Rational, irrational and their decimals

• ›the real numbers are the rationals together with the irrationals; every real number is a point on the number line, and every point is a real number
• ›a number is rational when it can be written p/q, p and q integers with q non-zero; its decimal terminates or is non-terminating recurring
• ›an irrational number has a decimal that is non-terminating and non-recurring — √2, √3, π

Which rationals terminate? — write p/q in lowest terms; the decimal terminates exactly when q has no prime factors other than 2 and 5, that is, q = 2ᵐ5ⁿ.

Recurring to fraction — x = 0.474747…; 100x = 47.4747…; subtracting, 99x = 47, so x = 47/99.

The trap: test the denominator after cancelling — 6/15 = 2/5 terminates, even though 15 contains a 3.

The 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 factors
• ›from the prime factorisations: the HCF is the product of the smallest powers of the common primes; the LCM is the product of the largest powers of all the primes
• ›for two numbers, HCF × LCM = product of the numbers — this does not hold for three

Example — 96 = 2⁵ × 3 and 404 = 2² × 101; HCF = 2² = 4; LCM = 2⁵ × 3 × 101 = 9696; check: 4 × 9696 = 38,784 = 96 × 404.

Euclid's division algorithm — a = bq + r with 0 ≤ r < b; repeat on (b, r) until the remainder is 0; the last non-zero remainder is the HCF.

• ›HCF(455, 42): 455 = 42 × 10 + 35; 42 = 35 × 1 + 7; 35 = 7 × 5 + 0, so the HCF is 7

A classic — can 6ⁿ end in 0 for some natural n? A number ending in 0 is divisible by 5, but 6ⁿ = 2ⁿ × 3ⁿ has no factor 5 — so never, by the uniqueness of factorisation.

The trap: HCF × LCM = product works for two numbers only — for three numbers it fails in general.

Proving irrationality

√2 is irrational — by contradiction

1. 1.suppose √2 = a/b with a, b integers having no common factor
2. 2.then 2b² = a², so a² is even, and hence a is even; write a = 2c
3. 3.then 2b² = 4c², so b² = 2c², and b is even too
4. 4.a and b share the factor 2 — contradicting step 1; so √2 is irrational

The lemma behind it — if a prime p divides a², then p divides a. The same argument proves √3, √5 and every √p irrational for a prime p.

Combining numbers

Between any two numbers — between two distinct rationals there are infinitely many rationals and infinitely many irrationals.

The trap: the sum or product of two irrationals is not always irrational — check the particular numbers.

Exponents and surds

Laws of exponents — for a > 0 and rational m, n:

• ›aᵐ × aⁿ = aᵐ⁺ⁿ · aᵐ ÷ aⁿ = aᵐ⁻ⁿ · (aᵐ)ⁿ = aᵐⁿ · aᵐbᵐ = (ab)ᵐ
• ›a⁰ = 1 · a⁻ⁿ = 1/aⁿ · a^(1/n) = ⁿ√a · a^(m/n) = (ⁿ√a)ᵐ
• ›examples: 64^(1/3) = 4, so 64^(−1/3) = 1/4; 32^(2/5) = (⁵√32)² = 2² = 4

Rationalising the denominator — multiply by the conjugate:

• ›1/(√5 − √3) = (√5 + √3)/[(√5)² − (√3)²] = (√5 + √3)/2
• ›1/(2 + √3) = (2 − √3)/(4 − 3) = 2 − √3

Identities with surds — (√a + √b)(√a − √b) = a − b; (√a + √b)² = a + b + 2√(ab).

Representing √x on the line — for x = 9.3: mark AB = 9.3 and BC = 1, draw a semicircle on AC as diameter, and erect a perpendicular at B; its length to the semicircle is √9.3.

The trap: √(a + b) is not √a + √b — √(9 + 16) = 5, while √9 + √16 = 7.

Points to Remember

• ›rational ⇔ terminating or recurring decimal · irrational ⇔ non-terminating, non-recurring
• ›p/q in lowest terms terminates ⇔ q = 2ᵐ5ⁿ · test after cancelling
• ›0.474747… = 47/99 — multiply by 10ᵏ for a k-digit period and subtract
• ›fundamental theorem — unique prime factorisation · HCF: smallest powers · LCM: largest powers
• ›HCF × LCM equals the product only for a pair of numbers · 96, 404 → HCF 4, LCM 9696
• ›Euclid: a = bq + r, 0 ≤ r < b · HCF(455, 42) = 7 · 6ⁿ never ends in 0
• ›√2 irrational by contradiction · p ∣ a² ⇒ p ∣ a for prime p
• ›rational ± irrational → irrational · two irrationals may combine to a rational
• ›a⁻ⁿ = 1/aⁿ · a^(m/n) = (ⁿ√a)ᵐ · 1/(√5 − √3) = (√5 + √3)/2
• ›Exam tip: in a terminating-decimal question, factorise the denominator fully first — a single 3, 7 or 11 left over settles it