Question

2.
Prove that rootp +rootq is irrational, where p. q are primes.​

Answer 1

let we assume that √p + √q is rational then they exist unique pair of integers a and b

now , √p + √q = a/b. [ a & b are co - primes ]

√p = a/b -  √q

squaring on both sides

(√p )² = ( a/b -  √q )² [ ( a+ b )² = a²- 2ab + b² ]

p = a²/ b² - 2( a/b ) √q + q

2( a/b ) √q = a²/b² + q - p

2 a/b  √q = a²+ b² ( q - p )

b²

 √q = a²+ b² ( q - p )

2b²

 √q is rational

This contradiction as  √q is irrational hence

√p + √q is irrational

Hence proved

Answer 2

let's assume that root p + root q is rational

so they can be written a/b form where b is not equal to zero where a and b are rational

√p+ √q= a/b

squaring on both sides

p + q + 2√pq = (a/b)^2

2√pq = (a/b)^2 + p + q

here are left hand side is an irrational number but right hand side is a rational number

this is a contradiction because rational number cannot be equal to an irrational number

this is due to our wrong assumption that root p + root q is irrational

Hence root p + root q is irrational.

it is a method called proof by contradiction