Chapter at a Glance

Here's the whole of Real Numbers on one page. Use this the night before your exam.

The Number System

  • Real numbers RR = all rational + all irrational numbers.
  • Chain: N⊂W⊂Z⊂Q⊂RN \subset W \subset Z \subset Q \subset R.
  • Rational: can be written pq\dfrac{p}{q}, q≠0q \neq 0; decimal is terminating or non-terminating recurring.
  • Irrational: cannot be written pq\dfrac{p}{q}; decimal is non-terminating non-recurring (e.g. 2\sqrt{2}, π\pi).

Remember: (rational) + (irrational) = irrational; (non-zero rational) × (irrational) = irrational. But (irrational) + (irrational) and (irrational) × (irrational) can be either.

Key Definitions and Theorems

Euclid's Division Lemma

For positive integers a,ba, b: there exist unique q,rq, r with a=bq+r,0≤r<b.a = bq + r, \quad 0 \le r < b.

Euclid's Division Algorithm (for HCF)

Apply the lemma repeatedly; when the remainder becomes 0, the last divisor is the HCF.

Fundamental Theorem of Arithmetic

Every composite number is a unique product of primes (apart from order).

Irrationality theorem

For a prime pp: if p∣a2p \mid a^2 then p∣ap \mid a. (Engine of all irrationality proofs.)

Exam Note: State the relevant theorem before you use it — markers expect it.

Must-Know Formulae

HCF and LCM by prime factorisation

  • HCF = product of the lowest powers of common primes.
  • LCM = product of the highest powers of all primes.

The product relation (two numbers only)

HCF(a,b)×LCM(a,b)=a×b\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b

  • Hence LCM=a×bHCF\text{LCM} = \dfrac{a \times b}{\text{HCF}} and HCF=a×bLCM\text{HCF} = \dfrac{a \times b}{\text{LCM}}.
  • HCF always divides LCM (handy validity check).

Word-problem cues

  • 'Largest / maximum equal size' → HCF.
  • 'Next time together / smallest common' → LCM.
  • 'Leaves remainder rr' (HCF type) → subtract rr, then HCF.
  • 'Same remainder rr on division' (LCM type) → LCM +r+ r.

Decimal Expansion Rule

For pq\dfrac{p}{q} in lowest terms:

Denominator qq Decimal
q=2n×5mq = 2^n \times 5^m (only 2's and 5's) Terminating
any other prime present Non-terminating recurring
  • Number of decimal places (when terminating) =max⁡(n,m)= \max(n, m).
  • To convert: supply the missing 2's or 5's to make the denominator a power of 10.

Common Mistake: Simplify the fraction first — an unreduced denominator can fool you (e.g. 615=25\dfrac{6}{15} = \dfrac{2}{5} terminates).

Last-Minute Tips and Common Traps

  • n\sqrt{n} is irrational exactly when nn is not a perfect square.
  • A number ends in 0 ⇔ its factorisation has at least one 2 and one 5. So 2n2^n, 4n4^n, 6n6^n never end in 0.
  • The product relation HCF × LCM = a×ba \times b holds for two numbers only — never for three.
  • In irrationality proofs, always begin by assuming the fraction is in co-prime form; the contradiction comes from finding a shared factor.
  • 'Non-terminating' is not the same as 'irrational' — recurring decimals are rational.

Final Word: Real Numbers is high-scoring. Nail the p\sqrt{p} proof, the HCF–LCM relation, and the terminating-decimal rule, and you have most of the chapter's marks secured. All the best!