Question
Prove that the following number is irrational: $\sqrt{3} + \sqrt{2}$
✓
Answer
$\sqrt{3}+\sqrt{2}aX$
Let $\sqrt{3}+\sqrt{2}$ be a rational number.
$\Rightarrow \sqrt{3 } +\sqrt{2}=x$
Squaring on both the sides, we get
$(\sqrt{3}+\sqrt{2})^2=x^2$
$ \Rightarrow 3+2+2 \times \sqrt{3 } \times \sqrt{ 2} =x^2$
$ \Rightarrow \mathrm{x}^2-5=2 \sqrt{6 } $
$ \Rightarrow \sqrt{ 6} =\frac{x^2-5}{2}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow x^2-5$ is a rational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a rational number.
But $\sqrt{ 6} $ is an irrational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a irrational number.
$\Rightarrow x^2-5$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $x$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{3 } +\sqrt{2}$ is a rational number is wrong.
$\therefore \sqrt{3}+\sqrt{2}$ is an irrational number.
Let $\sqrt{3}+\sqrt{2}$ be a rational number.
$\Rightarrow \sqrt{3 } +\sqrt{2}=x$
Squaring on both the sides, we get
$(\sqrt{3}+\sqrt{2})^2=x^2$
$ \Rightarrow 3+2+2 \times \sqrt{3 } \times \sqrt{ 2} =x^2$
$ \Rightarrow \mathrm{x}^2-5=2 \sqrt{6 } $
$ \Rightarrow \sqrt{ 6} =\frac{x^2-5}{2}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow x^2-5$ is a rational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a rational number.
But $\sqrt{ 6} $ is an irrational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a irrational number.
$\Rightarrow x^2-5$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $x$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{3 } +\sqrt{2}$ is a rational number is wrong.
$\therefore \sqrt{3}+\sqrt{2}$ is an irrational number.
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