NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

- Real Numbers Class 10 Ex 1.1
- Real Numbers Class 10 Ex 1.2
- Real Numbers Class 10 Ex 1.3
- Real Numbers Class 10 Ex 1.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Real Numbers |

Exercise |
Ex 1.3 |

Number of Questions Solved |
3 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

**Ex 1.3 Class 10 Maths Question 1.
**Prove that √5 is irrational.

**Solution:**

Let √5 = \(\frac { p }{ q }\) be a rational number, where p and q are co-primes and q ≠ 0.

Then, √5q = p => 5q

^{2}=p

^{2 }⇒ p

^{2}– Sq

^{2}… (i)

Since 5 divides p

^{2}, so it will divide p also.

Let p = 5r

Then p

^{2}– 25r

^{2}[Squaring both sides]

⇒ 5q

^{2}= 25r

^{2}[From(i)]

⇒ q

^{2}= 5r

^{2 }Since 5 divides q

^{2}, so it will divide q also. Thus, 5 is a common factor of both p and q.

This contradicts our assumption that √5 is rational.

Hence, √5 is irrational. Hence, proved.

**Ex 1.3 Class 10 Maths Question 2.**

Show that 3 + √5 is irrational.

**Solution:
**Let 3 + 2√5 = \(\frac { p }{ q }\) be a rational number, where p and q are co-prime and q ≠ 0.

Then, 2√5 = \(\frac { p }{ q }\) – 3 = \(\frac { p – 3q }{ q }\)

⇒ √5 = \(\frac { p – 3q }{ 2q }\)

since \(\frac { p – 3q }{ 2q }\) is a rational number,

therefore, √5 is a rational number. But, it is a contradiction.

Hence, 3 + √5 is irrational. Hence, proved.

**Ex 1.3 Class 10 Maths Question 3.
**Prove that the following are irrational.

**Solution:**

**(i)**Let \(\frac { 1 }{ \sqrt { 2 } }\) = \(\frac { p }{ q }\) be a rational number,

where p and q are co-prime and q ≠ 0.

Then, √2 = \(\frac { q }{ p }\)

Since \(\frac { q }{ p }\) is rational, therefore, √2 is rational.

But, it is a contradiction that √2 is rational, rather it is irrational.

Hence, \(\frac { 1 }{ \sqrt { 2 } }\) is irrational.

Hence, proved.

**(ii)** Let 7√5 = \(\frac { p }{ q }\) be a rational number, where p, q are co-primes and q ≠ 0.

Then, √5 = \(\frac { p }{ 7q }\)

Since \(\frac { p }{ 7q }\) is rational therefore, √5 is rational.

But, it is a contradiction that √5 is rational rather it is irrational.

Hence, 7√5 s is irrational.

Hence proved.

**(iii)** Let 6 + √2 = \(\frac { p }{ q }\) be a rational number, where p, q are co-primes and q ≠ 0.

Then, √2 = \(\frac { p }{ q }\) – 6 = \(\frac { p – 6q }{ q }\)

Since \(\frac { p – 6q }{ q }\) is rational therefore, √2 is rational.

But, it is a contradiction that √2 is rational, rather it is irrational.

Hence, 6 + √2 is irrational.

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