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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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