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Class 10

CBSE Class 10 Maths Chapter 1 Notes: Real Numbers

Complete local notes for CBSE Class 10 Maths Real Numbers: Euclid division lemma, HCF, Fundamental Theorem of Arithmetic, irrationality proofs, decimal expansions, examples, practice questions, and board exam tips.

Updated 23 May 20268 min read

Chapter Overview

Real Numbers is a foundation chapter for Class 10 Maths. It teaches how numbers are organised, how HCF is found using repeated division, why prime factorisation is unique, and how rational numbers behave in decimal form.

For board exams, this chapter is usually asked through:

  • HCF using Euclid division algorithm
  • LCM and HCF using prime factorisation
  • Proof that a number is irrational
  • Terminating or non-terminating recurring decimal questions
  • Word problems based on HCF or LCM

The chapter is short, but it needs neat step-by-step writing.

1. Euclid Division Lemma

Euclid division lemma says:

If a and b are two positive integers, where a > b, then there exist unique integers q and r such that:

a = bq + r, where 0 <= r < b

Here:

SymbolMeaning
aDividend
bDivisor
qQuotient
rRemainder

Example

Divide 135 by 28.

135 = 28 x 4 + 23

Here, a = 135, b = 28, q = 4, and r = 23.

Since 0 <= 23 < 28, this follows Euclid division lemma.

2. Euclid Division Algorithm for HCF

Euclid division algorithm is used to find the HCF of two positive integers.

Steps

1. Divide the larger number by the smaller number. 2. If the remainder is zero, the divisor is the HCF. 3. If the remainder is not zero, divide the previous divisor by the remainder. 4. Repeat until the remainder becomes zero. 5. The last non-zero remainder is the HCF.

Solved Example: Find HCF of 135 and 225

Start with the larger number:

225 = 135 x 1 + 90 135 = 90 x 1 + 45 90 = 45 x 2 + 0

The last non-zero remainder is 45.

So, HCF of 135 and 225 = 45.

Board Writing Tip

Do not write only the final HCF. Show every division line. Marks are awarded for the method.

3. Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic says:

Every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of the prime factors.

Example

140 = 2 x 2 x 5 x 7 140 = 2^2 x 5 x 7

You may write the factors in a different order, but the prime factors themselves remain the same.

4. HCF and LCM by Prime Factorisation

Rule for HCF

For HCF, take the common prime factors with the smallest powers.

Rule for LCM

For LCM, take all prime factors with the highest powers.

Solved Example: Find HCF and LCM of 72 and 120

Prime factorisation:

72 = 2^3 x 3^2 120 = 2^3 x 3 x 5

HCF:

HCF = 2^3 x 3 = 24

LCM:

LCM = 2^3 x 3^2 x 5 = 360

Check:

HCF x LCM = 24 x 360 = 8640 72 x 120 = 8640

So, HCF x LCM = product of the two numbers.

5. Important Formula

For two positive integers:

HCF x LCM = Product of the two numbers

This formula works for two numbers only in the direct form used above.

Example

If HCF of two numbers is 12 and their product is 6336, find LCM.

HCF x LCM = Product 12 x LCM = 6336 LCM = 6336 / 12 LCM = 528

6. Rational and Irrational Numbers

Rational Number

A number that can be written in the form p/q, where p and q are integers and q != 0, is called a rational number.

Examples:

  • 3/5
  • -7
  • 0.25
  • 2.333...

Irrational Number

A number that cannot be written in the form p/q is called an irrational number.

Examples:

  • sqrt(2)
  • sqrt(3)
  • pi

7. Standard Irrationality Proof Pattern

Many board questions ask you to prove that a number is irrational. The method is usually proof by contradiction.

Format

1. Assume the number is rational. 2. Write it as p/q, where p and q are co-prime. 3. Square both sides if needed. 4. Show that both p and q have a common factor. 5. This contradicts the assumption that p and q are co-prime. 6. Therefore, the number is irrational.

Solved Example: Prove that sqrt(2) is irrational

Assume sqrt(2) is rational.

Then:

sqrt(2) = p/q

where p and q are co-prime integers and q != 0.

