Question

prove
that √5-√13 is irrational​

Answer 1

Answer:

25-169

144

it is a answer I hoped it is help you

Answer 2

Step-by-step explanation:

Given:-

√5-√13

To find:-

Prove that √5-√13 is irrational

Solution:-

Given number = √5-√13

Let us assume that √5-√13 is a rational number

It must be in the p/q form

Let √5-√13 = a/b

Where a and b are co-primes.

=> √5 = (a/b)+√13

= > √13 = √5 -(a/b)

On squaring both sides then

=>[√13]^2 = [√5-(a/b)]^2

=> 13 = (√5)^2-2(√5)(a/b)+(a/b)^2

Since (a-b)^2 = a^2-2ab+b^2

=> 13 = 5-2(a/b)√5 +(a^2/b^2)

=> 13-5 = (a^2/b^2)-2(a/b)√5

=> 8 = (a^2/b^2)-2(a/b)√5

=> 8=(a^2-2ab√5)/b^2

=> 8b^2 = a^2-2ab√5

=> 8b^2-a^2 = -2ab√5

=> a^2-8b^2 = 2ab√5

=>√5 = (a^2-8b^2)/2ab

=>√5 is in the form of p/q

So, By the definition of rational numbers

=> √5 is a rational number.

But √5 is not a rational number.

This contradicts to our assumption that √5-√13 is a rational number.

√5-√13 is not a rational number.

√5-√13 is an irrational number.

Hence, Proved.

Used formulae:-

• The numbers are in the form of p/q ,where p and q are integers and q≠0 called. rational numbers.
• (a-b)^2 = a^2-2ab+b^2

Used Method:-

Method of Contradiction (Indirect method)