CBSE X Maths — Real Numbers: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit I: Number Systems

Real Numbers

This topic builds on your understanding of prime numbers to establish the Fundamental Theorem of Arithmetic, and uses it to prove that certain numbers can never be written as a fraction.

1. The Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

HCF × LCM = Product of the two numbers

Worked example — HCF and LCM by prime factorisation Find the HCF and LCM of 96 and 404.
96 = 2⁵ × 3 404 = 2² × 101
HCF = product of the smallest power of each common factor = 2² = 4
LCM = product of the greatest power of each factor involved = 2⁵ × 3 × 101 = 9696
Check: HCF × LCM = 4 × 9696 = 38784 = 96 × 404 ✓

Euclid’s Division Algorithm — an alternative way to find HCF To find HCF(455, 42), repeatedly apply a = bq + r:
455 = 42 × 10 + 35
42 = 35 × 1 + 7
35 = 7 × 5 + 0 → remainder is 0, so HCF = 7

2. Proving Irrationality

A number is irrational if it cannot be written as p/q where p, q are integers and q ≠ 0. Proofs of irrationality use proof by contradiction, together with the Fundamental Theorem of Arithmetic.

Worked example — proving √5 is irrational Assume, for contradiction, that √5 is rational. Then √5 = p/q, where p and q are coprime integers (no common factor other than 1) and q ≠ 0.
Squaring: 5q² = p², so p² is divisible by 5, which means p itself must be divisible by 5 (since 5 is prime). Write p = 5c.
Substituting: 5q² = 25c² → q² = 5c², so q² is also divisible by 5, meaning q is divisible by 5 too.
But then p and q share a common factor of 5 — contradicting our assumption that they are coprime. So √5 cannot be rational; √5 is irrational.

Worked example — proving 3 + 2√5 is irrational Assume, for contradiction, that 3 + 2√5 is rational, equal to some p/q.
Then 2√5 = p/q − 3, so √5 = (p − 3q)/(2q). Since p, q, and 3 are all integers, the right-hand side is a ratio of integers — meaning √5 would be rational.
This contradicts the fact that √5 is irrational (proved above). So 3 + 2√5 is irrational.

Common mistakes to watch for

Forgetting to state clearly that p and q are assumed coprime at the start of an irrationality proof — this assumption is essential to reach the contradiction.
Mixing up HCF and LCM in the “smallest power” / “greatest power” rule — HCF takes the smallest power of common factors only; LCM takes the greatest power of every factor that appears in either number.
Assuming 6ⁿ can end in 0 for some n — since 6ⁿ = 2ⁿ × 3ⁿ has no factor of 5, it can never end in 0, regardless of n.

Key Vocabulary Recap

Composite number — a number with more than two factors

Coprime — two numbers whose only common factor is 1

Irrational number — cannot be written as p/q for integers p, q≠0

Fundamental Theorem of Arithmetic — every composite number factorises uniquely into primes

Syllabus Reference

This note covers CBSE Class X Mathematics Unit I, Real Numbers: the Fundamental Theorem of Arithmetic and its applications, and proofs of irrationality of numbers such as √2, √3, √5, and 3 + 2√5.

Practice questions

CBSE Class X Mathematics · Unit I: Real Numbers — Practice Question Bank (no answers)

Real Numbers — Practice

Covers: prime factorisation, HCF & LCM, Euclid’s division algorithm, and proofs of irrationality. 10 questions.

Find the HCF and LCM of 96 and 404 using prime factorisation, and verify that HCF × LCM equals the product of the two numbers.

Find the HCF and LCM of 6 and 20 by prime factorisation, and verify the relationship between them.

Prove that √5 is irrational.

Show that 3 + 2√5 is irrational.

Find the HCF and LCM of 12 and 18 using prime factorisation.

If the HCF of two numbers a and b is 5, their LCM is 150, and a = 25, find b.

Explain why the product of a non-zero rational number and an irrational number is always irrational.

Find the prime factorisation of 1000 and 675, and hence find their HCF.

Use Euclid’s division algorithm to find the HCF of 455 and 42.

Q10

Check whether 6ⁿ can end with the digit 0 for any natural number n. Justify your answer.

More Practice

Q11

State Euclid’s Division Lemma.

Q12

Find the HCF of 4052 and 12576 using Euclid’s division algorithm.

Q13

Prove that √3 is irrational.

Q14

Show that 5 − √3 is irrational.

Q15

Find the HCF and LCM of 26 and 91 using prime factorisation.

Q16

Three bells ring at intervals of 6, 8, and 12 minutes. If they all ring together at 6:00 am, find when they will next ring together.

Q17

Two positive integers a and b are written as a = x³y² and b = xy³, where x and y are prime numbers. Find HCF(a,b) and LCM(a,b).

Q18

Explain why 7 × 11 × 13 + 13 is a composite number.

Q19

Find the largest number which divides 615 and 963, leaving remainder 6 in each case.

