N8 - Exact answers with pi and surds | AQA GCSE Mathematics

An exact answer keeps the full value, even when no short decimal can do that. You will preserve fractions and multiples of $\pi$ through calculations, then work with surds in the Higher parts. Knowing when to stop is as useful as knowing how to calculate.

Keeping fractions exact

If you cut a metre into three equal pieces and join them again, should the result be $0.999\,\mathrm m$ or $1\,\mathrm m$ ? The answer depends on whether you kept each piece's exact length or rounded it too early.

Each piece is exactly $\frac13\,\mathrm m$ . Writing $0.333\,\mathrm m$ loses a little length. Three copies of that rounded number give $0.999\,\mathrm m$ , but three exact thirds give the whole metre back.

You already know

Use common denominators to add fractions, and cancel common factors when multiplying or dividing. A terminating decimal is also exact when it is the whole value, for example $0.25=\frac14$ .

An exact fraction need not be untidy. Often cancellation leaves a very simple result. You only need a decimal approximation when the question asks for one.

Exact value

A value with no rounding or approximation. For example, $\frac13$ is exact, but $0.333$ is only an approximation to $\frac13$ . A fraction is exact even when its decimal expansion never terminates.

The fraction rules are not on the formulae sheet: learn them. Here we use those earlier rules to keep the full value throughout a calculation.

Worked example 1: subtracting before multiplying

Question. Work out $6\left(\frac56-\frac13\right)$ . Give an exact answer. Non-calculator.

Plan. Subtract the fractions exactly, then multiply.

\begin{aligned} 6\left(\frac56-\frac13\right) &=6\left(\frac56-\frac26\right) && \text{common denominator} \\ &=6\times\frac36 && \text{subtract numerators} \\ &=3 && \text{cancel the sixes} \end{aligned}

Why keep fractions? A rounded value of $\frac56$ would introduce an error before a calculation whose exact answer is an integer.

Answer. $3$ .

Check. Distribute the multiplication: $6\times\frac56-6\times\frac13=5-2=3$ .

Change only the operation outside the brackets: divide the same difference by $6$ .

Worked example 2: an exact non-terminating result

Question. Work out $\left(\frac56-\frac13\right)\div6$ . Give your answer as a fraction in its simplest form. Non-calculator.

Plan. Use the exact difference, then multiply by the reciprocal of $6$ .

\begin{aligned} \left(\frac56-\frac13\right)\div6 &=\frac12\div6 && \text{same exact difference} \\ &=\frac12\times\frac16 && \text{divide by }6 \\ &=\frac1{12} && \text{multiply fractions} \end{aligned}

Why not write $0.0833$ ? It is close, but it is not the requested exact fraction.

Answer. $\frac1{12}$ .

Check. $\frac1{12}\times6=\frac12$ , the value before division.

The same principle will protect calculations with $\pi$ and roots: carry the exact form, then round only a final answer if required.

Watch out

Wrong: $\frac13\times3=0.333\times3=0.999$ . The first equals sign replaces an exact value by an approximation. Right: $\frac13\times3=1$ . Use $\approx$ when a value has been rounded.

Use cancellation to keep the following answer exact, without converting any fraction to a decimal.

Calculating with multiples of pi

Keeping $\pi$ in an answer preserves a value that no finite decimal can capture exactly.

The number $\pi$ is irrational: it cannot be written as a ratio of two integers, and its decimal does not terminate or repeat. The familiar $3.14$ and $\frac{22}{7}$ are approximations, not alternative exact definitions. For example, $\frac{22}{7}-\pi$ is about $0.00126449$ .

A multiple of pi

A number obtained by multiplying $\pi$ by another number, for example $6\pi$ or $\frac{\pi}{4}$ . In “Give your answer in terms of $\pi$ ”, keep the symbol $\pi$ and simplify the number multiplying it.

You do not need circle methods yet to calculate with a supplied length such as $5\pi\,\mathrm{cm}$ . Treat $\pi$ as the same fixed number each time: $2\pi+3\pi=5\pi$ , just as two identical lengths plus three identical lengths make five.

