Question

Prove that 3+ √5 is an irrational number.
Please solve... this..

Answer 1

$lets \: assume \: 3 + \sqrt{5} \: as \: a \: rational \: number \\ therefore \: 3 + \sqrt{5} = \frac{a}{b} \: where \: b \: is \: not \: equal \: to \: 0 \: and \: p \: and \: q \: are \: integer \\ \sqrt{5} = \: \frac{a - 3}{b} \\ here \: a \: \: b \: and \: 3 \: are \: rational \: but \: \sqrt{5} \: is \: irrational \\ hence \: our \: assumption \: was \: wrong \\ therefore \: 3 + \sqrt{5} \: is \: irratinal. \\ hence \: proved.$
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Answer 2

$Here \: is \: the \: answer \: of \: your \: question$

 Let us assume that 3 + √5 is a rational number.

Now,

3 + √5 = (a ÷ b)

[Here a and b are co-prime numbers]

√5 = [(a ÷ b) - 3]

√5 = [(a - 3b) ÷ b]

Here, {(a - 3b) ÷ b} is a rational number.

But we know that √5 is a irrational number.

So, {(a - 3b) ÷ b} is also a irrational number.

So, our assumption is wrong.

3 + √5 is a irrational number.

Hence, proved.

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How √5 is a irrational number.?

→ √5 = a ÷ b [a and b are co-prime numbers]

b√5 = a

Now, squaring on both side we get,

5b² = a² ........(1)

b² = a² ÷ 5 

Here 5 divide a²

and 5 divide a also

Now,

a = 5c [Here c is any integer]

Squaring on both side

a² = 25c²

5b² = 25c² [From (1)]

b² = 5c²

c² = b² ÷ 5

Here 5 divide b²

and 5 divide b also

→ a and b both are co-prime numbers and 5 divide both of them.

So, √5 is a irrational number.

Hence, proved
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