Q20

Prove that √2 + √3 is irrational.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit I: Real Numbers — Solutions

Real Numbers — Practice: Solutions

HCF and LCM of 96 and 404.

Prime factorise both numbers

96 = 2⁵×3 404 = 2²×101

HCF = smallest power of common factors; LCM = greatest power of all factors

HCF = 2² = 4 LCM = 2⁵×3×101 = 9696

✓HCF = 4, LCM = 9696 (check: 4×9696 = 38784 = 96×404)

HCF and LCM of 6 and 20.

Prime factorise

6 = 2×3 20 = 2²×5

✓HCF = 2, LCM = 2²×3×5 = 60 (check: 2×60 = 120 = 6×20)

Prove √5 is irrational.

✓Assume √5 = p/q with p, q coprime integers, q≠0. Squaring: 5q²=p², so 5|p², hence 5|p. Write p=5c: 5q²=25c² → q²=5c², so 5|q too — contradicting p,q coprime. So √5 is irrational.

Show 3+2√5 is irrational.

✓Assume 3+2√5 = p/q (rational). Then √5 = (p−3q)/(2q), a ratio of integers, making √5 rational — contradicting Q3. So 3+2√5 is irrational.

HCF and LCM of 12 and 18.

Prime factorise

12 = 2²×3 18 = 2×3²

✓HCF = 2×3 = 6, LCM = 2²×3² = 36

HCF=5, LCM=150, a=25. Find b.

HCF × LCM = a × b

5 × 150 = 25 × b

Solve for b

750 ÷ 25

✓b = 30

Why is rational × irrational always irrational (for non-zero rational)?

✓If a non-zero rational r times an irrational x were rational, say equal to s, then x = s/r would be a ratio of two rational numbers — hence rational, contradicting that x is irrational. So the product must be irrational.

Prime factorise 1000 and 675; find HCF.

Factorise both

1000 = 2³×5³ 675 = 3³×5²

✓HCF = 5² = 25

HCF(455, 42) by Euclid’s algorithm.

Apply a = bq + r repeatedly

455=42×10+35; 42=35×1+7; 35=7×5+0

✓HCF = 7 (the last non-zero remainder)

Q10

Can 6ⁿ end in 0?

✓No. If 6ⁿ ended in 0, it would be divisible by 10 = 2×5, so 5 would have to appear in its prime factorisation. But 6ⁿ = 2ⁿ×3ⁿ, which has no factor of 5 at all — so 6ⁿ can never end in 0, for any natural number n.

More Practice — Solutions

Q11

State Euclid’s Division Lemma.

✓For any two positive integers a and b, there exist unique whole numbers q and r such that a = bq + r, where 0 ≤ r < b.

Q12

HCF(4052, 12576) by Euclid’s algorithm.

Apply a=bq+r repeatedly

12576=4052×3+420; 4052=420×9+272; 420=272×1+148; 272=148×1+124; 148=124×1+24; 124=24×5+4; 24=4×6+0

✓HCF = 4

Q13

Prove √3 is irrational.

✓Assume √3=p/q with p,q coprime, q≠0. Squaring: 3q²=p², so 3|p², hence 3|p. Write p=3c: 3q²=9c²→q²=3c², so 3|q too — contradicting p,q coprime. So √3 is irrational.

Q14

Show 5−√3 is irrational.

✓Assume 5−√3=p/q (rational). Then √3=5−p/q=(5q−p)/q, a ratio of integers, making √3 rational — contradicting Q13. So 5−√3 is irrational.

Q15

HCF and LCM of 26 and 91.

Prime factorise

26=2×13, 91=7×13

✓HCF = 13, LCM = 2×7×13 = 182

Q16

Bells at 6,8,12 min. Find next together time.

Find LCM(6,8,12)

6=2×3, 8=2³, 12=2²×3 → LCM=2³×3=24

✓They will next ring together after 24 minutes, at 6:24 am

Q17

a=x³y², b=xy³. Find HCF, LCM.

✓HCF = xy² (smallest power of each), LCM = x³y³ (greatest power of each)

Q18

Why is 7×11×13+13 composite?

Factor out 13

7×11×13+13 = 13×(7×11+1) = 13×78

✓Since it equals 13×78, it has factors other than 1 and itself — it is composite

Q19

Largest number dividing 615,963 leaving remainder 6.

Subtract the remainder first: find HCF(609, 957)

957=609×1+348; 609=348×1+261; 348=261×1+87; 261=87×3+0

✓The required number is 87

Q20

Prove √2+√3 is irrational.

✓Assume √2+√3=r (rational). Squaring: 2+3+2√6=r² → √6=(r²−5)/2, a ratio of integers — making √6 rational. But √6 is irrational (provable the same way as √2, √3, √5: if √6=p/q in lowest terms, 6q²=p² forces p and q to share a factor of 2 and 3, contradicting coprimality). This contradiction shows √2+√3 must be irrational.

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