Question

Prove that 3+2√5 is an Irrational.​

Answer 1

Question :–

Prove that 3+2√5 is an Irrational.

Solution :–

Let, 3 + 2√5 = $\sf \dfrac{a}{b}$ is a rational number.

where a & b are integers, q ≠ 0

$2 \sqrt{5} = \sf\dfrac{a}{b} - 3$

$2 \sqrt{5} = \sf\dfrac{a - 3}{b}$

$\sqrt{5} = \sf \dfrac{a - 3}{2b}$

Since, a & b are integers.

So, (a - 3b) & 2b are also integers. Therefore, $\sf \dfrac{a - 3b}{2b}$ is a rational number.

But we know that √5 is an Irrational.

its L.H.S. = R.H.S.

and R.H.S. = Rational.

It is not possible.

Hence, 3 + 2√5 is an Irrational.

Answer 2

Question :–

Prove that 3+2√5 is an Irrational.

Solution :–

Let, 3 + 2√5 = \sf \dfrac{a}{b}

b

a

is a rational number.

where a & b are integers, q ≠ 0

2 \sqrt{5} = \sf\dfrac{a}{b} - 32

5

=

b

a

−3

2 \sqrt{5} = \sf\dfrac{a - 3}{b}2

5

=

b

a−3

\sqrt{5} = \sf \dfrac{a - 3}{2b}

5

=

2b

a−3

Since, a & b are integers.

So, (a - 3b) & 2b are also integers. Therefore, \sf \dfrac{a - 3b}{2b}

2b

a−3b

is a rational number.

But we know that √5 is an Irrational.

its L.H.S. = R.H.S.

and R.H.S. = Rational.

It is not possible.

Hence, 3 + 2√5 is an Irrational.