Exercise 1.3 Class 10 Maths
exercise 1.3 class 10 maths ncert solutions
Class 10 Maths Exercise 1.3 solution : We are providing Class 10 Maths solutions for 1st chapter ex 1.3 class 10 maths “Real Numbers” textbook provided by NCERT for SEBA, CBSE etc complete exercise solutions with Additional Most Important Questions in our website – online, absolutely for FREE for everyone. No need to do any sign in process, just browse and enjoy all of the Class 10 Maths Chapter 1 exercise 1.3 solutions from us. From the First exercise “Real Numbers”, there are total 4 exercise given in bellow table. Click on the exercise number to browse the complete solutions from ex 1.3 class 10 ..
Go to Real Numbers Exercise 1.1
Go to Real Numbers Exercise 1.2
You are at Real Numbers Exercise 1.3
Go to Real Numbers Exercise 1.4
Sub-Contents of “Chapter 1 Real Numbers” : There are total 6 sub contents exists in Class 10 maths chapter 1 Real Numbers. We need to study all of the sub-contents first. The Sub Contents are-
- Introduction
- Euclid’s Division algorithm
- The Fundamental Theorem of Arithmetic
- Revisiting Irrational Numbers
- Revisiting Rational Numbers & Their Decimal Expansion
- Summary
FROM LATEST UPDATED NCERT SYLLABUS |
Exercise 1.3 Class 10 MathsReal Numbers Class 10 Maths Exercise 1.3 Question 1Prove that √5 is irrational. Proof: Let, √5 be a rational Then, √5 = \frac p q , where p, q are co-prime ⇒ p = √5.q ⇒ p² = 5q² —————-(1) Equation (1) means, p² is divisible by 5 So, p is divisible by 5 ——-(A) So, p = 5k , where k is any integers. ⇒ p² = 25k² ⇒ 5q² = 25k² [ from (1)] Implies, q² = 5k² Equation (2) means, q² is divisible by 5 So, q is divisible by 5 ———> (B) So, from (A) & (B), we see that, p & q both are divisible by 5. That means, they are not co-prime Number. So, √5 can’t be rational. Therefore, √5 is an irrational Number. Class 10 Maths Exercise 1.3 Question 2Prove that 3 + 2√5 is irrational. Solution: Let, 3 + 2√5 be a rational. So, 3 + 2√5 = \frac p q , where p, q are integers. ⇒ 2√5 = \frac p q – 3 ⇒ √5 = \cfrac { \cfrac p q - 3} {2} OR, √5 = \tfrac {p-3q} {2q} Implies, Irrational Number = Rational Numner Which is not possible. So, 3 + 2√5 can’t be rational Therefore, 3 + 2√5 is irrational Number. Class 10 Maths Exercise 1.3 Question 3Prove that the following numbers are irrational. (i) \cfrac {1} {√5} (ii) 7√5 (iii) 6 + √2 Solutions: (i) Let, \cfrac {1} {√5} be a rational number. So, \cfrac {1} {√5} = \frac p q , where p, q are co-prime ⇒ q = √2.p ⇒ q² = 2p² ———————-> (a) Means, q² is divisible by 2 Or, q is sdivisible by 2 ———> (1) ⇒ q = 2k, where k is any integer ⇒ q² = 4k² Or, 2p² = 4k² { from (a)} ⇒ p² = 2k² Means, p² is divisible by 2 So, p is divisible by 2 ————> (2) From (1) & (2) statements ⇒ p & q both are divisible by 2 So, p, q are not co-prime, means \cfrac {1} {√5} can’t be rational So, \cfrac {1} {√5} is an irrational number. (ii) Let, 7√5 be a rational number. So, 7√5 = \frac p q ⇒ √5 = \cfrac {p} {7q} ⇒ A irrational number = A rational number [ as p, 7, q all are rational] Which is not possible. So, 7√5 can’t be a rational number. Therefore, 7√5 is irrational. (iii) Let, 6 + √2 be a rational number So, 6 + √2 = \frac p q ⇒ √2 = \frac p q – 6 ⇒ √2 = \cfrac {p-6q} {q} Implies, a irrational number = a rational number as p, 6, q all are rational Which is not possible. So, 6 + √2 can’t be rational Therefore, 6 + √2 is a irrational number. |
ADDITIONAL QUESTIONS from Ex 1.3 class 10 Chapter 1:
We have listed some most important additional questions too from the exercise 1.3 class 10 maths chapter 1 “Real Numbers”. Try do solve them yourself and inform us by commenting bellow that you have solved these or NOT.
- Prove that √3 is irrational
- Show that 5 – √7 is a irrational number
- Prove that 3√5 is a irrational number.
- Prove that 3 – 2√5 is a irrational number.
- What is co-prime numbers?
WHAT IS CO PROIME NUMBERS?
Well! while solving the above problems, we often used a word “Co-Prime” numbers. So, Which type of numbers are known as the co-prime numbers? The defination of co-prime number is – ” Two or more numbers are said to be co-prime or relatively prime, if , the numbers are divisible by ONLY 1 at the same time together”. That means, if the common factor of some numbers is 1, then the numbers will be considered as co prime numbers.
For example, if we consider 2 and 5, then , at the same moment, together, they are only divisible by 1. So, 2 & 5 are coprime numbers. Sometimes, some composite numbers can also create some co-prime numbers together. For example, 4 and 9 are not prime numbers. But, if we consider them together, they are only divisible by 1 at the same time, isn’t it? So, though 4 and 9 are separately two composite numbers, they together create a pair of co-prime numbers.
Theorem USED in Ex 1.3 Class 10 maths :
Theorem 1: Suppose p is a prime number. Suppose p divides a² . Then automatically, p will also devide a
Rational Numbers:
Any number, that can be expressed in the form \frac p q , where p and q are integers (co-prime) and the denominator, means q ≠ 0 are known as a rational number. Examples: All integers, √4, √9 , \frac 22 7 , 3.14 etc….
Irrational Numbers:
Any numbers, which can’t be expressed in \frac p q , where p and q are integers (co-prime) and q ≠ 0 are known as a irrational numbers. For example: Pi (π), 3.57252….., √2, √5 etc…
NCERT SOLUTIONS FOR CLASS 10 MATHS :
We are providing complete Class 10 Maths Solution . You can get complete and updated solution from our website. Please click on your prefered chapter bellow from NCERT CLASS 10 MATHS and see the solutions. You can get complete ncert solutions for class 10 maths from here.
1 Chapter 1 Real Numbers |
2 Chapter 2 Polynomials |
3 Chapter 3 Pair of Linear Equations in Two Variables |
4 Chapter 4 Quadratic Equations |
5 Chapter 5 Arithmetic Progressions |
6 Chapter 6 Triangles |
7 Chapter 7 Co-ordinate Geometry |
8 Chapter 8 Introduction to trigonometry |
9 Chapter 9 Application of Trigonometry |
10 Chapter 10 Circle |
11 Chapter 11 Constructions |
12 Chapter 12 Area Related to Circles |
13 Chapter 13 Surface area and volumes |
14 Chapter 14 Statistics |
15 Chapter 15 Probability |
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