Real Numbers Exercise 1.2 Class 10

 

Real Numbers Exercise 1.2 Class 10

1. Prove that √5 is irrational

Solution:

Assume that √5 is rational.

Then it can be written in the form

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

where a and b are integers, $b \neq 0$ and HCF(a, b) = 1.

Squaring both sides,

$$5 = \frac{a^2}{b^2}$$ $$a^2 = 5b^2$$

This means $a^2$ is divisible by 5, so a must also be divisible by 5.

Let

$$a = 5k$$

Substitute in the equation:

$$(5k)^2 = 5b^2$$
$$25k^2 = 5b^2$$
$$b^2 = 5k^2$$

So b is also divisible by 5.

Thus a and b are both divisible by 5, which contradicts the fact that HCF(a, b) = 1.

Therefore our assumption is wrong.

Hence, √5 is irrational.

2. Prove that$3\; +$$2$$\sqrt{5}$ is irrational

Solution:

Assume that

$$3 + 2\sqrt5$$is rational.

Then

$$2\sqrt5 = (3 + 2\sqrt5) - 3$$

The right side is rational, so $2\sqrt5$becomes rational.

Dividing by 2,

$$$$5=252​

This means √5 is rational, which is a contradiction because √5 is irrational.

Therefore the assumption is false.

Hence $3 + 2\sqrt5$is irrational.

3. Prove that the following are irrational :

(i) $\frac{1}{\sqrt2}$

Solution:

Assume that

$$\frac{1}{\sqrt2}$$is rational.

Then

$$\sqrt2 = \frac{1}{\text{rational number}}$$

This implies √2 is rational, which is not true because √2 is irrational.

Therefore the assumption is false.

Hence $\frac{1}{\sqrt2}$ is irrational.

(ii)  $7\sqrt5$

Solution:

Assume that

$$7\sqrt5$$is rational.

Then

$$\sqrt5 = \frac{7\sqrt5}{7}$$This implies √5 is rational, which is false because √5 is irrational.

Therefore the assumption is wrong.

Hence $7\sqrt5$is irrational.

(iii)  $6 + \sqrt2$

Solution:

Assume that

$$6 + \sqrt2$$is rational.

Then

$$\sqrt{2}=(6+\sqrt{2})-6$$

The right side is rational, so √2 becomes rational.

But √2 is irrational, which is a contradiction.

Therefore the assumption is false.

Hence $6 + \sqrt2$ is irrational.

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