HookEvery sheet of A4 hides an irrational number
In 1922 a German engineer, Walter Porstmann, persuaded the DIN standards body to adopt a paper system — DIN 476, which the world later copied as ISO 216 — built on one demand: cutting a sheet in half must give a smaller sheet of exactly the same shape. That single rule fixes the proportions of every sheet of A4 in every printer in Britain. If the long side is \(l\) and the short side is \(s\), halving forces \(\frac{l}{s} = \frac{s}{l/2}\), which rearranges to \(l^2 = 2s^2\), so \(\frac{l}{s} = \sqrt{2}\). The ratio of A4 is not \(1.4\), not \(1.41\), not any fraction you can write down — it is \(\sqrt{2} = 1.41421356\ldots\), a number whose decimal never stops and never repeats.
Everything in this section is the machinery that makes a claim like that precise. A0 is defined to have an area of exactly \(1\) m², so its sides are \(2^{1/4}\) m by \(2^{-1/4}\) m — fractional indices doing real work. Each smaller size halves the area, so A4 has area \(2^{-4} = \frac{1}{16}\) m² and exactly \(2^4 = 16\) sheets of A4 tile one sheet of A0 — powers of two, counted. And \(\sqrt{2}\) can be written exactly only as a surd, because it is irrational. Ordering, the four operations, primes, powers, roots, surds and standard form are the nine skills the AQA Higher paper leans on more than any other, and almost every later topic quietly assumes you own them.
ModelOrdering, and the six symbols that carry the marks
A comparison is a number-line question, and the number line settles every argument: a value is larger the further right it sits. Negatives run backwards — \(-9 \lt -2\) because \(-9\) is further left, even though \(9 \gt 2\). Decimals are compared column by column from the left, not by length: \(0.4 \gt 0.375\) because the tenths digit \(4\) already beats \(3\), and the extra digits on \(0.375\) never get their turn. Fractions and surds are safest turned into decimals for comparing — but only for comparing, because \(\sqrt{2} = 1.41421\ldots\) becomes an approximation the moment you write it as a decimal.
The six relations each do a precise job: \(=\) equal, \(\neq\) not equal, \(\lt\) less than, \(\gt\) greater than, \(\le\) less than or equal to, \(\ge\) greater than or equal to. AQA expects you to write them, not just read them — 'n is at least 5' is \(n \ge 5\), and 'fewer than 200 people' is \(p \lt 200\). The knife-edge between \(\lt\) and \(\le\) is exactly the distinction that error intervals and inequalities are marked on later, so treat them as two different symbols, never as decoration.
Order these five numbers, smallest first: \(-\frac{3}{4}\), \(-0.7\), \(\frac{7}{5}\), \(1.41\) and \(\sqrt{2}\). Turn everything into a decimal: \(-\frac{3}{4} = -0.75\), \(\frac{7}{5} = 1.4\), and \(\sqrt{2} = 1.4142\ldots\). Deal with the negatives first — \(-0.75\) is further left than \(-0.7\), so \(-0.75 \lt -0.7\). Among the positives, compare column by column: \(1.4 \lt 1.41 \lt 1.4142\). The full chain is \(-\frac{3}{4} \lt -0.7 \lt \frac{7}{5} \lt 1.41 \lt \sqrt{2}\). Writing it as one chain of \(\lt\) signs, not a bare list, is what secures the final mark — and notice \(\frac{7}{5}\) and \(\sqrt{2}\) really are different numbers despite looking close.
MechanismThe four operations, place value and fractions
Paper 1 is non-calculator, and formal written methods are compulsory there. Column addition and subtraction work because place value lets you trade ten of one column for one of the next — 'carrying' and 'borrowing' are nothing more than that trade. Long multiplication splits a number by place value and multiplies each part; short division shares out each column and carries the remainder on. Decimals obey the same methods: to work out \(3.24 \times 2.6\), ignore the points and compute \(324 \times 26 = 8424\), then restore the three decimal places from the question to get \(8.424\) — legitimate because you scaled up by \(100\) and \(10\), a factor of \(1000\) you now divide back out.
Fractions add and subtract only over a common denominator, because you can count pieces only when they are the same size; they multiply straight across the top and bottom; and they divide by flipping the second fraction, because 'how many quarters in a half' has twice the answer of 'how many wholes'. Mixed numbers go improper first, every single time — carrying '2 and a bit' through the middle of a calculation is where marks quietly die.
