Question

2. Prove that 3 + 2√5 is irrational​

Answer 1

Step-by-step explanation:

Let us assume that 3 +

√

5

is a rational number.

⇒3 +

√

5

=

p

q

, where p and q are the integers and q ≠ 0.

⇒

√

5

=

p

q

-3=

p-3q

q

Since, p, q and 3 are integers. So,

p-3q

q

is a rational number.

⇒

√

5

is also a rational number.

but this contradicts the fact that

√

5

is an irrational number.

This contradiction has arisen due to the wrong assumption that 3 +

√

5

is a rational number.

Hence, 3 +

√

5

is an irrational number.

Answer 2

Step-by-step explanation:

Given: 3 + 2√5

To prove: 3 + 2√5 is an irrational number.

Proof: Let us assume that 3 + 2√5 is a rational number.

So, it can be written in the form a/b

3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving 3 + 2√5 = a/b we get,

=>2√5 = a/b – 3

=>2√5 = (a-3b)/b

=>√5 = (a-3b)/2b

This shows (a-3b)/2b is a rational number. But we know that √5 is an irrational number.

So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved