Standard form, estimation, rounding and bounds

Standard form writes very large and very small numbers compactly. Accuracy tells us what a rounded or measured number does, and does not, reveal. These ideas are essential whenever A-level Mathematics uses calculator output, experimental data or numerical models.

Prerequisites

You should be able to:

• multiply and divide by powers of $10$;
• use positive and negative integer indices;
• apply the order of operations;
• compare positive and negative decimals.

Review indices, roots and surds if negative powers are unfamiliar. Calculator fluency explains how to enter and interpret standard form on a scientific calculator.

Standard form

A non-zero number is in standard form when it is written as

$a\times 10^n,$

where

$1\leq |a|<10$

and $n$ is an integer. The coefficient $a$ may be negative, but its magnitude must be at least $1$ and less than $10$.

For example,

$4.82\times10^6,\qquad 7\times10^{-3},\qquad -1.09\times10^4$

are in standard form. These are not:

$48.2\times10^5,\qquad 0.7\times10^{-2},$

because the coefficients $48.2$ and $0.7$ are outside the required range.

Converting ordinary numbers to standard form

Move the decimal point until the coefficient has exactly one non-zero digit before it. The power of $10$ records the movement.

Worked example 1: a large number

Write $5\,730\,000$ in standard form.

The decimal point moves six places to the left:

$5\,730\,000=5.73\times10^6.$

The positive exponent is sensible because $10^6$ is large.

Worked example 2: a small number

Write $0.0000842$ in standard form.

The first non-zero digit is $8$. Moving the decimal point five places to the right gives $8.42$, so we compensate with $10^{-5}$:

$0.0000842=8.42\times10^{-5}.$

A negative exponent does not make the number negative. It means reciprocal:

$10^{-5}=\frac1{10^5}.$

Worked example 3: converting back

Write $-3.06\times10^4$ as an ordinary number.

Multiplication by $10^4$ moves the decimal point four places to the right:

$-3.06\times10^4=-30\,600.$

The negative sign belongs to the number and does not affect the direction of movement.

Self-check 1

1. Write $72\,900\,000$ in standard form.
2. Write $0.000000503$ in standard form.
3. Write $6.14\times10^{-4}$ as an ordinary number.
4. Explain why $0.92\times10^7$ is not in standard form, then correct it.

Answers

1. $7.29\times10^7$
2. $5.03\times10^{-7}$
3. $0.000614$
4. Its coefficient is less than $1$. Since $0.92\times10^7=9.2\times10^6$, the correct form is $9.2\times10^6$.

Calculating in standard form

The laws of indices do most of the work:

$10^m\times10^n=10^{m+n},\qquad \frac{10^m}{10^n}=10^{m-n}.$

After calculating, check that the coefficient is in the interval $[1,10)$ in magnitude.

Worked example 4: multiplication

Calculate $(3.2\times10^5)(4.5\times10^{-3})$, giving the answer in standard form.

Multiply the coefficients and powers separately:

$$\begin{aligned} (3.2\times10^5)(4.5\times10^{-3}) &=(3.2\times4.5)\times10^{5+(-3)}\\ &=14.4\times10^2. \end{aligned}$$

This is numerically correct but not in standard form. Replace $14.4$ by $1.44\times10$:

$14.4\times10^2=1.44\times10^3.$

Worked example 5: division

Calculate

$\frac{8.4\times10^{-6}}{3.5\times10^2}.$

$$\begin{aligned} \frac{8.4\times10^{-6}}{3.5\times10^2} &=\frac{8.4}{3.5}\times10^{-6-2}\\ &=2.4\times10^{-8}. \end{aligned}$$

Worked example 6: addition needs a common power

Calculate $4.7\times10^6+8.2\times10^5$.

The powers cannot simply be added. Rewrite both terms using the same power:

$$\begin{aligned} 4.7\times10^6+8.2\times10^5 &=4.7\times10^6+0.82\times10^6\\ &=5.52\times10^6. \end{aligned}$$

This mirrors adding like place values in ordinary decimal notation.

Misconception: using multiplication rules for addition

The index law applies to products, not sums:

$10^3\times10^4=10^7,$

but

$10^3+10^4=1000+10\,000=11\,000,$

not $10^7$.

Decimal places and significant figures

Rounding replaces a number by a nearby value with fewer recorded digits.

• Decimal places count digits after the decimal point.
• Significant figures count from the first non-zero digit.

Zeros between non-zero digits are significant. Leading zeros only locate the decimal point and are not significant. Trailing zeros after a decimal point may communicate accuracy.

For example:

$0.004050$

has four significant figures: $4$, $0$, $5$, $0$.

A reliable rounding method

1. Locate the last digit to be kept.
2. Inspect the next digit only.
3. If that digit is $5$ or more, increase the kept digit by $1$.
4. If it is $4$ or less, leave the kept digit unchanged.
5. Replace discarded place values with zeros where necessary.

Worked example 7: decimal places

Round $18.3764$ to two decimal places.

The second decimal digit is $7$. The next digit is $6$, so round the $7$ up:

$18.3764\approx18.38.$

Worked example 8: significant figures in a small number

Round $0.007946$ to three significant figures.

