CBSE Class 10 · Maths · board exam 2027
Chapter 1: Real Numbers — notes
The chapter’s 5 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.
1. Primes and the Fundamental Theorem of Arithmetic
First, two words we will use all through the chapter.
- • Prime: exactly two factors, 1 and itself: 2, 3, 5, 7, 11, 13, …
- • Composite: more than two factors: 4, 6, 8, 9, 12, …
- • 1: neither prime nor composite
- • Prime factorisation of 32760
- • 32760 ÷ 2 = 16380
- • 16380 ÷ 2 = 8190
- • 8190 ÷ 2 = 4095
- • 4095 ÷ 3 = 1365
- • 1365 ÷ 3 = 455
- • 455 ÷ 5 = 91
- • 91 ÷ 7 = 13
- • 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13
- • = 2³ × 3² × 5 × 7 × 13
- • Fundamental Theorem of Arithmetic
- • Every composite number can be written as a product of primes,
- • and this factorisation is unique, apart from the order of the primes.
- • Example: can 4ⁿ end with the digit 0?
- • Ends with 0 ⇒ divisible by 10
- • 10 = 2 × 5 ⇒ the prime 5 must be a factor
- • 4ⁿ = (2 × 2)ⁿ = 2²ⁿ
- • Only prime factor: 2, no 5
- • By uniqueness (FTA), 4ⁿ never ends with 0.
2. HCF and LCM by prime factorisation
We use prime factorisation to find the HCF and the LCM. HCF is the highest common factor, the biggest number that divides both numbers.
- • HCF: highest common factor
- • LCM: lowest common multiple
- • 6 = 2 × 3 = 2¹ × 3¹
- • 20 = 2 × 2 × 5 = 2² × 5¹
- • Common prime: 2
- • Smallest power of 2: 2¹
- • HCF(6, 20) = 2¹ = 2
- • Primes involved: 2, 3, 5
- • Greatest powers: 2², 3¹, 5¹
- • LCM(6, 20) = 2² × 3 × 5 = 4 × 3 × 5 = 60
- • HCF = product of the smallest power of each common prime factor
- • LCM = product of the greatest power of each prime factor involved
- • HCF × LCM = 2 × 60 = 120
- • 6 × 20 = 120
- • For two numbers: HCF(a, b) × LCM(a, b) = a × b
- • Example: HCF(26, 91) = 13, find LCM(26, 91)
- • LCM = (26 × 91)/13 = 2 × 91 = 182
- • Check: 26 = 2 × 13, 91 = 7 × 13
- • LCM = 2 × 7 × 13 = 182
- • Three numbers: 6, 72, 120
- • 6 = 2 × 3
- • 72 = 2³ × 3²
- • 120 = 2³ × 3 × 5
- • HCF = 2¹ × 3¹ = 6
- • LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360
- • HCF × LCM = 6 × 360 = 2160
- • 6 × 72 × 120 = 51840, not 2160
3. HCF or LCM? Word problems
In a word problem, the first question is always this.
- • Greatest, maximum, equal groups, largest piece ⇒ HCF
- • Least, smallest, together again ⇒ LCM
- • 36 apples, 60 bananas, shared equally: greatest number of children?
- • 36 = 2² × 3²
- • 60 = 2² × 3 × 5
- • HCF(36, 60) = 2² × 3 = 12
- • Apples each: 36 ÷ 12 = 3
- • Bananas each: 60 ÷ 12 = 5
- • Answer: 12 children, 3 apples and 5 bananas each
- • Rounds of 18 min and 12 min: when together again at the start?
- • 18 = 2 × 3²
- • 12 = 2² × 3
- • LCM(18, 12) = 2² × 3² = 4 × 9 = 36
- • Check: 36 ÷ 18 = 2, 36 ÷ 12 = 3
- • Answer: after 36 minutes
- • Greatest number dividing 85 and 72, remainders 1 and 2
- • 85 − 1 = 84, 72 − 2 = 70
- • 84 = 2² × 3 × 7
- • 70 = 2 × 5 × 7
- • HCF(84, 70) = 2 × 7 = 14
- • Check: 85 = 14 × 6 + 1, 72 = 14 × 5 + 2
- • Answer: 14
4. Irrational numbers: proving √2 is irrational
The second part of the chapter.
- • Rational: can be written as p/q, p and q integers, q ≠ 0
- • Irrational: cannot be written so: √2, √3, √5, π
- • Theorem: p prime, p divides a² ⇒ p divides a (a a positive integer)
- • Example: 3 divides 36 = 6², and 3 divides 6
- • a = p₁ × p₂ × … × pₙ
- • a² = (p₁ × p₁) × (p₂ × p₂) × … × (pₙ × pₙ)
- • Same primes as a, so p divides a² ⇒ p divides a
- • Only for a prime: 4 divides 36, but 4 does not divide 6
- • To prove: √2 is irrational
- • Assume, to the contrary, that √2 is rational.
- • √2 = p/q, p, q coprime integers, q ≠ 0
- • √2 q = p
- • Squaring both sides: 2q² = p²
- • 2 divides p², 2 is prime ⇒ 2 divides p … (i)
- • p = 2c, c an integer
- • 2q² = (2c)² = 4c²
- • q² = 2c²
- • 2 divides q² ⇒ 2 divides q … (ii)
- • From (i) and (ii): 2 is a common factor of p and q
- • This contradicts: p and q are coprime.
- • So our assumption is wrong. √2 is irrational. Hence proved.
5. Numbers like 5 − √3 and 3√2
Numbers made from a root, like 5 minus √3, or 3 √2.
- • rational ± irrational = irrational
- • non-zero rational × irrational = irrational
- • Prove: 5 − √3 is irrational (given √3 is irrational)
- • Assume, to the contrary, that 5 − √3 is rational.
- • 5 − √3 = a/b, a, b coprime integers, b ≠ 0
- • √3 = 5 − a/b
- • √3 = (5b − a)/b
- • a, b integers ⇒ (5b − a)/b is rational ⇒ √3 is rational
- • But √3 is irrational: a contradiction.
- • So 5 − √3 is irrational.
- • Show: 3√2 is irrational
- • Assume 3√2 = a/b, a, b coprime integers, b ≠ 0
- • √2 = a/3b
- • a/3b is rational ⇒ √2 rational: a contradiction
- • So 3√2 is irrational.
- • Write: assume → root alone → other side rational → contradiction
Summary
Very good! You learnt a lot today. First, the Fundamental Theorem of Arithmetic. Every composite number is a product of primes, in only one way, apart from the order. From it we saw that 4 to the power n can never end with zero. Then HCF and LCM by prime factorisation. Smallest powers of the common primes for the HCF, greatest powers of all the primes for the LCM. For two numbers, HCF into LCM is the product of the numbers. And in word problems, greatest equal groups means HCF, together again means LCM. Last, proof by contradiction. Root 2 is irrational, and so are numbers like 5 minus √3 and 3 √2. Write the proof of √2 once in your copy, without looking. Today's homework is on your screen. In the next chapter we will study polynomials. If anything was not clear, press the Ask button and ask me. Done, you worked very well today!
AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 1. Written from the textbook, checked twice.
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