Question
Prove that $\sqrt{3}+\sqrt{5}$ is irrational.
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Answer
Let us suppose $\sqrt{3}+\sqrt{5}$ is rational Let $\sqrt{3}+\sqrt{5}=\text{a},$ where a is rational. $\therefore\ \sqrt{3}=\text{a}-\sqrt{5}$ On squaring both sides, we get $(\sqrt{3})^2+(\text{a}-\sqrt{5})^2$ $\Rightarrow\ 3=\text{a}^2+5-2\text{a}\sqrt{5}\ \ \big[\because(\text{a}-\text{b})^2=\text{a}^2 +\text{b}^2-2\text{ab}\big]$ $\Rightarrow\ 2\text{a}\sqrt{5}=\text{a}^2+2$ $\therefore\ \sqrt{5}=\frac{\text{a}^2+2}{2\text{a}}$ which is contradiction.As the right hand side is rational number while V5 is irrational. Since, 3 and 5 are prime number. Hence, $\sqrt{3}+\sqrt{5}$ is irrational.
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