Squaring both sides:

2 = p^2/q^2 p^2 = 2q^2

So, p^2 is divisible by 2. Therefore, p is also divisible by 2.

Let:

p = 2m

Substitute in p^2 = 2q^2:

(2m)^2 = 2q^2 4m^2 = 2q^2 q^2 = 2m^2

So, q^2 is divisible by 2. Therefore, q is also divisible by 2.

This means both p and q are divisible by 2, which contradicts the assumption that they are co-prime.

Therefore, sqrt(2) is irrational.

8. Decimal Expansion of Rational Numbers

For a rational number p/q, where p and q are co-prime:

Terminating Decimal

The decimal expansion is terminating if the prime factorisation of q contains only the factors:

2, 5, or both 2 and 5

Examples:

7/8, because 8 = 2^3 13/20, because 20 = 2^2 x 5

Non-Terminating Recurring Decimal

The decimal expansion is non-terminating recurring if the denominator has any prime factor other than 2 or 5.

Examples:

5/6, because 6 = 2 x 3 7/11, because 11 is not 2 or 5

Important Step

Always reduce the fraction to lowest form first. Then check the denominator.

9. Solved Decimal Questions

Example 1: Will 13/1250 terminate?

First check denominator:

1250 = 2 x 5^4

Only prime factors are 2 and 5.

So, 13/1250 has a terminating decimal expansion.

Example 2: Will 17/30 terminate?

30 = 2 x 3 x 5

The denominator has 3 as a prime factor.

So, 17/30 has a non-terminating recurring decimal expansion.

10. Important Question Types

Question TypeHow to Answer
HCF by Euclid algorithmShow all division steps until remainder becomes 0
HCF and LCMPrime factorise both numbers clearly
Irrationality proofUse assumption, contradiction, conclusion
Decimal expansionReduce fraction, factorise denominator, decide
Word problemIdentify whether equal grouping means HCF or repeated cycle means LCM

11. Practice Questions

1. Find the HCF of 867 and 255 using Euclid division algorithm. 2. Find HCF and LCM of 96 and 404 by prime factorisation. 3. Prove that sqrt(3) is irrational. 4. Check whether 43/125 has terminating decimal expansion. 5. Check whether 17/48 has terminating decimal expansion. 6. If HCF of two numbers is 18 and their product is 12960, find their LCM. 7. Find the least number divisible by 12, 18, and 30. 8. Explain why 7 x 11 x 13 + 13 is a composite number.

12. Quick Answers

1. HCF of 867 and 255 = 51. 2. 96 = 2^5 x 3, 404 = 2^2 x 101; HCF = 4, LCM = 9696. 3. Use the same contradiction pattern as sqrt(2). 4. 125 = 5^3, so it terminates. 5. 48 = 2^4 x 3, so it is non-terminating recurring. 6. LCM = 12960 / 18 = 720. 7. LCM of 12, 18, and 30 = 180. 8. 7 x 11 x 13 + 13 = 13(7 x 11 + 1) = 13 x 78, so it is composite.

Common Mistakes to Avoid

  • Forgetting the condition 0 <= r < b in Euclid division lemma.
  • Stopping Euclid algorithm before the remainder becomes zero.
  • Writing prime factorisation with composite factors like 4 or 6.
  • Checking decimal expansion without reducing the fraction to lowest form.
  • In irrationality proofs, saying "therefore irrational" without showing contradiction.

Final Revision Checklist

  • I can state Euclid division lemma correctly.
  • I can find HCF using repeated division.
  • I can find HCF and LCM by prime factorisation.
  • I can prove sqrt(2) and sqrt(3) are irrational.
  • I can decide whether a rational number has terminating decimal expansion.
  • I can solve word problems using HCF or LCM.

Internal Links

  • [CBSE Class 10 Maths guide](/school/cbse/class/10/maths)
  • [CBSE Class 10 Science guide](/school/cbse/class/10/science)
  • [Practice aptitude questions](/aptitude)

Frequently Asked Questions

The most important topics are Euclid division lemma, HCF by Euclid algorithm, Fundamental Theorem of Arithmetic, proof of irrationality, and decimal expansion of rational numbers.

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