Prove that the following are irrational:(i) (ii) (iii)

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that the following are irrational:
(i) $\frac{1}{\sqrt{2}}$ (ii) $7 \sqrt{5}$ (iii) $6 + \sqrt{2}$

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(i) $\frac{1}{\sqrt{2}}$

Let us assume

\frac{1}{\sqrt{2}}

is rational.

So we can write this number as

\frac{1}{\sqrt{2}} = \frac{a}{b}

---- (1)

Here, a and b are two co-prime numbers and b is not equal to zero.

Simplify the equation (1) multiply by

\sqrt{2}

both sides, we get

1 = \frac{a \sqrt{2}}{b}

Now, divide by b, we get

b = a \sqrt{2}

\frac{b}{a} = \sqrt{2}

Here, a and b are integers so,

\frac{b}{a}

is a rational number,

\sqrt{2}

should be a rational number.

But

\sqrt{2}

is a irrational number, so it is contradictory.

Therefore,

\frac{1}{\sqrt{2}}

is irrational number.

(ii)

7 \sqrt{5}

Let us assume

7 \sqrt{5}

is rational.

So, we can write this number as

7 \sqrt{5} = \frac{a}{b}

---- (1)

Here, a and b are two co-prime numbers and b is not equal to zero.

Simplify the equation (1) divide by 7 both sides, we get

\sqrt{5} = \frac{a}{7 b}

Here, a and b are integers, so

\frac{a}{7 b}

is a rational

number, so

\sqrt{5}

should be a rational number.

But

\sqrt{5}

is a irrational number, so it is contradictory.

Therefore,

7 \sqrt{5}

is irrational number.
(iii)

6 + \sqrt{2}

Let us assume

6 + \sqrt{2}

is rational.

So we can write this number as

6 + \sqrt{2} = \frac{a}{b}

---- (1)

Here, a and b are two co-prime number and b is not equal to zero.

Simplify the equation (1) subtract 6 on both sides, we get

\sqrt{2} = \frac{a}{b} - 6

\sqrt{2} = \frac{a - 6 b}{b}

Here, a and b are integers so,

\frac{a - 6 b}{b}

is a rational

number, so

\sqrt{2}

should be a rational number.

But

\sqrt{2}

is a irrational number, so it is contradictory.

Therefore,

6 + \sqrt{2}

is irrational number.