Class 10 Chapter 1 Maths Real Numbers

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real numbers

Table of Contents

Class 10 – Chapter 1: Real Numbers in Mathematics. This chapter covers important topics such as Euclid’s Division Lemma, fundamental theorem of arithmetic, and various properties of irrational numbers, rational numbers, and real numbers. Let’s break it down step by step.

Introduction

In Chapter 1 of Class 10 Mathematics, Real Numbers, we deal with the different types of numbers such as:

  • Natural numbers (N\mathbb{N}N)
  • Whole numbers (W\mathbb{W}W)
  • Integers (Z\mathbb{Z}Z)
  • Rational numbers (Q\mathbb{Q}Q)
  • Irrational numbers
  • Real numbers (R\mathbb{R}R)

We will also study key theorems and principles such as Euclid’s Division Lemma and the Fundamental Theorem of Arithmetic.

Exercise 1.1: Basic Concepts and Operations on Real Numbers

Q1. Find whether the following numbers are rational or irrational:

  • 57\frac{5}{7}75​
  • 25\sqrt{25}25​
  • 40\frac{4}{0}04​
  • 2\sqrt{2}2​

Solution:

  1. 57\frac{5}{7}75​ is a rational number because it can be expressed as the ratio of two integers.
  2. 25=5\sqrt{25} = 525​=5, which is a rational number.
  3. 40\frac{4}{0}04​ is undefined because division by zero is not possible.
  4. 2\sqrt{2}2​ is an irrational number because it cannot be expressed as a ratio of two integers and has a non-terminating, non-repeating decimal expansion.

Q2. Express the following numbers as the sum of a rational and an irrational number:

  • 5+3\sqrt{5} + 35​+3
  • π+4\pi + 4π+4

Solution:

  1. 5+3\sqrt{5} + 35​+3 is already expressed as the sum of a rational number 333 and an irrational number 5\sqrt{5}5​.
  2. π+4\pi + 4π+4 is the sum of a rational number 444 and an irrational number π\piπ.

Q3. Prove that the sum of a rational number and an irrational number is always irrational.

Solution:

Let rrr be a rational number and iii be an irrational number. Suppose that r+ir + ir+i is a rational number. Then, we can write:r+i=q(where q is a rational number)r + i = q \quad \text{(where \(q\) is a rational number)}r+i=q(where q is a rational number)

Now, subtract rrr from both sides:i=q−ri = q – ri=q−r

Since both qqq and rrr are rational, their difference q−rq – rq−r must also be rational. But this contradicts the assumption that iii is irrational. Therefore, the sum of a rational number and an irrational number is always irrational.

Exercise 1.2: Euclid’s Division Lemma

Q1. Use Euclid’s Division Lemma to find the HCF of 56 and 96.

Solution:

Euclid’s Division Lemma states that for any two positive integers aaa and bbb, there exist unique integers qqq (quotient) and rrr (remainder) such that:a=bq+rwhere0≤r<ba = bq + r \quad \text{where} \quad 0 \leq r < ba=bq+rwhere0≤r<b

We use this method to find the HCF of 56 and 96.

  1. First, divide 96 by 56:

96=56×1+4096 = 56 \times 1 + 4096=56×1+40

  1. Now divide 56 by 40:

56=40×1+1656 = 40 \times 1 + 1656=40×1+16

  1. Next, divide 40 by 16:

40=16×2+840 = 16 \times 2 + 840=16×2+8

  1. Now divide 16 by 8:

16=8×2+016 = 8 \times 2 + 016=8×2+0

Since the remainder is 0, the HCF is the last non-zero remainder, which is 888.

Answer: The HCF of 56 and 96 is 888.

Q2. Apply Euclid’s Division Lemma to find the HCF of 72 and 120.

Solution:

  1. Divide 120 by 72:

120=72×1+48120 = 72 \times 1 + 48120=72×1+48

  1. Now divide 72 by 48:

72=48×1+2472 = 48 \times 1 + 2472=48×1+24

  1. Now divide 48 by 24:

48=24×2+048 = 24 \times 2 + 048=24×2+0

Since the remainder is 0, the HCF is the last non-zero remainder, which is 242424.

Answer: The HCF of 72 and 120 is 242424.

Exercise 1.3: Fundamental Theorem of Arithmetic

Q1. Find the prime factorization of 72 and express it as the product of prime numbers.

