> [!NOTE] Lemma > $\sqrt{ 2 }+\sqrt{ 3 }+\sqrt{ 5 }$ is irrational ###### Proof Suppose $\sqrt{ 2 }+\sqrt{ 3 }+\sqrt{ 5 }=r\in \mathbb{Q}.$ Then $\begin{align} (\sqrt{ 2 } +\sqrt{ 3 })^2 &= (r-\sqrt{ 5 })^2 \\ 5+ 2\sqrt{ 6 } &= r^2 - 2r\sqrt{ 5}+5 \\ r^2 &= 2\sqrt{ 6 } +2r \sqrt{ 5 } \end{align}$that is, $2\sqrt{ 6 }+2r\sqrt{ 5 }=q\in \mathbb{Q}.$ Squaring both sides gives $\begin{align} 24+4r\sqrt{ 30 }+20r^2 &=q^2 \\ \sqrt{ 30 } &= (q^2 -20r^2 -24) \end{align}$that is, $\sqrt{ 30 }$ is rational. Let $\sqrt{ 30 }=\frac{m}{n}$ where $m,n\in \mathbb{Z}$ and are coprime. Then $m^2=30n^2$ which gives that $30$ divides $m$ (if $m$ is not divisible 30 then so is $m^2$). Let $m=30a.$ Then $90a^2=30n^2$ which shows that $n$ is also divisible by $30$ - a contradiction.