Work out \(3\frac{1}{4} - 1\frac{5}{6}\) and \(2\frac{2}{3} \div 1\frac{1}{5}\). Improper first: \(3\frac{1}{4} = \frac{13}{4}\) and \(1\frac{5}{6} = \frac{11}{6}\). Common denominator \(12\): \(\frac{39}{12} - \frac{22}{12} = \frac{17}{12} = 1\frac{5}{12}\). For the division, \(2\frac{2}{3} = \frac{8}{3}\) and \(1\frac{1}{5} = \frac{6}{5}\); flip and multiply: \(\frac{8}{3} \times \frac{5}{6} = \frac{40}{18} = \frac{20}{9} = 2\frac{2}{9}\). Every line is evidence for a method mark — an unsupported answer risks one mark where full working scores three.
ModelPriority of operations, inverses and reciprocals
When operations meet in one expression, order decides the answer: brackets first, then indices (powers and roots), then multiplication and division together left to right, then addition and subtraction together left to right. Under that rule \(5 + 2 \times 3^2 = 5 + 2 \times 9 = 23\), not \(63\) — the power binds tighter than the multiplication, which binds tighter than the addition. Your calculator applies these rules whether you want it to or not, so \(12 \div 3 + 1\) returns \(5\); if you meant \(\frac{12}{3+1}\) you must bracket the denominator: \(12 \div (3 + 1) = 3\).
Underneath sits the idea of inverse operations: addition undoes subtraction, multiplication undoes division, squaring undoes square-rooting. Inverses are how you check work — \(8424 \div 26\) should hand back \(324\) — and later they solve every equation you meet. The reciprocal of a number is \(1\) divided by it: the reciprocal of \(4\) is \(\frac{1}{4}\), and the reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\). Any number times its reciprocal gives \(1\); zero alone has no reciprocal, because nothing multiplied by \(0\) makes \(1\).
Evaluate \(20 - 2 \times (3 + 4)^2 \div 7\). Brackets first: \(3 + 4 = 7\). Indices next: \(7^2 = 49\). Now multiplication and division, left to right: \(2 \times 49 = 98\), then \(98 \div 7 = 14\). Finally the subtraction: \(20 - 14 = 6\). Change one step — divide before you square, say — and the answer collapses; the order is the whole question.
MechanismPrimes, HCF and LCM — the factor-tree engine
A prime has exactly two factors, itself and \(1\), so \(1\) is not prime (one factor) and \(2\) is the only even prime. Primes matter because every whole number is built from them in exactly one way — the fundamental theorem of arithmetic. A factor tree breaks a number down: split it into any factor pair, keep splitting until every branch ends in a prime, then collect the leaves in index notation. AQA wants the index form for full marks: \(360 = 2^3 \times 3^2 \times 5\), not the bare list of primes.
HCF and LCM read straight off the prime factorisations. The highest common factor takes each shared prime at its lowest power; the lowest common multiple takes every prime that appears anywhere at its highest power. In disguise, LCM problems are buses leaving together or lights flashing in step, and HCF problems are 'largest equal groups' or the biggest square tile that fits a floor with no cutting.
Find the HCF and LCM of \(168\) and \(180\). Factor trees give \(168 = 2^3 \times 3 \times 7\) and \(180 = 2^2 \times 3^2 \times 5\). The HCF — shared primes at their lowest powers — is \(2^2 \times 3 = 12\). The LCM — every prime at its highest power — is \(2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520\). Free check: \(\text{HCF} \times \text{LCM} = 12 \times 2520 = 30240\), and \(168 \times 180 = 30240\) too. The product of any two numbers always equals their HCF times their LCM — a verification almost nobody uses.
CaseSystematic listing and the product rule for counting
When Britain reformed number plates in September 2001, it chose the format two letters, two digits, three letters — like 'AB12 CDE'. The three trailing letters alone give \(26 \times 26 \times 26 = 26^3 = 17576\) combinations for each area-and-age block, before the DVLA holds a few back. That multiplication is the product rule for counting: if one choice can be made \(a\) ways and an independent next choice \(b\) ways, the pair can be made \(a \times b\) ways. It is a Higher-tier named skill, and it turns 'how many possible…' questions into a single line.
When the choices are few, list them systematically instead — hold the first item fixed, cycle the rest in order, then move the first on. The order is not fussiness; it is the visible proof that nothing has been missed and nothing counted twice, and the mark scheme rewards exactly that.
A café meal deal is one of \(4\) sandwiches, one of \(3\) snacks and one of \(5\) drinks. By the product rule there are \(4 \times 3 \times 5 = 60\) different meal deals — and a table of \(4\) rows by \(3\) columns, repeated for each of the \(5\) drinks, would show all \(60\) without a gap. Contrast a four-digit PIN, where each digit is chosen from \(10\) and repeats are allowed: \(10 \times 10 \times 10 \times 10 = 10^4 = 10000\) possible codes. Same rule whether or not repeats are permitted — you just adjust how many options each stage really has.
