Question
Prove that $4-5\sqrt{2}$ is an irrational number.
✓
Answer
Let $4-5\sqrt{2}$ is not are irrational number.
and let $4-5\sqrt{2}$ is a rational number.
and $4-5\sqrt{2}=\frac{\text{a}}{\text{b}}$ where a and b are positive prime integers,
$\Rightarrow\ 4-\frac{\text{a}}{\text{b}}=5\sqrt{2}$
$\Rightarrow\ \frac{4\text{b}-\text{a}}{\text{b}}=5\sqrt{2}$
$\Rightarrow\ \frac{4\text{b}-\text{a}}{5\text{b}}=\sqrt{2}$
$\sqrt{2}$ is a rational number.
But $\sqrt{2}$ is an irrational number.
Our supposition is wrong.
$4-5\sqrt{2}$ is an irrational number.
and let $4-5\sqrt{2}$ is a rational number.
and $4-5\sqrt{2}=\frac{\text{a}}{\text{b}}$ where a and b are positive prime integers,
$\Rightarrow\ 4-\frac{\text{a}}{\text{b}}=5\sqrt{2}$
$\Rightarrow\ \frac{4\text{b}-\text{a}}{\text{b}}=5\sqrt{2}$
$\Rightarrow\ \frac{4\text{b}-\text{a}}{5\text{b}}=\sqrt{2}$
$\sqrt{2}$ is a rational number.
But $\sqrt{2}$ is an irrational number.
Our supposition is wrong.
$4-5\sqrt{2}$ is an irrational number.
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