Question

Show that (\sqrt{3}-\sqrt{5})^{2} is a rational no. 49 points for free

Answer 1

hi mate,

solution:

we need this formula which is

( a + b )² = a² + 2ab + b²

now,

( √5 + √3 )²

= ( √5 )² + 2 √5 √3 + ( √3 )²

= 5 + 2√15 + 3

= 8 + 2√15

hence we can say that it is a rational number.

i hope it helps you.

Answer 2

Explanation:

Let us assume to the contrary that (√3+√5)² is a rational number,then there exists a and b co-prime integers such that,

(√3+√5)²=a/b

3+5+2√15=a/b

8+2√15=a/b

2√15=a/b-8

2√15=(a-8b)/b

√15=(a-8b)/2b

(a-8b)/2b is a rational number.

Then √15 is also a rational number

But as we know √15 is an irrational number.

This is a contradiction.

This contradiction has arisen as our assumption is wrong.

Hence (√3+√5)² is an irrational number.