Question

6. Given that v5 is irrational, prove that 2 5 - 3 is an irrational number.​

Answer 1

Answer:

(1) Let us assume that √5 is a rational number.

we know that the rational numbers are in the form of p/q form where p,q are intezers.

so, √5 = p/q

    p = √5q

we know that 'p' is a rational number. so √5 q must be rational since it equals to p

but it doesnt occurs with √5 since its not an intezer

therefore, p =/= √5q

this contradicts the fact that √5 is an irrational number

hence our assumption is wrong and √5 is an irrational number.

(2)Let us assume that 5+2√3 is rational

5+2√3 = p/q ( where p and q are co prime)

2√3 = p/q-5

2√3 = p-5q/q

√3 = p-5q/2q

now p , 5 , 2 and q are integers  

∴ p-5q/2q is rational

∴ √3 is rational

but we know that √3 is irrational . This is a contradiction which has arisen due to our wrong assumption.

∴ 5+2√3 is irrational

brainlest to banta hai

Answer 2

$\maltese \: \: {\underline{\bold{Question :}}}$

Given that √5 is irrational , prove that 2 √5 – 3 is an irrational number.

$\maltese \: \: {\underline{\bold{Solution :}}}$

We have to prove that 2√5 - 3 is an irrational number.

Let us assume the contrary the 2√5 - 3 is a rational number.

So we can represent 2√5 - 3 in the form of $\frac{p}{q}$ where p and q are co-prime integers and q ≠ 0.

2√5 - 3 = $\frac{p}{q}$

⇒ 2√5 = 3 + $\frac{p}{q}$

⇒ 2√5 = $\frac{3q \: + \: p}{q}$

⇒ √5 = $\frac{1}{2} \times \frac{3q \: + \: p}{q}$

⇒ √5 =$\frac{3q \: + \: q}{2q}$

Here 3 , 2 , p and q are integers so $\frac{3q \: + \: q}{2q}$ is a rational number.

But we are given that √5 is irrational number so our assumption that 2√5 - 3 is an rational number was wrong.

So we conclude that 2√5 - 3 is an irrational number.