Real Numbers — chapter notes

A prime number has exactly two factors, $1$ and itself: $2, 3, 5, 7, 11, 13, \dots$ A composite number has more than two factors. The number $1$ is neither prime nor composite, and $2$ is the only even prime.

To factorise a composite number, split off primes repeatedly — by a factor tree or by repeated division — until only primes remain. Write the result with powers in increasing order of primes, e.g. $360 = 2^3 \times 3^2 \times 5$.

Divisibility shortcuts help: $2$ (last digit even), $3$ (digit sum divisible by 3), $5$ (last digit 0 or 5), $11$ (alternating digit sum divisible by 11).

Theorem. Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the primes occur.

Consequences you will use:

• A number ends in $0$ exactly when it is divisible by $10 = 2 \times 5$, so its factorisation must contain both $2$ and $5$. Hence $12^n = 2^{2n} \times 3^n$ never ends in $0$.
• Perfect square: every exponent even. Perfect cube: every exponent a multiple of $3$. To make $1188 = 2^2 \times 3^3 \times 11$ a square, multiply by $3 \times 11 = 33$.
• "Factorisations" such as $2520 = 4 \times 630$ are not prime factorisations; reduce every factor to primes.

Not in the 2026–27 syllabus: Euclid's division lemma and algorithm, and decimal expansions of rational numbers.

• HCF (highest common factor) $=$ product of the common primes, each with its smaller power.
• LCM (lowest common multiple) $=$ product of all primes that occur, each with its greater power.

Working: $336 = 2^4 \times 3 \times 7$ and $180 = 2^2 \times 3^2 \times 5$, so $\text{HCF} = 2^2 \times 3 = 12$ and $\text{LCM} = 2^4 \times 3^2 \times 5 \times 7 = 5040$.

For three numbers, use primes common to all three for the HCF: $\text{HCF}(48, 72, 108) = 2^2 \times 3 = 12$, $\text{LCM} = 2^4 \times 3^3 = 432$.

An exponent table (one row per prime, power $0$ when a prime is missing) prevents slips.

For any two natural numbers $a$ and $b$, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$, because for each prime $\min(m, n) + \max(m, n) = m + n$.

Use it to find a missing number: if $\text{HCF} = 23$, $\text{LCM} = 1449$ and one number is $161$, the other is $\dfrac{23 \times 1449}{161} = 207$.

It fails for three numbers: for $4, 6, 10$, $\text{HCF} \times \text{LCM} = 2 \times 60 = 120$, but $4 \times 6 \times 10 = 240$.

Useful facts: the HCF always divides the LCM; co-prime numbers have $\text{HCF} = 1$ and $\text{LCM} =$ product; consecutive numbers are co-prime.

HCF problems ask for the largest size that divides several quantities exactly: cutting ribbons or rods, packing sweets, tiling floors with square tiles, arranging groups in equal rows, spacing poles.

LCM problems ask for the smallest quantity or the first time at which events coincide: bells ringing together, buses or trains leaving together, runners meeting at the start, lights blinking together.

Remainder variants: "largest number dividing 245 and 1029 leaving remainders 5 and 9" → HCF of $240$ and $1020$ $= 60$. "Smallest number leaving remainder 5 when divided by 12, 16, 24, 36" → LCM $+ 5$.

Always finish with what is asked: the number of pieces, a clock time, or a cost.

A number is rational if it can be written as $\dfrac pq$, where $p$ and $q$ are integers and $q \ne 0$. A real number that is not rational is irrational. Examples of irrationals: $\sqrt2, \sqrt3, \sqrt5, \sqrt6, \pi$. The square root of a natural number that is not a perfect square is irrational; $\sqrt{49} = 7$ is rational.

Irrational numbers are real lengths: the diagonal of a $1$ m square tile is $\sqrt2$ m, and $\sqrt2$ can be marked on the number line with a compass.

$\pi$ is irrational; $\dfrac{22}{7}$ and $3.14$ are only approximations.

Key fact. If a prime $p$ divides $a^2$, then $p$ divides $a$. (By uniqueness of factorisation, $a^2$ has the same primes as $a$.) It is false for non-primes: $4$ divides $36$ but not $6$.

Proof that $\sqrt5$ is irrational. Suppose $\sqrt5 = \dfrac ab$ with $a, b$ co-prime and $b \ne 0$. Then $5b^2 = a^2$, so $5 \mid a^2$ and hence $5 \mid a$; write $a = 5c$. Then $5b^2 = 25c^2$, i.e. $b^2 = 5c^2$, so $5 \mid b^2$ and $5 \mid b$. Thus $5$ divides both $a$ and $b$, contradicting co-primality. So $\sqrt5$ is irrational.

The same steps with $2$ or $3$ prove $\sqrt2$ and $\sqrt3$ irrational.

If $r$ is rational and $x$ is irrational:

• $r + x$ and $r - x$ are irrational;
• $rx$ and $\dfrac xr$ are irrational when $r \ne 0$.

Method for "prove $4 - 3\sqrt2$ is irrational": suppose it equals a rational $r$; then $\sqrt2 = \dfrac{4 - r}{3}$ would be rational, a contradiction.

Two irrationals can give a rational: $\sqrt2 \times \sqrt{18} = 6$, $(3 + \sqrt2) + (3 - \sqrt2) = 6$, $(3 + \sqrt2)(3 - \sqrt2) = 7$. Simplify before classifying.