Question

Prove that √2 + √7 is an irrational number​

Answer 1

Answer:

Step-by-step explanation:

√2 + √7 = a/b              (where a and b are co-prime)

√7 = a/b - √2

Squaring both sides

a²/b² - 2√2a/b + 2 = 7

a²/b² - 5 = 2√2a/b

(a² - 5b²) / 2ab  = √2

√2 is an irrational number .

This contradicts the fact that √2 is irrational . So , our assumption is wrong.

Hence ,  

                 √2 + √7 is irrational

Answer 2

Answer:

Let us assume that √2 + √7 is rational.

Then their exist two positive integers p and q , where

q is not = 0.

√2 + √7 = p/q ,where p and q are co-primes.

√2 = p/q - √7

√2 = p - √7 q / q

squaring both sides

(√2) ^2 = (p - √7q) ^2 / q^2

2 = p^2 - 2 × p × √7q + (√7q) ^2

2 = p^2 - 2√7pq + 7q^2

2-2√7pq = p^2 + 7q^2

2√7pq = p^2 + 7q^2 +2

√7 = p^2 + 7q^2 +2 / 2pq ---- eq 1

In equation 1, p and q are integers and

p^2 + 7q^2 +2 / 2pq is rational

that implies √7 is also rational

But we know that √7 is irrational. This is a contradiction to our assumption √2 + √7 is rational.

So we conclude that √2 + √7 is irrational.