Solution:

To find the prime factorization of 72, start by dividing it by the smallest prime number:72÷2=3672 \div 2 = 3672÷2=36 36÷2=1836 \div 2 = 1836÷2=18 18÷2=918 \div 2 = 918÷2=9 9÷3=39 \div 3 = 39÷3=3 3÷3=13 \div 3 = 13÷3=1

Thus, the prime factorization of 72 is:72=23×3272 = 2^3 \times 3^272=23×32

Answer: The prime factorization of 72 is 23×322^3 \times 3^223×32.

Q2. Find the LCM and HCF of 48 and 180 using their prime factorizations.

Solution:

First, find the prime factorizations of 48 and 180:

  • Prime factorization of 48:

48=24×348 = 2^4 \times 348=24×3

  • Prime factorization of 180:

180=22×32×5180 = 2^2 \times 3^2 \times 5180=22×32×5

Now, to find the HCF, take the lowest powers of the common prime factors:HCF=22×3=12\text{HCF} = 2^2 \times 3 = 12HCF=22×3=12

To find the LCM, take the highest powers of all prime factors:LCM=24×32×5=720\text{LCM} = 2^4 \times 3^2 \times 5 = 720LCM=24×32×5=720

Answer: The HCF of 48 and 180 is 121212, and the LCM is 720720720.

Exercise 1.4: Irrational Numbers

Q1. Show that 7\sqrt{7}7​ is an irrational number.

Solution:

Assume that 7\sqrt{7}7​ is a rational number. Then, we can express it as:7=pq\sqrt{7} = \frac{p}{q}7​=qp​

where ppp and qqq are coprime integers (i.e., gcd⁡(p,q)=1\gcd(p, q) = 1gcd(p,q)=1). Now square both sides:7=p2q27 = \frac{p^2}{q^2}7=q2p2​ p2=7q2p^2 = 7q^2p2=7q2

This implies that p2p^2p2 is divisible by 7, so ppp must also be divisible by 7. Let p=7kp = 7kp=7k, where kkk is an integer. Substituting this into the equation:(7k)2=7q2(7k)^2 = 7q^2(7k)2=7q2 49k2=7q249k^2 = 7q^249k2=7q2 7k2=q27k^2 = q^27k2=q2

This shows that q2q^2q2 is also divisible by 7, which means qqq is divisible by 7. But this contradicts the assumption that ppp and qqq are coprime. Therefore, 7\sqrt{7}7​ is irrational.

Answer: 7\sqrt{7}7​ is irrational.

Additional Questions for Chapter 1: Real Numbers

1. Find the HCF of 84 and 126 using Euclid’s Division Lemma.

Solution: Use Euclid’s Division Lemma to find the HCF of 84 and 126.

  1. Divide 126 by 84:126=84×1+42126 = 84 \times 1 + 42126=84×1+42
  2. Divide 84 by 42:84=42×2+084 = 42 \times 2 + 084=42×2+0

Since the remainder is 0, the HCF is the last non-zero remainder, which is 42.

2. Use Euclid’s Division Lemma to find the HCF of 56 and 72.

Solution:

  1. Divide 72 by 56:72=56×1+1672 = 56 \times 1 + 1672=56×1+16
  2. Divide 56 by 16:56=16×3+856 = 16 \times 3 + 856=16×3+8
  3. Divide 16 by 8:16=8×2+016 = 8 \times 2 + 016=8×2+0

The remainder is 0, so the HCF is 8.

3. Show that 3\sqrt{3}3​ is an irrational number.

Solution:

Assume 3\sqrt{3}3​ is a rational number, and it can be expressed as:3=pq\sqrt{3} = \frac{p}{q}3​=qp​

where ppp and qqq are coprime integers (i.e., gcd⁡(p,q)=1\gcd(p, q) = 1gcd(p,q)=1). Squaring both sides:3=p2q23 = \frac{p^2}{q^2}3=q2p2​ p2=3q2p^2 = 3q^2p2=3q2

This implies p2p^2p2 is divisible by 3, so ppp must also be divisible by 3. Let p=3kp = 3kp=3k, where kkk is an integer.

Substitute p=3kp = 3kp=3k into the equation:(3k)2=3q2(3k)^2 = 3q^2(3k)2=3q2 9k2=3q29k^2 = 3q^29k2=3q2 3k2=q23k^2 = q^23k2=q2

This shows that q2q^2q2 is divisible by 3, so qqq must also be divisible by 3. But this contradicts the assumption that ppp and qqq are coprime. Hence, 3\sqrt{3}3​ is irrational.

4. Express 0.626262… as a rational number.

Solution:

Let x=0.626262…x = 0.626262…x=0.626262…. This is a repeating decimal with a repeating block “62”.

