Question
Show that the following numbers are irrational.
$6+\sqrt{2}$
$6+\sqrt{2}$
✓
Answer
Let us assume that $6+\sqrt{2}$ is a rational number.
$\therefore\ 6+\sqrt{2}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-primes numbers.
$\Rightarrow\sqrt{2}=\frac{\text{a}}{\text{b}}-6$
$\Rightarrow\sqrt{2}=\frac{\text{a}-6\text{b}}{\text{b}}$
We know that $\sqrt{2}$ is an irrational number.
This contradicts our assumption that $6+\sqrt{2}$ is a rational number.
Hence, $\sqrt{2}$ must be irrational.
$\therefore\ 6+\sqrt{2}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-primes numbers.
$\Rightarrow\sqrt{2}=\frac{\text{a}}{\text{b}}-6$
$\Rightarrow\sqrt{2}=\frac{\text{a}-6\text{b}}{\text{b}}$
We know that $\sqrt{2}$ is an irrational number.
This contradicts our assumption that $6+\sqrt{2}$ is a rational number.
Hence, $\sqrt{2}$ must be irrational.
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