Calculating with pi

For numbers $a$ , $b$ ,

a\pi+b\pi=(a+b)\pi

For $b\ne0$ ,

\frac{a\pi}{b\pi}=\frac ab

Multiplication retains every factor: $(a\pi)(b\pi)=ab\pi^2$ .

Not on the formulae sheet: learn it.

The 2027 formulae sheet does give circumference and area of a circle. Those geometry formulae will be taught in the circle lesson; the exact-number rules here apply whenever a question supplies a multiple of $\pi$ .

Worked example 3: adding exact lengths

Question. Two lengths are $5\pi\,\mathrm{cm}$ and $3\pi\,\mathrm{cm}$ . Work out their total length in terms of $\pi$ . Non-calculator.

Plan. Add the numbers multiplying the common $\pi$ .

\begin{aligned} 5\pi+3\pi&=(5+3)\pi && \text{collect multiples of }\pi \\ &=8\pi && \text{add coefficients} \end{aligned}

Why keep $\pi$ ? The question asks for an exact multiple, so replacing it by $3.14$ would change the answer's value and form.

Answer. $8\pi\,\mathrm{cm}$ .

Check. Eight copies of $\pi$ are the same as five copies and three copies.

Change only the operation: compare the first length with the second by division instead of finding a total.

Worked example 4: cancelling a common pi

Question. Work out $5\pi\div3\pi$ , meaning $\frac{5\pi}{3\pi}$ . Give an exact answer. Non-calculator.

Plan. Cancel the common non-zero factor $\pi$ .

\begin{aligned} \frac{5\pi}{3\pi}&=\frac53\times\frac\pi\pi && \text{separate the factors} \\ &=\frac53 && \pi\div\pi=1 \end{aligned}

Why does no $\pi$ remain? It is a factor of both the numerator and the denominator.

Answer. $\frac53$ .

Check. $\frac53\times3\pi=5\pi$ .

Cancellation applies to factors. A plain $3$ in $2\pi+3$ has no $\pi$ factor to collect, so this sum does not become $5\pi$ .

Watch out

Wrong: $2\pi\times3\pi=6\pi$ . Right: both factors of $\pi$ remain, giving $6\pi^2$ . Addition collects matching units; multiplication multiplies them.

A fraction multiplying $\pi$ is just as exact as a whole-number multiple. Finish the arithmetic on its coefficient.

Simplifying surds (Higher)

A square root can be exact even when it has no exact fraction or terminating decimal.

For example, $\sqrt2$ is irrational, so its root sign is useful information rather than unfinished arithmetic. A root can sometimes be simplified: $\sqrt{12}=\sqrt{4\times3}=2\sqrt3$ . The square factor $4$ comes out as its root $2$ .

Surd

An irrational root expressed exactly with a root sign, such as $\sqrt2$ or $\sqrt7$ . A root such as $\sqrt9=3$ is not a surd, because it simplifies to a rational number.

To simplify a square root of a positive integer, find a square factor greater than $1$ . The largest square factor gives the shortest route. A fully simplified square-root surd has no such square factor left inside its root.

Taking out a square factor

For $a\geqslant0$ , $b\geqslant0$ ,

\sqrt{a^2b}=a\sqrt b

The number $a$ outside the root is the square root of the square factor $a^2$ .

Not on the formulae sheet: learn it.

A smaller square factor still works; you may simply need another simplification afterwards.

Worked example 5: the largest square factor

Question. Simplify $\sqrt{72}$ . Non-calculator.

Plan. Write $72$ as a square factor times the remaining factor.

\begin{aligned} \sqrt{72}&=\sqrt{36\times2} && 36\text{ is a square factor} \\ &=\sqrt{36}\times\sqrt2 && \text{root of a product} \\ &=6\sqrt2 && \text{square root of }36 \end{aligned}

Why choose $36$ ? It leaves $2$ , which has no square factor greater than $1$ .

Answer. $6\sqrt2$ .

Check. $(6\sqrt2)^2=36\times2=72$ , and $6\sqrt2$ is positive.

