In each of the following determine rational numbers $a$ and $b$:$ \frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3} $

Given:

\( \frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3} \)

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 $=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$

$=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$

$=\frac{3+1-2 \sqrt{3}}{3-1}$

$=\frac{4-2 \sqrt{3}}{2}$

$=2-\sqrt{3}$

Therefore,

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

Comparing both sides, we get,

$a=2$ and $b=1$

Hence, $a=2$ and $b=1$.

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Updated on: 10-Oct-2022

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