Question
Prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.
✓
Answer
If possible, let $\sqrt{3}+\sqrt{5}$ be a rational number equal to $x.$
Then,
$\text{x}=\sqrt{3}+\sqrt{5}$
$\text{x}^2=\Big(\sqrt{3}+\sqrt{5}\Big)^2$
$\text{x}^2=8+2\sqrt{15}$
$\frac{\text{x}^2-8}{2}=\sqrt{15}$
Now, $\sqrt{\frac{\text{x}^2-8}{2}}$ is rational
$\sqrt{15}$ is rational
Thus, we arrive at a contradiction.
Hence, $\sqrt{3}+\sqrt{5}$ is an irrational number.
Then,
$\text{x}=\sqrt{3}+\sqrt{5}$
$\text{x}^2=\Big(\sqrt{3}+\sqrt{5}\Big)^2$
$\text{x}^2=8+2\sqrt{15}$
$\frac{\text{x}^2-8}{2}=\sqrt{15}$
Now, $\sqrt{\frac{\text{x}^2-8}{2}}$ is rational
$\sqrt{15}$ is rational
Thus, we arrive at a contradiction.
Hence, $\sqrt{3}+\sqrt{5}$ is an irrational number.
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