Question

given that root P is an irrational number where p is an prime number prove that root 3 + root 5 is irrational​

Answer 1

Answer:

If possible,let √p be a rational number.

also a and b is rational.

then,√p = a/b

on squaring both sides,we get,

(√p)²= a²/b²

→p = a²/b²

→b² = a²/p [p divides a² so,p divides a]

Let a= pr for some integer r

→b² = (pr)²/p

→b² = p²r²/p

→b² = pr²

→r² = b²/p [p divides b² so, p divides b]

Thus p is a common factor of a and b.

But this is a contradiction, since a and b have no common factor.

This contradiction arises by assuming √p a rational number.

Hence,√p is irrational.

Answer 2

Answer:

Let$\sqrt{p}$be a rational number and

$\sqrt{p} = \frac{a}{b}$

$p = \frac{ {a}^{2} }{ {b}^{2} }$

${a}^{2} = p \times {b}^{2}$

Therefore p divides ${a}^{2}$

But when a prime number divides the product of two numbers, it must divide atleast one of them.

Here ${a}^{2} = a \times a$

p divides a

Let $a = pk......(1)$

${p}^{2} {k}^{2} =p {b}^{2}$

So,

$\: \: {b}^{2} = p {k}^{2}$

p divides ${b}^{2}$

But

${b}^{2} = b \times b$

Therefore p divides b

Thus, a and b have atleast one common multiple p But it arises the contradiction to our assumption that a and b are coprime.

Thus, our assumption is wrong and

$\sqrt{p}$ is irrational number.

So we can say $\sqrt{3}$

and $\sqrt{5}$are irrational number.

We know that sum of two irrational number is also a irrational number.

So,$\sqrt{3} + \sqrt{5}$is a irrational number [proved].