Question
Prove that the following number is irrational: $\sqrt{5 }- 2$
✓
Answer
$\sqrt{5}-2$
Let $\sqrt{5}-2$ be a rational number.
$\Rightarrow \sqrt{5}-2=x$
Squaring on both the sides, we get
$(\sqrt{5}-2)^2=x^2$
$ \Rightarrow 5+4-2 \times 2 \times \sqrt{5}=x^2$
$ \Rightarrow 9-x^2=4 \sqrt{5}$
$ \Rightarrow \sqrt{5}=\frac{9-x^2}{4}$
Here, $\mathrm{x}$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow 9-x^2$ is a rational number.
$\Rightarrow \frac{9-x^2}{4}$ is also a rational number.
$\Rightarrow \sqrt{2 } =\frac{9-x^2}{4}$ is a rational number
But $\sqrt{2 } $ is an irrational number.
$\Rightarrow \sqrt{5}=\frac{9-x^2}{4}$ is an irrational number.
$\Rightarrow 9-x^2$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $\mathrm{x}$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{5}-2$ is a rational number is wrong.
$\therefore \sqrt{5}-2$ is an irrational number.
Let $\sqrt{5}-2$ be a rational number.
$\Rightarrow \sqrt{5}-2=x$
Squaring on both the sides, we get
$(\sqrt{5}-2)^2=x^2$
$ \Rightarrow 5+4-2 \times 2 \times \sqrt{5}=x^2$
$ \Rightarrow 9-x^2=4 \sqrt{5}$
$ \Rightarrow \sqrt{5}=\frac{9-x^2}{4}$
Here, $\mathrm{x}$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow 9-x^2$ is a rational number.
$\Rightarrow \frac{9-x^2}{4}$ is also a rational number.
$\Rightarrow \sqrt{2 } =\frac{9-x^2}{4}$ is a rational number
But $\sqrt{2 } $ is an irrational number.
$\Rightarrow \sqrt{5}=\frac{9-x^2}{4}$ is an irrational number.
$\Rightarrow 9-x^2$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $\mathrm{x}$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{5}-2$ is a rational number is wrong.
$\therefore \sqrt{5}-2$ is an irrational number.
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