Question
Prove that $\big(2\sqrt3-1\big)$ is an irrational number.
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Answer
Let $\text{x}=2\sqrt3-1$ be a rational number.
$\text{x}=2\sqrt3-1$
$\Rightarrow\text{x}^2=\big(2\sqrt3-1\big)^2$
$\Rightarrow\text{x}^2=\big(2\sqrt3\big)^2+1^2-2\big(2\sqrt3\big)(1)$
$\Rightarrow\text{x}^2=12+1-4\sqrt3$
$\Rightarrow\text{x}^2-13=-4\sqrt3$
$\Rightarrow\frac{13-\text{x}^2}{4}=\sqrt3$
Since x is a rational number, $x^2$ is also a rational number.
$\Rightarrow 13 − x^2$is a rational number
$\Rightarrow\frac{13-\text{x}^2}{4}$ is a rational number
$\Rightarrow\sqrt3$ is a rational number
But $\sqrt3$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(2\sqrt3-1\big)$ is an irrational number.
$\text{x}=2\sqrt3-1$
$\Rightarrow\text{x}^2=\big(2\sqrt3-1\big)^2$
$\Rightarrow\text{x}^2=\big(2\sqrt3\big)^2+1^2-2\big(2\sqrt3\big)(1)$
$\Rightarrow\text{x}^2=12+1-4\sqrt3$
$\Rightarrow\text{x}^2-13=-4\sqrt3$
$\Rightarrow\frac{13-\text{x}^2}{4}=\sqrt3$
Since x is a rational number, $x^2$ is also a rational number.
$\Rightarrow 13 − x^2$is a rational number
$\Rightarrow\frac{13-\text{x}^2}{4}$ is a rational number
$\Rightarrow\sqrt3$ is a rational number
But $\sqrt3$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(2\sqrt3-1\big)$ is an irrational number.
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