Question
Prove that the following number is irrational: $3 - \sqrt{2}$
✓
Answer
$3-\sqrt{2}$
Let $3-\sqrt{2}$ be a rational number.
$\Rightarrow 3-\sqrt{2 } =x$
Squaring on both the sides, we get
$(3-\sqrt{2})^2=x^2$
$ \Rightarrow 9+2-2 \times 3 \times \sqrt{2 } =x^2$
$ \Rightarrow 11-x^2=6 \sqrt{2 } $
$ \Rightarrow \sqrt{2 } =\frac{11-x^2}{6}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow 11-x^2$ is a rational number.
$\Rightarrow \frac{11-x^2}{6}$ is also a rational number.
$\Rightarrow \sqrt{2}=\frac{11-x^2}{6}$ is a rational number.
But $\sqrt{ 2} $ is an irrational number.
$\Rightarrow \frac{11-x^2}{6}=\sqrt{2}$ is an irrational number.
$\Rightarrow 11-x^2$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow \mathrm{x}$ 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 $3-\sqrt{ 2} $ is a rational number is wrong.
$\therefore 3-\sqrt{ 2} $ is an irrational number.
Let $3-\sqrt{2}$ be a rational number.
$\Rightarrow 3-\sqrt{2 } =x$
Squaring on both the sides, we get
$(3-\sqrt{2})^2=x^2$
$ \Rightarrow 9+2-2 \times 3 \times \sqrt{2 } =x^2$
$ \Rightarrow 11-x^2=6 \sqrt{2 } $
$ \Rightarrow \sqrt{2 } =\frac{11-x^2}{6}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow 11-x^2$ is a rational number.
$\Rightarrow \frac{11-x^2}{6}$ is also a rational number.
$\Rightarrow \sqrt{2}=\frac{11-x^2}{6}$ is a rational number.
But $\sqrt{ 2} $ is an irrational number.
$\Rightarrow \frac{11-x^2}{6}=\sqrt{2}$ is an irrational number.
$\Rightarrow 11-x^2$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow \mathrm{x}$ 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 $3-\sqrt{ 2} $ is a rational number is wrong.
$\therefore 3-\sqrt{ 2} $ is an irrational number.
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