Question

Prove that √p+√Q us an irrational where p and Q are primes

Answer 1

let us assume to our contradiction that √p +√q is rational
so it can be written in form a÷b where a and b are Co prime
√p+√q=a/b
squaring both side
(√p+√q)²=a²/b²
p+q+2√pq=a²/b²
2√pq=a²/b²-p-q
2√pq=(a²-b²p-b²q) /b²
√pq=(a²-b²p-b²q) /2b²
here a,b,p and q are integers so the R. H. S is rational. this contradicts the fact that√pq is irrational..
this contradiction has arisen due to our wrong assumption that √p+√q is rational
so √p+√q is rational..

HOPE IT WORKS
If you are satisfied mark it as brainliest... :)

Answer 2

$The square root of every prime integers are irrational. By proving this, we can find the answer to the question. \\ \\ Let's prove whether \sqrt{p}\ \ is irrational. \\ \\ Let \ \sqrt{p}\ \ be a rational number and can be written in the form of \ \frac{a}{b}.\ Here \ a\ and \ b\ are coprime integers, and \ \frac{a}{b}\ is in its least form. \\ \\ \sqrt{p} = \frac{a}{b}\ \ \implies \ \ p = \frac{a^2}{b^2} \\ \\ a^2 = pb^2$

$\\ \\ Here it seems that \ a^2\ is a multiple of \ p. \\ \\ If \ a^2\ is a multiple of \ p,\ then \ a\ is also multiple of \ p. \\ \\ Let \ a = mp \\ \\ a^2 = pb^2\ \ \implies \ \ (mp)^2 = pb^2\ \ \implies\ \ m^2p^2 = pb^2 \\ \\ Dividing \ p\ from both sides, we get, \\ \\ m^2p = b^2$

$Here it seems that \ b^2\ is also multiple of \ p , and thereby is \ b.\ \\ \\ It is first considered that \ a\ and \ b\ are coprime integers, i. e., \ a\ and \ b\ have only 1 as common factor. Here it seems that \ a\ and \ b\ are multiples of \ p.\ How is this possible?! \\ \\ When we continue this process by taking \ b\ as multiple of \ p , like \ a, it'll be endless. So this is a contradiction.$

$\therefore\ \sqrt{p}\ is irrational. So do \sqrt{Q}. \\ \\ The sum of two irrational numbers is always irrational. \\ \\ \therefore\ \sqrt{p} + \sqrt{Q}\ is irrational. \\ \\ Thank you.$