Class 10 Maths Real Numbers: Prime Factorisation, HCF, LCM and Irrational Numbers (Notes + Solved Examples)

Real Numbers is the first chapter of Class 10 Maths and one of the easiest to score full marks in. These notes follow the NCERT approach used by both CBSE and BSEB, with solved examples for every question type.

1. The Fundamental Theorem of Arithmetic#

Example: 360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5

Use the factor tree or repeated division by the smallest prime:

So 360 = 2³ × 3² × 5.

2. HCF and LCM by prime factorisation#

• HCF = product of the smallest power of each common prime factor.
• LCM = product of the greatest power of each prime factor present.

Example: Find the HCF and LCM of 96 and 404.

96 = 2⁵ × 3 and 404 = 2² × 101

• HCF = 2² = 4
• LCM = 2⁵ × 3 × 101 = 9696

Check with the key relation (two numbers only):

4 × 9696 = 38784 and 96 × 404 = 38784 ✔

Three numbers example: HCF and LCM of 12, 15 and 21.

12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7 → HCF = 3, LCM = 2² × 3 × 5 × 7 = 420.

3. Word problems on HCF and LCM#

HCF type ("largest", "maximum", "greatest that divides"):

An army contingent of 616 members marches behind a band of 32 members. Both groups must march in the same number of columns. What is the maximum number of columns?

HCF(616, 32): 616 = 2³ × 7 × 11, 32 = 2⁵ → HCF = 2³ = 8 columns.

LCM type ("smallest", "together again", "at the same time"):

Two bells ring every 18 minutes and 24 minutes. If they ring together at 8:00 a.m., when will they ring together next?

LCM(18, 24): 18 = 2 × 3², 24 = 2³ × 3 → LCM = 2³ × 3² = 72 minutes → 9:12 a.m.

4. Irrational numbers#

A number is irrational if it cannot be written as p/q (p, q integers, q ≠ 0). Examples: √2, √3, √5, π.

Key theorem: if a prime p divides a², then p divides a.

Proof that √2 is irrational (by contradiction)#

1. Assume √2 is rational, so √2 = a/b where a and b are co-prime integers (no common factor other than 1) and b ≠ 0.
2. Squaring: 2 = a²/b², so a² = 2b². Hence 2 divides a², so 2 divides a.
3. Let a = 2c. Then 4c² = 2b², so b² = 2c². Hence 2 divides b², so 2 divides b.
4. Now 2 divides both a and b, which contradicts that they are co-prime.
5. So our assumption was wrong: √2 is irrational.

The same steps prove √3 and √5 irrational (replace 2 with 3 or 5).

Sums and products with irrational numbers#

Prove that 5 − √3 is irrational.

Assume 5 − √3 = r, a rational number. Then √3 = 5 − r. Since 5 and r are rational, 5 − r is rational, so √3 would be rational: a contradiction. Hence 5 − √3 is irrational.

5. Quick practice#

1. Express 3825 as a product of prime factors.
2. Find the LCM and HCF of 6, 72 and 120.
3. Check whether 6ⁿ can end with the digit 0 for any natural number n.
4. Prove that 3 + 2√5 is irrational.

Answers:

1. 3825 = 3² × 5² × 17
2. HCF = 6, LCM = 360
3. No. For a number to end in 0 its prime factorisation must contain both 2 and 5. 6ⁿ = 2ⁿ × 3ⁿ has no factor 5.
4. Assume it is rational r. Then √5 = (r − 3)/2 is rational, a contradiction.