Change only the square factor inside the root. The remaining surd will again be $\sqrt2$ .

Worked example 6: a larger square factor

Question. Simplify $\sqrt{200}$ . Non-calculator.

Plan. Use $100$ as the square factor.

\begin{aligned} \sqrt{200}&=\sqrt{100\times2} && \text{factor the number} \\ &=10\sqrt2 && \sqrt{100}=10 \end{aligned}

Why not stop at $2\sqrt{50}$ ? That is equal to the original root, but $50$ still contains the square factor $25$ .

Answer. $10\sqrt2$ .

Check. $(10\sqrt2)^2=100\times2=200$ .

The simplification changes the appearance, not the value. A calculator approximation may check the size, but it cannot replace an exact surd answer.

Watch out

Wrong: $\sqrt{72}=36\sqrt2$ . Right: $\sqrt{36}=6$ , so $\sqrt{72}=6\sqrt2$ . Take the root of the square factor before bringing it outside.

Look for the largest square factor in a new number.

Calculating with surds (Higher)

Simplify surds first when that reveals matching root terms.

For instance, $2\sqrt3+5\sqrt3=7\sqrt3$ : both terms contain the same $\sqrt3$ . But $\sqrt2+\sqrt3$ cannot be collected into one multiple of a common root. Multiplication is different: $\sqrt2\times\sqrt3=\sqrt6$ .

Collecting and multiplying surds

For $c\geqslant0$ ,

a\sqrt c+b\sqrt c=(a+b)\sqrt c

For $c\geqslant0$ , $d\geqslant0$ ,

(a\sqrt c)(b\sqrt d)=ab\sqrt{cd}

For division, use the corresponding quotient rule with a non-zero denominator.

Not on the formulae sheet: learn it.

A number outside a root multiplies the root; it is not part of the number inside. In $3\sqrt5$ , for example, the coefficient is $3$ .

Worked example 7: adding after simplification

Question. Simplify $\sqrt{18}+\sqrt{50}$ . Non-calculator.

Plan. Take out square factors, then collect matching roots.

\begin{aligned} \sqrt{18}+\sqrt{50}&=3\sqrt2+5\sqrt2 && \text{use square factors }9,\ 25 \\ &=8\sqrt2 && \text{add coefficients} \end{aligned}

Why simplify first? The original roots look unlike, but both are multiples of $\sqrt2$ .

Answer. $8\sqrt2$ .

Check. Replacing the simplified terms by $\sqrt{9\times2}$ and $\sqrt{25\times2}$ recovers the original sum.

Change only the operation from addition to multiplication. Matching roots now multiply to a whole number.

Worked example 8: multiplying the same roots

Question. Work out $\sqrt{18}\times\sqrt{50}$ . Non-calculator.

Plan. Use the same simplified roots, then multiply every factor.

\begin{aligned} \sqrt{18}\times\sqrt{50}&=(3\sqrt2)(5\sqrt2) && \text{simplify roots} \\ &=15\times2 && \sqrt2\times\sqrt2=2 \\ &=30 && \text{multiply} \end{aligned}

Why does the root disappear? Two factors of $\sqrt2$ make $2$ , by the definition of square root.

Answer. $30$ .

Check. Directly, $\sqrt{18\times50}=\sqrt{900}=30$ .

Division can also simplify a root away: $\frac{\sqrt{48}}{\sqrt3}=\sqrt{16}=4$ . For subtraction, collect coefficients just as for addition: $7\sqrt5-2\sqrt5=5\sqrt5$ .

Watch out

Wrong: $\sqrt{18}+\sqrt{50}=\sqrt{68}$ . Roots do not split or combine over addition. Right: simplify them separately to get $3\sqrt2+5\sqrt2=8\sqrt2$ .

Watch for a coefficient already outside a root. It multiplies any new coefficient taken out during simplification.

Rationalising denominators (Higher)

Rationalising writes an equivalent fraction with a rational denominator.

For $\frac1{\sqrt3}$ , multiply top and bottom by $\sqrt3$ . This changes nothing because $\frac{\sqrt3}{\sqrt3}=1$ , but it turns the denominator into $3$ . The result $\frac{\sqrt3}{3}$ is still exact.

