Question

given that root5

is irrational prove that 2 root5-3 is irrational


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Answer 1

Question:

Given that √5 is an irrational number, prove that ;

(2√5 - 3) is an irrational number.

Note:

• Irrational number : The number which can't be written in the form of p/q where p & q are any integer and q ≠ 0 .

Prove:

Let's us assume that (2√5 - 3) is a rational number.

Thus,

=> 2√5 - 3 = p/q

=> 2√5 = p/q + 3

=> 2√5 = (p+3q)/q

=> √5 = (p+3q)/2q -------(1)

Since ,

p/q is rational,

Thus,

=> (p/q + 3) is rational.

=> (p+3q)/q is rational.

=> (p+3q)/2q is rational.

=> √5 is rational. { using eq-(1) }

We get that √5 is a rational number which is a contradiction to the fact that √5 is irrational.

Hence, our assumption is wrong.

And hence, (2√5 - 3) is an irrational number.

Hence proved.

Answer 2

To prove :-

2√5 - 3 is an irrational number.

Given :-

√5 is an irrational number .

Proof :-

Let's assume that 2√5 - 3 is a rational number. so we can write it in the p/q form, where q≠0

→ 2√5 - 3 = p/q

→ 2√5 =( p/q ) - 3

$2 \sqrt{5} = \frac{p - 3q}{q}$

$\sqrt{5} = \frac{p - 3q}{2q}$

→ As 3q , 2q and p are some integers . But it is given that √5 is an irrational .

Hence it contradict the fact that√5 is an irrational number . So our hypothesis was wrong .

2√5 - 3 is an irrational number .