Question

Prove that √p +√q is irrational if p and q are prime numbers.

Answer 1

$Let \: \sqrt{p} + \sqrt{q} \: is \: rational$

$\sqrt{p} + \sqrt{q} = \frac{a}{b} [a,b \: are \: co-primes \: b \: is \\ \\not \: equal \: to \: 0]$ [Squaring on both the sides]

${( \sqrt{p} + \sqrt{q})}^{2} = {( \frac{a}{b}) }^{2}$

${( \sqrt{p})}^{2} + { (\sqrt{q})}^{2} + 2. \sqrt{p}. \sqrt{q} = \frac{ {a}^{2} }{ {b}^{2} }$

$p + q + 2 \sqrt{pq} = \frac{ {a}^{2} }{ {b}^{2} }$

$2 \sqrt{pq} = \frac{ {a}^{2} }{ {b}^{2} } - p - q$

$2 \sqrt{pq} = \frac{ {a}^{2} - p {b}^{2} - q {b}^{2} }{ {b}^{2} }$

$\sqrt{pq} = \frac{ {a}^{2} - p {b}^{2} - y {b}^{2} }{2 {b}^{2} }$

$\frac{ {a}^{2} - p {b}^{2} - q {b}^{2} }{2 {b}^{2} } \: is \: rational \: , \: so\: \sqrt{pq} \: is \\also \: rational$

$But \: \sqrt{pq} \: is \: irrational \: [ because\: p , q \: are \: primes ]$

It contradicts the fact that our assumption √p+√q is rational is false.

Therefore , √p+√q is irrational.

Answer 2

Proof :

Let us prove that, √p +√q is irrational if p and q are prime numbers by Method of contradiction

To start with, Consider that √p +√q is rational.

Since it's rational, We can express it in the form of m/n where m, n are co - primes and m, n are Integers, n ≠ 0

√p + √q = m/n

Squaring on both sides,

(√p + √q)² = m²/n²

p + q + 2√pq = m²/n²

2√pq = m²/n² - p - q

√pq = m² - pn² - qn² / 2n²

Now,

√pq is irrational since p, q are primes.

But, RHS of the equation is rational because p, q, m, n all are Integers.

Hence, We see that RHS which is rational became equal to LHS which is irrational, This contradiction is because of our faulty assumption that √p +√q is rational.

Therefore, We conclude √p +√q is irrational if p and q are prime numbers.