Question

prove that root 5 + root 7 is an irrational no.​

Answer 1

Step-by-step explanation:

Proof :-

Given number = √5 + √7

Let us assume that √5+√7 is a rational number.

We know that

A rational number is in the form of p/q

Let √5+√7 = a/b , where a and b are co - primes

=> √5 = (a/b)-√7

=> √5 = (a-b√7)/b

=> √5 is in the form of p/q

=> √5 is a rational number.

But √5 is not a rational number.

It contradicts to our assumption i.e. √5+√7 is a rational number.

Therefore, √5+√7 is not a rational number.

√5+√7 is an irrational number.

Hence, Proved.

Used Method:-

Method of contradiction (Indirect method)

Points to know:-

→ The sum of two irrational numbers is also an irrational number.

√5 is an irrational number

√7 is an irrational number.

Their sum = √5+√7 is also an irrational number.

Answer 2

Answer:

Given:

√2+√5

Prove that

We need to prove√2+√5 is an irrational number.

Proof:

Let us assume that √2+√5 is a rational number.

A rational number can be written in the form of p/q where p,q are integers and q≠0

√2+√5 = p/q

On squaring both sides we get,

(√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 assumption is incorrect

√2+√5 is an irrational number.

Hence proved