Question

prove that √7+√5 is irrational

Answer 1

Hello friends!!

Here is your answer :

$Let \: \sqrt{7} + \sqrt{5} \: \: be \: \: rational \: \: number$

$\sqrt{7} + \sqrt{5} = \frac{x}{y} \: \: where \: \: x \: \: and \: \: y \: \: are \: \: coprime.$

Squaring both sides,

${( \sqrt{7} + \sqrt{5}) }^{2} = { (\frac{x}{y} )}^{2}$

Using identity :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

$7 + 5 + 2 \sqrt{35} = \frac{ {x}^{2} }{ {y}^{2} }$

$2 \sqrt{35} = \frac{ {x}^{2} }{ {y}^{2} } - 7 - 5$

$2 \sqrt{35} = \frac{ {x}^{2} }{ {y}^{2} } - 7 - 5$

$2 \sqrt{35} = \frac{ {x}^{2} - 7{y}^{2} - 5{y}^{2}}{ {y}^{2} }$

$\sqrt{35} = \frac{ {x}^{2} - 7 {y}^{2}-5{y}^{2} }{2 {y}^{2} }$

Hence, x, y ,7, 5and 2 has no common factor.

$\frac{ {x}^{2} - 7 {y}^{2} -5{y}^{2}}{ {2y}^{2} } \: is \: a \: \: rational \: \: number.$

$therefore \: \sqrt{35} \: \: is \: \: a \: \: rational \: \: number.$

$But \: \: \sqrt{35} \: \: is \: \: an \: \: irrational \: \: number.$

$So,our \: \: assumption \: \: is \: \: wrong.$

Therefore,

$\sqrt{7} + \sqrt{5} \: \: is \: \: an \: \: irrational \: \: number.$

Hope it helps you.. ☺️☺️☺️☺️

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Answer 2

Answer :

If possible, let us consider that (√7 + √5) is rational.

Let, √7 + √5 = a/b, where a and b are integers with non-zero b.

⇒ √7 = a/b - √5

Now, squaring both sides, we get

7 = ( a/b - √5 )²

⇒ 7 = a²/b² - 2√5 a/b + 5

⇒ 2√5 a/b = a²/b² + 5 - 7

⇒ 2√5 a/b = a²/b² - 2

⇒ √5 = a/2b - b/a

Since, a/b is rational, b/a is also rational, then (a/2b - b/a) is also rational which leads to a contradiction to the fact that √5 is irrational.

Thus, (√7 + √5) is irrational. [Proved]

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