ModelPowers, roots and indices — including fractional powers
A power is repeated multiplication: \(2^5 = 32\), emphatically not \(2 \times 5\). Higher candidates are expected to recognise the powers of small bases on sight — that \(64 = 2^6 = 4^3 = 8^2\), that \(81 = 3^4\), that \(625 = 5^4\) — and to estimate roots by trapping them between known ones: \(\sqrt{50}\) sits between \(7\) and \(8\) because \(7^2 = 49\) and \(8^2 = 64\). Roots run the machine backwards: \(\sqrt[3]{64} = 4\) because \(4^3 = 64\).
The index laws make it quick — multiplying powers of the same base adds the indices, dividing subtracts them, a power of a power multiplies them. Higher pushes past Foundation into fractional and negative indices: a negative index means reciprocal, so \(2^{-3} = \frac{1}{8}\); a unit fraction means a root, so \(x^{1/2} = \sqrt{x}\) and \(x^{1/3} = \sqrt[3]{x}\); and a general fraction does both, \(x^{m/n} = \left(\sqrt[n]{x}\right)^m\). This is where the A0 sheet from the introduction comes from: its sides are \(2^{1/4}\) and \(2^{-1/4}\) metres.
Evaluate \(8^{2/3}\) and \(16^{-3/4}\). For \(8^{2/3}\), the denominator says cube-root, the numerator says square: \(\sqrt[3]{8} = 2\), then \(2^2 = 4\). For \(16^{-3/4}\), deal with the sign first — the negative index means reciprocal — then the fraction: \(16^{-3/4} = \frac{1}{16^{3/4}} = \frac{1}{\left(\sqrt[4]{16}\right)^3} = \frac{1}{2^3} = \frac{1}{8}\). Rooting before powering keeps the numbers small; powering first (\(16^3 = 4096\)) works too but wastes time on Paper 1.
DataSurds, multiples of π and standard form — staying exact
'Give your answer exactly' or 'in terms of \(\pi\)' means stop before the decimal. A circle of radius \(5\) cm has area \(25\pi\) cm² exactly; push it through a calculator and \(78.539\ldots\) is only an approximation that rounds. Surds are the same discipline for roots: \(\sqrt{2}\) is exact, its decimal is not. Simplify a surd by pulling out the largest square factor — \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\) — and add surds only when the root matches, exactly like collecting like terms. To rationalise a denominator, multiply top and bottom by the surd to clear the root from the bottom: \(\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\).
Standard form writes any number as \(A \times 10^{n}\) with \(1 \le A \lt 10\) and \(n\) an integer — indispensable for the very large and very small. The mass of an electron is about \(9.11 \times 10^{-31}\) kg; the Sun is about \(1.99 \times 10^{30}\) kg. To multiply, multiply the fronts and add the powers; to divide, divide and subtract; to add or subtract, match the powers first or drop back to ordinary numbers — the powers do not combine just because you are adding. The condition \(1 \le A \lt 10\) is marked: \(45 \times 10^{6}\) is the right size but the wrong form, and must be rewritten as \(4.5 \times 10^{7}\).
Simplify \(\sqrt{50} + \sqrt{8}\), then rationalise \(\frac{6}{\sqrt{3}}\). First \(\sqrt{50} = 5\sqrt{2}\) and \(\sqrt{8} = 2\sqrt{2}\), so \(\sqrt{50} + \sqrt{8} = 7\sqrt{2}\) — the roots matched, so they added like terms. Then \(\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\). Now a standard-form product: \((3 \times 10^4) \times (2 \times 10^5) = 6 \times 10^{9}\) (fronts multiply, powers add), while \((3 \times 10^4) + (2 \times 10^3)\) is not \(5 \times 10^7\) — convert instead, \(30000 + 2000 = 32000 = 3.2 \times 10^4\).
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Paper 1 is non-calculator and this section is its spine: expect long multiplication or division, fraction arithmetic, an HCF/LCM problem and a surd simplification, all marked method-first. AQA award M marks for a correct method even when a digit slips, so write every stage — improper fractions before you add, the factor trees before an HCF, the index-notation line \(2^3 \times 3^2 \times 5\) before the final answer. A bare wrong answer keeps nothing; a complete method with one arithmetic slip usually keeps most of its marks.
Exactness is a form of answer, not a level of precision. If a question says 'exactly', 'in surd form' or 'in terms of \(\pi\)', a decimal loses the final mark even on a calculator paper — leave \(2\sqrt{3}\), \(7\sqrt{2}\) or \(25\pi\) as they stand, and always rationalise so no surd sits in a denominator. On Papers 2 and 3, use the \(\times 10^{x}\) key for standard form rather than typing '× 10 ^' with stray brackets, and write the full display down before you round.
Two habits worth a grade: check any division by multiplying back (the inverse sits on the spec precisely because it is the checking tool), and when a question mixes fractions, decimals, percentages and surds, convert everything into one form before you compare — the mark scheme routinely credits the conversions alone.