Maths : Ch 01 : Real Numbers

Class 10 Mathematics · NCERT Chapter 1 · Real Numbers

1. The Fundamental Theorem of Arithmetic (FTA)

Key terms

Prime number: a natural number greater than 1 with exactly two factors, 1 and itself (2, 3, 5, 7, 11 …). There are infinitely many primes.
Composite number: a natural number greater than 1 that has more than two factors (4, 6, 9, 12 …).
Factor tree: a diagram that keeps splitting a number into factors until only primes remain.

Theorem 1.1 – Fundamental Theorem of Arithmetic Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

What the theorem says (two parts)

Existence: every composite number can be written as a product of primes.
Uniqueness: there is only one way to do it. 2 × 3 × 5 × 7 and 3 × 5 × 7 × 2 count as the same factorisation.

Standard form

Write the primes in ascending order, p₁ ≤ p₂ ≤ … ≤ p_n, and combine repeats into powers. Then the factorisation is unique.

Example: 32760

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2³ × 3² × 5 × 7 × 13
Another: 123456789 = 3² × 3803 × 3607.

Making a factor tree

Write the number and split it into any two factors.
Keep splitting every factor that is not prime.
Stop when every branch ends in a prime.
Multiply the end primes, then write in index form.

Try: 140 = 2² × 5 × 7 | 156 = 2² × 3 × 13 | 3825 = 3² × 5² × 17 | 5005 = 5 × 7 × 11 × 13 | 7429 = 17 × 19 × 23

Application 1: checking whether a number can end in 0

RuleA number ends with the digit 0 only if it is divisible by 10, that is, only if its prime factorisation contains both 2 and 5. By uniqueness, if 5 is not in the factorisation, it can never appear.

Example 1: Can 4ⁿ end with the digit 0?

4ⁿ = (2²)ⁿ = 2²ⁿ. The only prime factor is 2. There is no factor of 5, so 4ⁿ never ends in 0 for any natural number n.

Practice: Can 6ⁿ end with 0?

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. There is no factor of 5, so it never ends in 0.

Application 2: showing a number is composite

MethodTake out a common factor. If the number becomes (common factor) × (a number greater than 1), it has more than two factors, so it is composite.

7 × 11 × 13 + 13

= 13 × (7 × 11 + 1) = 13 × 78. It has factors other than 1 and itself, so it is composite.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009. So it is composite.

2. HCF and LCM by Prime Factorisation

Term	Meaning	Prime factorisation rule
HCF (Highest Common Factor)	Largest number that divides all the given numbers	Product of the smallest power of each common prime factor
LCM (Least Common Multiple)	Smallest number that is a multiple of all the given numbers	Product of the greatest power of every prime factor that appears

Steps

Write each number as a product of prime powers.
List every prime that appears.
For HCF: use only primes found in all the numbers, each with its smallest power.
For LCM: use all primes, each with its greatest power.

Example 2: HCF and LCM of 6 and 20

6 = 2¹ × 3¹ | 20 = 2² × 5¹

Prime	In 6	In 20	HCF uses	LCM uses
2	2¹	2²	2¹	2²
3	3¹	–	–	3¹
5	–	5¹	–	5¹

HCF = 2 | LCM = 2² × 3 × 5 = 60

Product of two numbers For any two positive integers a and b:
HCF(a, b) × LCM(a, b) = a × b
So LCM = (a × b) ÷ HCF and HCF = (a × b) ÷ LCM. Check: 2 × 60 = 120 = 6 × 20 ✓

Example 3: HCF of 96 and 404, hence their LCM

96 = 2⁵ × 3 | 404 = 2² × 101
HCF = 2² = 4
LCM = (96 × 404) ÷ 4 = 38784 ÷ 4 = 9696

Three numbers

Example 4: HCF and LCM of 6, 72 and 120

6 = 2 × 3 | 72 = 2³ × 3² | 120 = 2³ × 3 × 5
HCF = 2¹ × 3¹ = 6 | LCM = 2³ × 3² × 5 = 360

Remark: only for two numbers! HCF × LCM = product of the numbers holds only for two numbers. For three numbers, 6 × 72 × 120 ≠ HCF × LCM.

Note to the reader: formulas for three numbers p, q, r LCM(p, q, r) = [ p · q · r · HCF(p, q, r) ] ÷ [ HCF(p, q) · HCF(q, r) · HCF(p, r) ]
HCF(p, q, r) = [ p · q · r · LCM(p, q, r) ] ÷ [ LCM(p, q) · LCM(q, r) · LCM(p, r) ]
Check with 6, 72, 120: HCF(6,72) = 6, HCF(72,120) = 24, HCF(6,120) = 6, so LCM = 51840 × 6 ÷ (6 × 24 × 6) = 360 ✓

More solved examples (Exercise 1.1 types)

Numbers	Prime factorisation	HCF	LCM
26, 91	2 × 13; 7 × 13	13	182
510, 92	2 × 3 × 5 × 17; 2² × 23	2	23460
336, 54	2⁴ × 3 × 7; 2 × 3³	6	3024
12, 15, 21	2² × 3; 3 × 5; 3 × 7	3	420
17, 23, 29	all primes	1	11339
8, 9, 25	2³; 3²; 5²	1	1800

Given HCF(306, 657) = 9, find the LCM

LCM = (306 × 657) ÷ 9 = 201042 ÷ 9 = 22338

Word problem: LCM in real life

When to use LCMQuestions about events that repeat together again (meeting at the start, bells ringing together, traffic lights changing) use LCM. Questions about the largest equal groups or pieces use HCF.

