In each of the following determine rational numbers a and b:
(i) $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3}$
(ii) $\frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}$
(iii) $\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2}$
(iv) $\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$
(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b \sqrt{77}$
(vi) $\frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=a+b \sqrt{5}$
Given,
(i) $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$\sqrt{3}-1$
$=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{3-2 \sqrt{3}+1}{3-1}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}$
$2-\sqrt{3}=a-b \sqrt{3}$
On comparing the rational and irrational parts of the above equation, we get, a = 2 and b = 1
(ii) $\frac{4+\sqrt{2}}{2+\sqrt{2}}=\mathrm{a}-\sqrt{\mathrm{b}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$2-\sqrt{2}$
$=\frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(8-4 \sqrt{2}+2 \sqrt{2}-2)}{4-2}$
$=\frac{(6-2 \sqrt{2})}{2}$
$=3-\sqrt{2}$
$3-\sqrt{2}=a-\sqrt{b}$
On comparing the rational and irrational parts of the above equation, we get, $a=3 a n d b=2$
(iii) $\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$3+\sqrt{2}$
$=\frac{(3+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(9+6 \sqrt{2}+2)}{9-2}$
$=\frac{(11+6 \sqrt{2})}{7}$
$=\frac{11}{7}+\frac{6 \sqrt{2}}{7}$
$\frac{11}{7}+\frac{6 \sqrt{2}}{7}=a+b \sqrt{2}$
On comparing the rational and irrational parts of the above equation, we get,
$7-4 \sqrt{3}$
$=\frac{(5+3 \sqrt{3})(7-4 \sqrt{3})}{(7+4 \sqrt{3})(7-4 \sqrt{3})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(35-20 \sqrt{3}+21 \sqrt{3}-36)}{49-48}=-1+\sqrt{3}$
$-1+\sqrt{3}=a+b \sqrt{3}$
On comparing the rational and irrational parts of the above equation, we get, $a=-1$ and $b=1$
(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b \sqrt{77}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$\sqrt{11}-\sqrt{7}$
$=\frac{(\sqrt{11}-\sqrt{7})(\sqrt{11}-\sqrt{7})}{(\sqrt{11}+\sqrt{7})(\sqrt{11}-\sqrt{7})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(11-\sqrt{77}-\sqrt{77}+7)}{11-7}$
$=\frac{(18-2 \sqrt{77})}{4}$
$=\frac{9}{2}-\frac{\sqrt{77}}{2}$
$\frac{9}{2}-\frac{\sqrt{77}}{2}=a-b \sqrt{77}$
On comparing the rational and irrational parts of the above equation, we get, a = 92 and b = 12
(vi) $\frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=\mathrm{a}+\mathrm{b} \sqrt{5}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$4+3 \sqrt{5}$
$=\frac{(4+3 \sqrt{5})(4+3 \sqrt{5})}{(4-3 \sqrt{5})(4+3 \sqrt{5})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(16+24 \sqrt{5}+45)}{-29}$
$=\frac{(61+24 \sqrt{5})}{-29}$
$=\frac{-61}{29}-\frac{(24 \sqrt{5})}{29}$
$\frac{-61}{29}-\frac{(24 \sqrt{5})}{29}=a+b \sqrt{5}$
On comparing the rational and irrational parts of the above equation, we get,
$a=\frac{-61}{29}$, and $b=\frac{-24}{29}$
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