Question

prove that
$3 + \sqrt{5}$
is an irrational number ​

Answer 1

Answer:

explanation is given below

explanation

Let √3+√5 be a rational number.

A rational number can be written in the form of p/q where p,q are integers.

√3+√5 = p/q

√3 = p/q-√5

Squaring on both sides,

(√3)² = (p/q-√5)²

3 = p²/q²+√5²-2(p/q)(√5)

√5×2p/q = p²/q²+5-3

√5 = (p²+2q²)/q² × q/2p

√5 = (p²+2q²)/2pq

p,q are integers then (p²+2q²)/2pq is a rational number.

Then √5 is also a rational number.

But this contradicts the fact that √5 is an irrational number.

So,our supposition is false.

Therefore, √3+√5 is an irrational numbe

Answer 2

$let \: 3 + \sqrt{5} = x \: is \: a \: rational \: number$

Now,by squaring both sides ,we get-----

$= > (3 + \sqrt{5} )^{2} = {x}^{2}$

$= > {3}^{2} + ( \sqrt{5} ) ^{2} + 2 \times 3 \times \sqrt{5} = {x}^{2}$

$= > 9 + 5 + 6\sqrt{5} = {x}^{2}$

$= > 6 \sqrt{5} = {x}^{2} - 14$

$= > \sqrt{5} = \frac{ {x}^{2} - 14 }{6} - - - - (1)$

Since,x is a rational number

=>Then,x² is also a rational number

=>(x²-14) is also rational

=>(x²-14/6) is also rational

But,since_/5 is a irrational number.

So,here our conclusion contradicts with our initial consideration.

=>x cannot be rational

Thus-----

$3 + \sqrt{5} \: is \: irrational$