Counting begins at $7$, not at the zeros:

$0.00\underline{794}6.$

The next digit is $6$, so the $4$ rounds up:

$0.007946\approx0.00795.$

Worked example 9: place value in a large number

Round $385\,721$ to two significant figures.

Keep $3$ and $8$. The next digit is $5$, so $38$ becomes $39$. The discarded place values must become zeros:

$385\,721\approx390\,000.$

Writing $39$ would change the size of the number.

Self-check 2

Round each number as stated.

1. $7.2851$ to three decimal places.
2. $0.00098271$ to two significant figures.
3. $64\,950$ to three significant figures.
4. $12.04$ to three significant figures.

Answers

1. $7.285$
2. $0.00098$
3. $65\,000$
4. $12.0$. The zero is significant and records the requested accuracy.

Estimation and plausibility

An estimate is a deliberately simple approximation. It is useful before or after a calculator calculation because it can expose a wrong operation, mistyped exponent or misplaced decimal point.

Unless a question specifies otherwise, rounding each quantity to one significant figure usually produces manageable arithmetic.

Worked example 10: estimate a calculation

Estimate

$\frac{19.7\times0.482}{3.08}.$

Round each value to one significant figure:

$19.7\approx20,\qquad0.482\approx0.5,\qquad3.08\approx3.$

Then

$\frac{19.7\times0.482}{3.08}\approx\frac{20\times0.5}{3}=\frac{10}{3}\approx3.3.$

The calculator value is about $3.08$, which is consistent with the estimate. An answer such as $30.8$ would not be plausible.

Estimation is not guaranteed to give an upper or lower bound. In this example, some values were rounded up and another was rounded down. Use bounds when a guaranteed limit is required.

Error intervals

If a value has been rounded, the original value is usually not known exactly. An error interval describes every value that would round to the stated result.

Suppose $x=7.4$ correct to the nearest $0.1$. The largest possible rounding error is half of $0.1$:

$\frac{0.1}{2}=0.05.$

Therefore

$7.35\leq x<7.45.$

The lower endpoint is included because $7.35$ rounds to $7.4$. The upper endpoint is excluded because $7.45$ rounds to $7.5$. This half-open interval is the standard convention:

$\text{lower bound}\leq x<\text{upper bound}.$

General rule

If $x$ rounds to $r$ to the nearest unit $u$, then

$r-\frac{u}{2}\leq x<r+\frac{u}{2}.$

Worked example 11: nearest whole number

A mass is recorded as $62$ kg to the nearest kilogram. If the exact mass is $m$ kg, then $u=1$ and

$62-0.5\leq m<62+0.5.$

Hence

$61.5\leq m<62.5.$

Worked example 12: significant figures

The population of a town is $48\,000$ correct to two significant figures. Find its error interval.

The last significant digit is the $8$ in the thousands column, so the number was rounded to the nearest $1000$. Half of $1000$ is $500$:

$47\,500\leq P<48\,500.$

Do not use $47\,999.5\leq P<48\,000.5$. The precision is determined by the place value of the last significant digit, not by whether the displayed number looks whole.

Worked example 13: a small value

A length is $0.0062$ m correct to two significant figures. Find its error interval.

The final significant digit is in the $0.0001$ place, so the rounding unit is $0.0001$ and the half unit is $0.00005$:

$0.00615\leq L<0.00625.$

Self-check 3

Write an error interval for each exact value.

1. $x=3.7$ correct to the nearest $0.1$.
2. $t=120$ seconds correct to the nearest $10$ seconds.
3. $d=8.40$ cm correct to two decimal places.
4. $N=6.3\times10^5$ correct to two significant figures.

Answers

1. $3.65\leq x<3.75$
2. $115\leq t<125$
3. $8.395\leq d<8.405$
4. The rounding unit is $10\,000$, so $625\,000\leq N<635\,000$. Equivalently, $6.25\times10^5\leq N<6.35\times10^5$.

Upper and lower bounds in calculations

For a positive measured quantity $x$, write its lower and upper bounds as $x_L$ and $x_U$. Then

$x_L\leq x<x_U.$

To bound a result, choose input bounds that make the expression as small or as large as possible.

For positive quantities:

$$\begin{array}{c|c|c} \text{expression} & \text{lower bound} & \text{upper bound}\\ \hline x+y & x_L+y_L & x_U+y_U\\ x-y & x_L-y_U & x_U-y_L\\ xy & x_Ly_L & x_Uy_U\\ \dfrac{x}{y} & \dfrac{x_L}{y_U} & \dfrac{x_U}{y_L} \end{array}$$

The subtraction and division rows are the common traps. To make $x-y$ small, make $x$ small and $y$ large. To make $x/y$ large, make the numerator large and the denominator small.

Worked example 14: perimeter bounds

A rectangle has length $8.2$ cm and width $3.6$ cm, each correct to the nearest $0.1$ cm. Find bounds for its perimeter $P$.

