Real Numbers Exercise 1.2 Class 10 Solution.

Introduction:

        यह अभ्यास (Exercise 1.2) कक्षा 10 गणित के Real Numbers अध्याय का एक महत्वपूर्ण भाग है, जिसमें rational और irrational numbers से जुड़े सिद्धांतों को समझाया गया है। इस अभ्यास में विद्यार्थियों को यह सिखाया जाता है कि किन संख्याओं को rational और किन्हें irrational कहा जाता है। सभी प्रश्नों को सरल और step-by-step तरीके से हल किया गया है, ताकि हर विद्यार्थी इसे आसानी से समझ सके।

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,

$$\sqrt5 = \frac{2\sqrt5}{2}$$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.

Conclusion:

इस अभ्यास में हमने rational और irrational numbers के बीच अंतर को समझा और उनके गुणों का अध्ययन किया। यह विषय गणित की नींव को मजबूत करने में बहुत महत्वपूर्ण भूमिका निभाता है। नियमित अभ्यास से विद्यार्थी इस concept को आसानी से समझ सकते हैं और परीक्षा में अच्छे अंक प्राप्त कर सकते हैं।

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