 3$ as well as $q \mid 3$, therefore $\sqrt 3$ is not [[Rational Numbers|Rational]].