Sonia takes 18 min per round, Ravi takes 12 min. When do they meet again at the start?

18 = 2 × 3², 12 = 2² × 3. LCM = 2² × 3² = 36. They meet again after 36 minutes.

3. Revisiting Irrational Numbers

Definitions

Rational number: can be written as p/q, where p and q are integers and q ≠ 0.
Irrational number: cannot be written as p/q. Examples: √2, √3, √15, π, 2/3 is not irrational, but 0.10110111011110… is.
Rational and irrational numbers together make up the real numbers.
Coprime numbers: two integers whose only common factor is 1.

Theorem 1.2

StatementLet p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Idea of proof (this proof is not from the examination point of view): write a = p₁p₂…p_n, so a² = p₁²p₂²…p_n². By uniqueness in FTA, the only primes in a² are p₁, …, p_n. Since p divides a², p must be one of them, so p divides a.

Use: 2 divides a² ⇒ 2 divides a. 3 divides a² ⇒ 3 divides a. This theorem is the key step in every irrationality proof below.

Proof by contradiction: the plan

Assume the opposite: the number is rational.
Write it as a/b with a, b coprime, b ≠ 0.
Square and rearrange, then use Theorem 1.2 twice.
Show a and b share a common factor. This contradicts “coprime”.
So the assumption was wrong, and the number is irrational.

Theorem 1.3: √2 is irrational

Assume √2 is rational: √2 = a/b, with a, b coprime, b ≠ 0.
Then a = √2 b, so a² = 2b². So 2 divides a², hence 2 divides a (Theorem 1.2).
Let a = 2c. Then 2b² = 4c², so b² = 2c². So 2 divides b², hence 2 divides b.
Now 2 is a common factor of a and b. This contradicts that they are coprime.
Therefore √2 is irrational. ∎

Example 5: √3 is irrational

Same method with 3 in place of 2. Assume √3 = a/b (coprime). Then 3b² = a², so 3 divides a, so a = 3c. Then 3b² = 9c², so b² = 3c², so 3 divides b. So 3 divides both a and b, which contradicts that they are coprime. Hence √3 is irrational. ∎
The same proof works for √5, √7 and in general for √p, where p is any prime.

Sums, differences, products with irrationals

Facts (from Class 9) • Rational ± irrational = irrational
• Non-zero rational × irrational = irrational
• Non-zero rational ÷ irrational (or irrational ÷ non-zero rational) = irrational

CarefulThe sum, difference or product of two irrationals can be rational: √2 + (−√2) = 0 and √2 × √2 = 2. So always check what is being combined.

Example 6: 5 − √3 is irrational

Assume 5 − √3 = a/b (a, b coprime, b ≠ 0). Then √3 = 5 − a/b = (5b − a)/b, which is rational. But √3 is irrational. Contradiction, so 5 − √3 is irrational. ∎

Example 7: 3√2 is irrational

Assume 3√2 = a/b. Then √2 = a/(3b), which is rational. But √2 is irrational. Contradiction, so 3√2 is irrational. ∎

Exercise 1.2 types: same pattern

To prove	Step after assuming it equals a/b	Contradiction with
√5 irrational	5b² = a² ⇒ 5 \| a ⇒ 5 \| b	a, b coprime
3 + 2√5	√5 = (a − 3b)/(2b)	√5 is irrational
1/√2	√2 = b/a (invert)	√2 is irrational
7√5	√5 = a/(7b)	√5 is irrational
6 + √2	√2 = (a − 6b)/b	√2 is irrational

4. Quick Revision

Formula sheet

FTA: every composite number = product of primes, uniquely (order ignored)
HCF = product of smallest powers of common primes
LCM = product of greatest powers of all primes
HCF(a, b) × LCM(a, b) = a × b (two numbers only)
Theorem 1.2: prime p divides a² ⇒ p divides a
Irrational: √p (p prime), rational ± irrational, non-zero rational × irrational
Ends with 0 ⇔ prime factorisation contains both 2 and 5

Common mistakes

Using the “product = HCF × LCM” rule for three numbers.
Taking the largest power for HCF or the smallest for LCM. HCF is smallest of the common primes, and LCM is greatest of all primes.
Counting 1 as a prime number. 1 is neither prime nor composite.
Forgetting “a and b are coprime” at the start of a proof, so the contradiction is missed.
Writing “√2 = 1.414…” as proof of anything. 1.414 is only an approximation.
Assuming the sum of two irrationals is always irrational.

Did you know?

An equivalent version of the Fundamental Theorem of Arithmetic appears as Proposition 14 of Book IX in Euclid’s Elements. The first correct proof was given by Carl Friedrich Gauss (1777–1855), the “Prince of Mathematicians”, in Disquisitiones Arithmeticae.