Question

Prove that √2-√3 irrational

Answer 1

Let √2 - √3 be equal to a rational number x.

→  x = √2 - √3

On squaring both the sides we get,

→  x² = (√2 - √3)²

→  x² = √2² + √3² - 2 • √2 • √3

→  x² = 2 + 3 - 2√6

→  x² - 5 = - 2√6

Multiplying - 1 at both the sides,

→  5 - x² = 2√6

→ (5 - x²)/2 = √6

Thus, √6 is in p/q form which is a rational number but this contradicts the fact that √6 is an irrational number.

Therefore, √2 - √3 is an irrational number.

→ Q.E.D

Answer 2

Let us assume that $\sqrt{2} \: - \: \sqrt{3}$ is an Irrational Number.

$\sqrt{2} \: - \: \sqrt{3}$ = $\dfrac{a}{b}$

Squaring on both sides;

Using identify (a - b)² = a² - 2ab + b²

${(\sqrt{2} \: - \: \sqrt{3} )}^{2}$ = ${ (\dfrac{a}{b} )}^{2}$

2 - $2 \sqrt{6}$ + 3 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

5 - $2 \sqrt{6}$ = $\dfrac{ {a}^{2} }{ {b}^{2} }$

- $2 \sqrt{6}$ = $\dfrac{ {a}^{2} }{ {b}^{2} }$ - 5

- $2 \sqrt{6}$ = $\dfrac{ {a}^{2} \: - \: 5 {b}^{2} }{ {b}^{2} }$

$2 \sqrt{6}$ = $\dfrac{ {- \: a}^{2} \: + \: 5 {b}^{2} }{ {b}^{2} }$

$\sqrt{6}$ = ${( \dfrac{1}{2}) (\dfrac{ { - \: a}^{2} \: + \: 5 {b}^{2} }{ {b}^{2} })}$

Here;

${( \dfrac{1}{2}) (\dfrac{ { - \: a}^{2} \: + \: 5 {b}^{2} }{ {b}^{2} })}$ is a Rational Number.

$\sqrt{6}$ is also a rational number. But we know that $\sqrt{6}$ is an Irrational Number.

Which means that our whole assumption is wrong.

$\sqrt{2} \: - \: \sqrt{3}$ is an Irrational Number.

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