CBSE BoardEnglish MediumSTD 10MathsReal Numbers5 Marks
Question
Prove that $\sqrt{5}+\sqrt{3}$ is an irrational number.
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Answer
Let us assume that $\sqrt{5}+\sqrt{3}$ is a rational number.
$\therefore\ \sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-prime numbers.
$\sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \sqrt{5}=\frac{\text{a}}{\text{b}}-\sqrt{3}$
Squaring both sides,
$\Rightarrow(\sqrt{5})^2=\Big(\frac{\text{a}}{\text{b}}-\sqrt{3}\Big)^2$
$\Rightarrow5=\Big(\frac{\text{a}}{\text{b}}\Big)^2+3-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow5-3=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow2=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow\frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-2$
$\Rightarrow\ \frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)$
$\Rightarrow\ \sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)\frac{\text{b}}{2\text{a}}$
$\Rightarrow\sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{2\text{ab}}\Big)$
We know that $\sqrt{3}$ is an irrational number.
This contradicts our assumption that $\sqrt{5}+\sqrt{3}$ is a rational number.
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