First find the measurement intervals:

$8.15\leq l<8.25,\qquad3.55\leq w<3.65.$

Since $P=2(l+w)$ and all coefficients are positive,

$$\begin{aligned} P_L&=2(8.15+3.55)=23.4,\\ P_U&=2(8.25+3.65)=23.8. \end{aligned}$$

Therefore

$23.4\leq P<23.8\text{ cm}.$

Worked example 15: speed bounds

A car travels $150$ km, correct to the nearest $10$ km, in $2.4$ hours, correct to the nearest $0.1$ hour. Find the upper bound for its average speed.

The intervals are

$145\leq d<155,qquad2.35\leq t<2.45.$

Average speed is $v=d/t$. To maximise it, use the upper bound for distance and lower bound for time:

$v_U=\frac{155}{2.35}=65.9574\ldots$

Thus the speed is less than approximately $65.96$ km/h. The unrounded quotient is the actual upper bound; rounding it down could produce a number that is no longer safely above every possible speed. If a decimal upper bound is required, round upwards to a stated number of decimal places, for example $65.96$ km/h to two decimal places.

Worked example 16: a difference

Two lengths are $a=12.3$ cm and $b=5.8$ cm, each correct to the nearest $0.1$ cm. Find bounds for $a-b$.

$12.25\leq a<12.35,qquad5.75\leq b<5.85.$

For the smallest difference, subtract the largest possible $b$ from the smallest possible $a$:

$12.25-5.85=6.40.$

For the largest difference, subtract the smallest possible $b$ from the largest possible $a$:

$12.35-5.75=6.60.$

Hence

$6.40<a-b<6.60.$

Strict endpoint details can depend on which original endpoints are attainable. In most exam questions, the requested numerical lower and upper bounds are $6.40$ and $6.60$; keep the original half-open intervals visible if interval notation itself is required.

Bounds with powers

If $x$ is positive and the expression increases as $x$ increases, substitute the lower bound for the minimum and the upper bound for the maximum.

Worked example 17: area of a circle

A radius is $4.2$ cm correct to the nearest $0.1$ cm. Find the lower and upper bounds for the area.

$4.15\leq r<4.25.$

Since $A=\pi r^2$ increases for positive $r$,

$$\begin{aligned} A_L&=\pi(4.15)^2=17.2225\pi,\\ A_U&=\pi(4.25)^2=18.0625\pi. \end{aligned}$$

Therefore

$17.2225\pi\leq A<18.0625\pi\text{ cm}^2.$

Keeping $\pi$ exact avoids unnecessary rounding. For expressions involving negative ranges or turning points, inspect how the function behaves rather than applying a memorised substitution rule.

Accuracy in multi-step calculations

Keep all available calculator digits during working and round once, at the end. Premature rounding can alter the final stated answer.

Worked example 18: premature rounding

Let

$Q=\frac{7.83^2}{2.47}.$

Using full values,

$Q=24.8206\ldots=24.8\text{ to }3\text{ significant figures}.$

If $7.83^2=61.3089$ is prematurely rounded to $61.3$, then

$\frac{61.3}{2.47}=24.8178\ldots.$

This happens to give the same three significant figure answer, but that agreement is not guaranteed. Store intermediate values or enter the whole expression at once.

Distinguish three kinds of output:

• an exact value, such as $\frac{5\pi}{3}$;
• a calculator approximation, such as $5.235987\ldots$;
• a stated rounded value, such as $5.24$ to three significant figures.

Use $=$ only for equality and $\approx$ when digits have been discarded.

Mixed self-check

1. Calculate $(6\times10^{-4})(3.5\times10^7)$ in standard form.
2. Calculate $7.1\times10^5-3.8\times10^4$ in standard form.
3. Estimate $\dfrac{48.7\times19.6}{0.203}$ by rounding each value to one significant figure.
4. A length is $36$ mm to the nearest millimetre. State its lower and upper bounds.
5. A rectangle has length $10$ cm to the nearest centimetre and width $4.0$ cm to the nearest $0.1$ cm. Find the upper bound for its area.
6. Explain why $2.9996$ should not be rounded to $2.1000$ to four significant figures.

Answers

1. $(6\times3.5)\times10^3=21\times10^3=2.1\times10^4$.
2. $7.1\times10^5-0.38\times10^5=6.72\times10^5$.
3. $48.7\approx50$, $19.6\approx20$ and $0.203\approx0.2$, so the estimate is $\frac{50\times20}{0.2}=5000$.
4. $35.5\leq L<36.5$, so the lower bound is $35.5$ mm and the upper bound is $36.5$ mm.
5. $l_U=10.5$ and $w_U=4.05$, so $A_U=10.5\times4.05=42.525\text{ cm}^2$.
6. Rounding propagates through place value. Keeping $2,9,9,9$ and rounding up produces $3.000$, not $2.1000$.

Checklist and next steps

You should now be able to:

• convert between ordinary numbers and $a\times10^n$;
• calculate with numbers in standard form;
• round to decimal places or significant figures;
• use estimation to test plausibility;
• construct an error interval from stated accuracy;
• select appropriate bounds in sums, differences, products and quotients;
• retain full precision until the final answer.

Next, study exact arithmetic with fractions and pi and calculator fluency. These accuracy skills also support units and compound measures, statistics and checking answers.