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 => 5q2=p2
⇒ p2 – Sq2 … (i)
Since 5 divides p2, so it will divide p also.
Let p = 5r
Then p2 – 25r 2 [Squaring both sides]
⇒ 5q2 = 25r2 [From(i)]
⇒ q2 = 5r2
Since 5 divides q2, 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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