Question
Find rational numbers $a$ and $b$ such that: $\frac{\sqrt{2}-1}{\sqrt{2}+1}=\text{a}+\text{b}\sqrt{2}$
✓
Answer
$\frac{\sqrt{2}-1}{\sqrt{2}+1}=\text{a}+\text{b}\sqrt{2}$
we have, $\frac{\sqrt{2}-1}{\sqrt{2}+1}$
$=\frac{\sqrt{2}-1}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}$
$=\frac{\big(\sqrt{2}-1\big)^2}{\big(\sqrt{2}\big)^2-(1)^2}$
$=\frac{\big(\sqrt{2}\big)^2-2\sqrt{2}+1}{2-1}$
$=\frac{2-2\sqrt{2}+1}{1}$
$=3-2\sqrt{2}$
Now, $\frac{\sqrt{2}-1}{\sqrt{2}+1}=\text{a}+\text{b}\sqrt{2}$
$\Rightarrow3-2\sqrt{2}=\text{a}+\text{b}\sqrt{2}$
$\Rightarrow\text{a}=3$ and $\text{b}=-2$
we have, $\frac{\sqrt{2}-1}{\sqrt{2}+1}$
$=\frac{\sqrt{2}-1}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}$
$=\frac{\big(\sqrt{2}-1\big)^2}{\big(\sqrt{2}\big)^2-(1)^2}$
$=\frac{\big(\sqrt{2}\big)^2-2\sqrt{2}+1}{2-1}$
$=\frac{2-2\sqrt{2}+1}{1}$
$=3-2\sqrt{2}$
Now, $\frac{\sqrt{2}-1}{\sqrt{2}+1}=\text{a}+\text{b}\sqrt{2}$
$\Rightarrow3-2\sqrt{2}=\text{a}+\text{b}\sqrt{2}$
$\Rightarrow\text{a}=3$ and $\text{b}=-2$
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