Question

given that root 5 is irrational number prove that 2 root 5-3 is an irrational number​

Answer 1

Answer:

Given that √5 is irrational, prove that 2√5-3 is an irrational number. Since p and q are integers ∴p+3q2q is a rational number. ∴√5 is a rational number which is a contradiction as √5 is an irrational number . Hence our assumption is wrong and hence 2√5-3 is an irrational number

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.