Question

If P,Q are prime positive integers, prove that $\sqrt{p} + \sqrt{q}$ is an irrational number

Answer 1

SOLUTION :  

Let us assume that √p +√q is rational.  Then,it will be of the form a/b where a, b are co primes integers and b ≠0.

√p + √q = a/b

√p = a/b −√q

on squaring both sides,

(√p)² = (ab−√q)²

p = (a/b)² −(2a√q)/b + q

p − q =(a/b)² −(2a√q)/b

(2a√q)/b = (a/b)² - (p - q)

2a√qb =( a² −b²(p - q)) /b²

√q  =( a² −b²(p - q)) /b² × (b/2a)

√q  = (a² −b²(p - q)) / 2ab  

since, a & b is an integer so, (a² −b²(p - q)) / 2ab  is a rational number.  

∴ √q is rational .

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

Hence,√p + √q  is an irrational .

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

hello....

Let us assume that √p +√q is rational.
so therefore we can say that a, b are co-primes integers and b no equal to 0.

✔️√p + √q = a/b

✔️√p = a/b −√q

on squaring...

✔️(√p)² = (ab−√q)²

✔️p = (a/b)² −(2a√q)/b + q

✔️(2a√q)/b = (a/b)² - (p - q)

✔️2a√qb =( a² −b²(p - q)) /b²

✔️√q  = (a² −b²(p - q)) / 2ab  

so... √q is rational .

This contradicts the fact that √q is an irrational number.....

so we can say that...

✔️✔️√p + √q  is an irrational .

thank you...