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Ch 1Real Numbers

Number Systems — 6 marks

Chapter 1: Real Numbers

This chapter deals with the Fundamental Theorem of Arithmetic, finding HCF and LCM by prime factorisation, and proving that certain numbers are irrational. In the rationalised syllabus, only these topics are examined.

1. The Fundamental Theorem of Arithmetic (FTA)

Statement: 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.

  • Example: 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7. No other set of primes can multiply to give 140.
  • A prime number has exactly two factors: 1 and itself. A composite number has more than two factors. The number 1 is neither prime nor composite.
  • The prime factorisation is usually written with primes in ascending order and repeated primes as powers.

Showing a number is composite (board favourite)

To show an expression like 7 × 11 × 13 + 13 is composite, take out the common factor:

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78 = 13 × 78. Since it has a factor other than 1 and itself, it is composite.

Can a number end with the digit 0 (or 5)?

A number ends with digit 0 only if it is divisible by 10 = 2 × 5, i.e. both 2 and 5 must appear in its prime factorisation.

  • 6n = (2 × 3)n = 2n × 3n — the prime 5 never occurs, so 6n can never end with 0 for any natural number n.
  • Similarly 4n = 22n contains no 5, so 4n never ends with 0.

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 involved in the numbers.

Worked example: Find HCF and LCM of 96 and 404.

  1. 96 = 25 × 3 and 404 = 22 × 101.
  2. Common prime: 2. Smallest power = 22. So HCF = 4.
  3. LCM = 25 × 3 × 101 = 9696.

Relation between HCF and LCM

For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b

Exam tip: This relation is valid for two numbers only. For three numbers, HCF × LCM ≠ product of the three numbers in general — a common trick question. Also remember: LCM is always a multiple of HCF, so if HCF does not divide the given "LCM" exactly, such a pair of numbers cannot exist.

GivenTo findMethod
Two numbersHCF, LCMPrime factorisation, then apply the two rules above
HCF and both numbersLCMLCM = (a × b) / HCF
HCF, LCM, one numberOther numberOther number = (HCF × LCM) / given number

Standard word-problem patterns

  • LCM type ("again together / minimum quantity"): bells tolling together, traffic lights changing together, runners meeting at the start point, least number of items — find the LCM of the given intervals or sizes. If the answer must leave remainder r, required number = LCM + r.
  • HCF type ("maximum / largest / greatest"): greatest capacity of a container measuring quantities exactly, longest tape measuring dimensions exactly, maximum number of items in equal groups, largest number dividing a and b leaving remainders r1 and r2 — find HCF (after subtracting the remainders, i.e. HCF of a − r1 and b − r2).

3. Irrational Numbers

  • A number is rational if it can be written as p/q where p, q are integers and q ≠ 0.
  • A number that cannot be written in the form p/q is irrational, e.g. √2, √3, √5, π.

Theorem (used in proofs): If p is a prime number and p divides a2, then p divides a, where a is a positive integer.

Proof that √2 is irrational (by contradiction) — learn every step

  1. Assume, to the contrary, that √2 is rational. Then √2 = a/b, where a and b are coprime integers (HCF = 1) and b ≠ 0.
  2. So b√2 = a. Squaring both sides: 2b2 = a2.
  3. Therefore 2 divides a2, and by the theorem above, 2 divides a. Write a = 2c.
  4. Substituting: 2b2 = 4c2, so b2 = 2c2. Hence 2 divides b2, so 2 divides b.
  5. Thus 2 is a common factor of a and b, contradicting that a and b are coprime.
  6. The contradiction arose from our wrong assumption. Hence √2 is irrational.

The proofs for √3 and √5 are identical — replace 2 by 3 or 5 throughout. All three proofs are in the syllabus.

Numbers of the form a + b√p

To prove numbers like 3 + 2√5 or 5 − √3 are irrational:

  1. Assume the number is rational, say 3 + 2√5 = a/b (a, b integers, b ≠ 0).
  2. Rearrange to isolate the surd: √5 = (a − 3b) / 2b.
  3. The right side is rational (integers under the operations +, −, ×, ÷), so √5 would be rational — a contradiction, since √5 is irrational.
  4. Hence the given number is irrational.

Exam tip: In such proofs you may state "√5 is irrational" only if it is given or already proved. If the question says "Given that √5 is irrational, prove 3 + 2√5 is irrational", quote that line. Useful facts: (rational) + (irrational) = irrational; (non-zero rational) × (irrational) = irrational. But the sum or product of two irrationals may be rational, e.g. √2 × √2 = 2.

Common board question patterns

  • 1 mark: HCF/LCM of small numbers by prime factorisation; identify rational/irrational; number of factors of a prime.
  • 2-3 marks: HCF × LCM relation problems; "can 6n end with 0"; showing an expression is composite; bells/lights/track word problems.
  • 3 marks: full proof of irrationality of √2, √3 or √5, or of a + b√p forms.
  • Case study: stacking books/planting trees in equal rows (HCF), or events recurring together (LCM).

Deleted from the rationalised syllabus (NOT to be studied)

  • Euclid's division lemma and Euclid's division algorithm.
  • Decimal expansions of rational numbers — theorems on terminating and non-terminating recurring decimals (old sections on when p/q terminates based on the form 2n5m).

Retained: Fundamental Theorem of Arithmetic, HCF and LCM by prime factorisation, HCF × LCM = product of two numbers, and proofs of irrationality of √2, √3, √5.

Ch 2: Polynomials →