Question

prove √2-√5 is irrational

Answer 1

Heya!

Here is yr answer.....


Let us assume √2 - √5 is rational.

√2-√5 = a/b ( a, b are any integers)


=> √2 = a/b + √5

by squaring on both sides.....


=> 2 = a²/b² + 5 + 2√5a/b

=> 2√5a/b = 2 - 5 - a²/b²

=> 2√5a/b = -3 - a²/b²

=> 2√5a/b = -3b²-a²/b²

=> 2√5a = -3b² - a²/b

=> √5 = -3b²-a²/2ab


For any two integers RHS is rational

But, LHS is irrational.

A rational and irrational are never equal!

Since, our assumption is false.

Hence, √2 - √5 is irrational!

Hope it hlpz....

Answer 2

Answer:

Step-by-step explanation:

Let √2+√5 be a rational number.

A rational number can be written in the form of p/q where p,q are integers.

√2+√5 = p/q

Squaring on both sides,

(√2+√5)² = (p/q)²

√2²+√5²+2(√5)(√2) = p²/q²

2+5+2√10 = p²/q²

7+2√10 = p²/q²

2√10 = p²/q² - 7

√10 = (p²-7q²)/2q

p,q are integers then (p²-7q²)/2q is a rational number.

Then √10 is also a rational number.

But this contradicts the fact that √10 is an irrational number.

.°. Our supposition is false.

√2+√5 is an irrational number.

Hence proved