Rationalising a denominator

Rewriting a fraction so that its denominator contains no surd. The numerator and denominator must be multiplied by the same non-zero quantity, so the value of the fraction stays unchanged.

When the denominator contains a sum, a matching expression with the opposite sign is useful. For example,

\begin{aligned} (2+\sqrt3)(2-\sqrt3) &=4-2\sqrt3+2\sqrt3-3 \\ &=1 \end{aligned}

The two middle terms cancel. The pair $2+\sqrt3$ and $2-\sqrt3$ are called conjugates.

Removing a surd from a denominator

For $b>0$ ,

\frac a{\sqrt b}=\frac{a\sqrt b}{b}

For a denominator $c+\sqrt b$ , use its conjugate $c-\sqrt b$ , provided the original denominator and the conjugate are non-zero:

(c+\sqrt b)(c-\sqrt b)=c^2-b

For $c-\sqrt b$ , use $c+\sqrt b$ .

Not on the formulae sheet: learn it.

Choose the multiplying expression from the denominator. A single root needs the same root; a two-term denominator needs the sign-changing partner.

Worked example 9: a single-root denominator

Question. Rationalise the denominator and simplify $\frac6{\sqrt3}$ . Non-calculator.

Plan. Multiply top and bottom by $\sqrt3$ , then cancel.

\begin{aligned} \frac6{\sqrt3}&=\frac{6\sqrt3}{\sqrt3\times\sqrt3} && \text{multiply by }1 \\ &=\frac{6\sqrt3}{3} && \text{denominator is rational} \\ &=2\sqrt3 && \text{cancel }3 \end{aligned}

Why multiply the numerator too? Changing only the denominator would change the fraction's value.

Answer. $2\sqrt3$ .

Check. $2\sqrt3\times\sqrt3=6$ , recovering the numerator.

Change only the denominator: it now contains $2+\sqrt3$ , so use its conjugate instead of a single root.

Worked example 10: a two-term denominator

Question. Rationalise the denominator and simplify $\frac6{2+\sqrt3}$ . Non-calculator.

Plan. Multiply top and bottom by $2-\sqrt3$ , whose product with the denominator is $1$ .

\begin{aligned} \frac6{2+\sqrt3} &=\frac{6(2-\sqrt3)}{(2+\sqrt3)(2-\sqrt3)} && \text{use the conjugate} \\ &=\frac{6(2-\sqrt3)}{4-3} && \text{middle terms cancel} \\ &=12-6\sqrt3 && \text{denominator equals }1 \end{aligned}

Why change the sign? It makes the two mixed root terms cancel, leaving the difference $4-3$ .

Answer. $12-6\sqrt3$ .

Check. Multiplying the answer by $2+\sqrt3$ gives $24-18=6$ , with the root terms cancelling.

Here is the surprising exact fact behind this method: $\sqrt2+1$ and $\sqrt2-1$ are reciprocals, because their product is $2-1=1$ . Neither is rational, but together they make a whole number.

Watch out

Wrong: multiplying $\frac6{2+\sqrt3}$ by $\frac{\sqrt3}{\sqrt3}$ and claiming the denominator is $2+3$ . The denominator actually becomes $2\sqrt3+3$ . Right: use $2-\sqrt3$ to cancel the mixed terms.

The opening metre is exactly $1\,\mathrm m$ . That same habit of preserving the exact value has now carried us through fractions, $\pi$ and surds without rounding.

Quick Recap

Keep fractions exact until a decimal approximation is requested.
In an answer “in terms of $\pi$ ”, simplify the coefficient and keep $\pi$ .
Keep the full value through intermediate steps, then round only if a final approximation is requested.

Three exact thirds restore the opening metre: $\frac13+\frac13+\frac13=1$ .

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N8 - Exact answers with pi and surds

Keeping fractions exact

Calculating with multiples of pi

Simplifying surds (Higher)

Calculating with surds (Higher)

Rationalising denominators (Higher)

Quick Recap

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