Multiply both sides by 100 to shift the decimal point:100x=62.626262…100x = 62.626262…100x=62.626262…

Now subtract the original equation from this equation:100x−x=62.626262…−0.626262…100x – x = 62.626262… – 0.626262…100x−x=62.626262…−0.626262… 99x=6299x = 6299x=62 x=6299x = \frac{62}{99}x=9962​

So, 0.626262…=62990.626262… = \frac{62}{99}0.626262…=9962​, which is a rational number.

5. Find the LCM and HCF of 24 and 36 using their prime factorizations.

Solution:

  • Prime factorization of 24:

24=23×324 = 2^3 \times 324=23×3

  • Prime factorization of 36:

36=22×3236 = 2^2 \times 3^236=22×32

Now, for HCF, take the lowest powers of the common prime factors:HCF=22×3=12\text{HCF} = 2^2 \times 3 = 12HCF=22×3=12

For LCM, take the highest powers of all prime factors:LCM=23×32=72\text{LCM} = 2^3 \times 3^2 = 72LCM=23×32=72

6. Prove that 12\frac{1}{\sqrt{2}}2​1​ is an irrational number.

Solution:

Assume that 12\frac{1}{\sqrt{2}}2​1​ is a rational number. Then it can be expressed as:12=pq\frac{1}{\sqrt{2}} = \frac{p}{q}2​1​=qp​

where ppp and qqq are coprime integers.

Squaring both sides:12=p2q2\frac{1}{2} = \frac{p^2}{q^2}21​=q2p2​ p2=q22p^2 = \frac{q^2}{2}p2=2q2​

This implies that p2p^2p2 is an even number, so ppp must also be even. Let p=2kp = 2kp=2k, where kkk is an integer.

Substitute p=2kp = 2kp=2k into the equation:(2k)2=q22(2k)^2 = \frac{q^2}{2}(2k)2=2q2​ 4k2=q224k^2 = \frac{q^2}{2}4k2=2q2​ q2=8k2q^2 = 8k^2q2=8k2

This shows that q2q^2q2 is divisible by 8, so qqq is divisible by 2. But this contradicts the assumption that ppp and qqq are coprime. Hence, 12\frac{1}{\sqrt{2}}2​1​ is irrational.

7. Find the prime factorization of 180 and express it as a product of prime numbers.

Solution:

To find the prime factorization of 180:180÷2=90180 \div 2 = 90180÷2=90 90÷2=4590 \div 2 = 4590÷2=45 45÷3=1545 \div 3 = 1545÷3=15 15÷3=515 \div 3 = 515÷3=5 5÷5=15 \div 5 = 15÷5=1

Thus, the prime factorization of 180 is:180=22×32×5180 = 2^2 \times 3^2 \times 5180=22×32×5

8. Prove that the product of a rational number and an irrational number is always irrational.

Solution:

Let rrr be a rational number and iii be an irrational number. Suppose the product r×ir \times ir×i is rational. Then:r×i=q(where q is a rational number)r \times i = q \quad \text{(where \(q\) is a rational number)}r×i=q(where q is a rational number)

Now, divide both sides by rrr:i=qri = \frac{q}{r}i=rq​

Since rrr is non-zero, and qqq and rrr are rational, qr\frac{q}{r}rq​ is rational. But this contradicts the assumption that iii is irrational. Therefore, the product of a rational number and an irrational number is always irrational.

9. Find the HCF of 45 and 75 using the prime factorization method.

Solution:

Prime factorization of 45:45=32×545 = 3^2 \times 545=32×5

Prime factorization of 75:75=3×5275 = 3 \times 5^275=3×52

The HCF is the product of the lowest powers of common prime factors:HCF=31×51=15\text{HCF} = 3^1 \times 5^1 = 15HCF=31×51=15

10. Express 5+3\sqrt{5} + \sqrt{3}5​+3​ as a sum of a rational and an irrational number.

Solution:

5\sqrt{5}5​ and 3\sqrt{3}3​ are irrational numbers. Therefore, the expression 5+3\sqrt{5} + \sqrt{3}5​+3​ is already written as a sum of an irrational number (5+3\sqrt{5} + \sqrt{3}5​+3​) and a rational number 000.

Answer: 5+3\sqrt{5} + \sqrt{3}5​+3​ is the sum of the irrational numbers 5\sqrt{5}5​ and 3\sqrt{3}3​, with a rational number 0.

Conclusion

These additional questions will further help in deepening your understanding of real numbers, Euclid’s Division Lemma, and the Fundamental Theorem of Arithmetic. By solving these problems, you’ll not only become more confident with concepts like HCF, LCM, irrational numbers, and prime factorization but will also strengthen your problem-solving skills for upcoming exams.

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