Question

prove that 2 - root 3 is an irrational​

Answer 1

Here Is Your Ans ⤵

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$let \: 2 - \sqrt{3} \: is \: an \: rational \: number \\ \\ = > 2 - \sqrt{3} = \frac{a}{b} \\ \\ = > - \sqrt{3} = \frac{a}{b} - 2 \\ \\ = > \sqrt{3} = \frac{ - a + 2b}{b}$

We Know That Irrational Number Always Not Equal To Rational Number

Irrational ≠ Rational number

So , Our Assumptions Is Wrong

Hence , 2 - √3 Is an Irrational Number

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Answer 2

• Let us assume that 2 - √3 is an rational number.

→ 2 - √3 = $\dfrac{a}{b}$

Here ... a and b are co-prime numbers

→ - √3 = $\dfrac{a}{b}$ - 2

→ - √3 = $\dfrac{a\:-\:2b}{b}$

→ √3 = $\dfrac{2b\:-\:a}{b}$

Here ..

$\dfrac{2b\:-\:a}{b}$ is rational number.

So, √3 is also a rational number.

But we know that √3 is irrational number.

So, our assumption is wrong.

2 - √3 is irrational number.

_____ [ HENCE PROVED ]

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