In each of the following determine rational numbers $a$ and $b$:$ \frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b} $

Given:

\( \frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b} \)

To do: 

We have to determine rational numbers $a$ and $b$.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

LHS $=\frac{4+\sqrt{2}}{2+\sqrt{2}}=\frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$

$=\frac{8-4 \sqrt{2}+2 \sqrt{2}-2}{(2)^{2}-(\sqrt{2})^{2}}$

$=\frac{6-2 \sqrt{2}}{4-2}$

$=\frac{6-2 \sqrt{2}}{2}$

$=3-\sqrt{2}$

Therefore,

$a-\sqrt{b}=3-\sqrt{2}$

Comparing both sides, we get,

$a=3$ and $b=2$

Hence, $a=3$ and $b=2$.