Question
Prove that $\big(4-5\sqrt2\big)$ is an irrational number.

Answer

Let $\text{x}=4-5\sqrt2$ be a rational number.
$\text{x}=4-5\sqrt2$
$\Rightarrow\text{x}^2=\big(4-5\sqrt2\big)^2$
$\Rightarrow\text{x}^2=(4)^2+\big(2\sqrt3\big)^2-2(4)\big(5\sqrt2\big)$
$\Rightarrow\text{x}^2=16+50-40\sqrt2$
$\Rightarrow\text{x}^2-66=-40\sqrt2$
$\Rightarrow\frac{66-\text{x}^2}{40}=\sqrt2$
Since x is a rational number, $x ^2$ is also a rational number. $\Rightarrow 66- x ^2$ is a rational number
$\Rightarrow\frac{66-\text{x}^2}{40}$ is a rational number
$\Rightarrow\sqrt2$ is a rational number
But $\sqrt2$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(4-5\sqrt2\big)$ is an irrational number.

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