Question

prove that √2-√5 is an irrational number​

Answer 1

Solution :-

Let us assume that √2 - √5 is a rational number.

As we know that,

• A rational number can be written in the form of p/q where p, q are integers and q ≠ 0

so, we can say that,

→ √2 - √5 = p/q

Squaring both sides we get,

→ (√2 - √5)² = (p/q)²

→ 2 + 5 - 2(√5)(√2) = (p²/q²)

→ 7 - 2√10 = (p²/q²)

→ 2√10 = 7 - (p²/q²)

→ 2√10 = (7q² - p²) / q²

→ √10 = (7q² - p²) / 2q²

since we assume that, p and q are integers then we can conclude that, {(7q² - p²) / 2q²} is a rational number.

Therefore we can say that, √10 is also a rational number.

But this contradicts the fact that √10 is an irrational number.

∴

→ our assumption that (√2 - √5) is a rational number is wrong.

Hence, (√2 - √5) is an irrational number.

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Answer 2

SOLUTION

TO PROVE

√2 - √5 is an irrational number

PROOF

We prove it by method of contradiction

If possible let √2 - √5 is a rational number

$\displaystyle \sf{ \implies \: \frac{1}{ \sqrt{2} - \sqrt{5} } \: \: is \: a \: rational \: number }$

$\displaystyle \sf{ \implies \: \frac{( \sqrt{2} + \sqrt{5}) }{ (\sqrt{2} - \sqrt{5} )( \sqrt{2} + \sqrt{5} )} \: \: is \: a \: rational \: number }$

$\displaystyle \sf{ \implies \: \frac{( \sqrt{2} + \sqrt{5}) }{ (4 - 5)} \: \: is \: a \: rational \: number }$

$\displaystyle \sf{ \implies \: - ( \sqrt{2} + \sqrt{5}) \: \: is \: a \: rational \: number }$

$\displaystyle \sf{ \implies \: - \sqrt{2} - \sqrt{5} \: \: is \: a \: rational \: number }$

Now it is assumed that √2 - √5 is an irrational number

Again we know that sum of two rational numbers is a rational number

$\displaystyle \sf{ \implies \: - \sqrt{2} - \sqrt{5} + \sqrt{2} - \sqrt{5} \: \: is \: a \: rational \: number }$

$\displaystyle \sf{ \implies \: - 2\sqrt{5} \: \: is \: a \: rational \: number }$

$\displaystyle \sf{ \implies \: \sqrt{5} \: \: is \: a \: rational \: number }$

Which is a contradiction

∴ √2- √5 is an irrational number

Hence proved

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