RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.3 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Exercise 2.3.

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Real Numbers |

E×ercise |
Exercise 2.3 |

Number of Questions Solved |
3 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Ex 2.3

Question 1.

Prove that 5 – √3 is (RBSESolutions.com) irrational.

Solution

Let 5 – √3 is a rational number.

we have to find out two integers a and b such as

a, b and 5 all are integers

\(\frac { 5b-a }{ b }\) is a rational number.

√3 will be also rational number.

But this contradicts the fact that √3 is an irrational number.

So our hypothesis is wrong.

Hence, 5 – √3 is an irrational number.

Hence proved.

Question 2.

Prove that following (RBSESolutions.com) numbers are ir-rational

(i) \(\frac { 1 }{ \surd 2 }\)

(ii) 6 + √2

(iii) 3√2

Solution

(i) \(\frac { 1 }{ \surd 2 }\) is a rational number

we find two integers b such as \(\frac { 1 }{ \surd 2 }\) = \(\frac { a }{ b }\)

where a and b co-prime integers (b ≠ 0)

Square both sides

So, b^{2} divides by 2.

.’. So, b also divides by 2.

Now let b = 2c, where c is any integer

b^{2} = (2c)^{2}

⇒ b^{2} = 4c^{2}

⇒ 2a^{2} = 4c^{2} (∴ b^{2} = 2a^{2})

⇒ a^{2} = 2c^{2}

Hence, a^{2}, divides by 2.

a also divides by 2.

Hence, 2 is a common (RBSESolutions.com) factor of a and b.

But, this contradicts the fact that a and b have no common factor other than 1.

This contradiction arises by assuming that \(\frac { 1 }{ \surd 2 }\) is rational,

Hence \(\frac { 1 }{ \surd 2 }\) is a irrational number.

Hence proved

(ii) Let 6 + √2 is a rational number

we can find two integers a and b (b ≠ 0)

Such as

a, b and 6 all are integers.

\(\frac { a-6b }{ b }\) is a rational numbers

√2 is also a rational number

But this contradicts the fact that √2 is an irrational number.

Our hypothesis is wrong.

Hence, 6 + √2 is an irrational number.

Hence proved.

(iii) Let 3√2 is a rational number we (RBSESolutions.com) find two integers a and b such as

3√2 = \(\frac { a }{ b }\) (where a and b co-prime integers)

⇒ √2 = \(\frac { a }{ 3b }\)

⇒ a, b and 3 are integers.

\(\frac { a }{ 3b }\) is a rational number

√2 is a rational number

3√2 will be also a rational numbers.

But this contradicts the fact that √2 is an irrational number.

So, our hypothesis is wrong.

So, 3√2 is an irrational number.

Hence proved.

Question 3.

If p and q are a positive prime number (RBSESolutions.com) then prove that √p + √q is irrational.

Solution

Let √p + √q is a rational number

√q is also a (RBSESolutions.com) rational number.

But this result is contradicted.

So our hypothesis is wrong.

Hence, √p + √q is an irrational number.

Hence proved.

We hope the given RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Exercise 2.3, drop a comment below and we will get back to you at the earliest.

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