# Simplify: $\frac {\sqrt2+\sqrt3+\sqrt4}{\sqrt2+\sqrt3+\sqrt6+\sqrt8+4}$

I have tried breaking up the fraction into two parts so the first part would cancel to 1, but to no avail. Rewriting the expression also doesn't seem to work, so how can I start?

• One observation: The denominator is of the form $a\sqrt2+b\sqrt3$ for some $a, b.$ Nov 4, 2017 at 23:56

$\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4 =\\ \sqrt{2} + \sqrt{3}+\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{4}+2\sqrt{4}=\\ \sqrt{2}+\sqrt{3}+\sqrt{4} + \sqrt{2}(\sqrt{3} + \sqrt{4} + \sqrt{2})$
Let $$\frac {\sqrt2+\sqrt3+\sqrt4}{\sqrt2+\sqrt3+\sqrt6+\sqrt8+4} = \frac{x}{y}$$. By componendo and dividendo (Brilliant):
$$\frac {\sqrt2+\sqrt3+\sqrt6+\sqrt8+4}{\sqrt2+\sqrt3+\sqrt4} = \frac{y}{x}$$ $$\Rightarrow \frac {\sqrt2+\sqrt3+\sqrt6+\sqrt8+4 -(\sqrt{2} + \sqrt{3} + \sqrt{4})}{\sqrt2+\sqrt3+\sqrt4} = \frac{y-x}{x}$$ $$\Rightarrow \frac {\sqrt4 + \sqrt6+\sqrt8}{\sqrt2+\sqrt3+\sqrt4} = \frac{y-x}{x}$$ $$\Rightarrow \frac {\sqrt{2} (\sqrt 2 + \sqrt3+\sqrt4)}{\sqrt2+\sqrt3+\sqrt4} = \frac{y}{x} - 1$$ $$\Rightarrow 1 + \sqrt{2} = \frac{y}{x} \Rightarrow \frac{x}{y} = \frac{1}{1 + \sqrt2} = \sqrt{2} - 1$$
• So ${x\over y}=\sqrt{2}-1$, brilliant! Aug 24, 2020 at 6:37