Rounding and significant figures — Mathematics

Before this

This leads to

Estimation

Why keep three digits?

The speed of light is exactly 299 792 458 m/s. In the last lecture, we used $3.00 \times 1 0^{8}$ m/s. The Earth's mass is given by NASA as $5.9722 \times 1 0^{24}$ kg. We used $5.97 \times 1 0^{24}$ kg. Both times, the long number was cut down to three digits.

Why three? Why not two, or all nine? And why write $3.00$ , with two zeros that seem to add nothing? The last lecture promised to explain how many digits are worth keeping. This lecture keeps that promise. The idea is called significant figures, and it answers a practical question: which digits in a number actually tell you something?

Which digits are significant?

A digit is significant if it tells you about the size of the number, rather than only holding a place. Look at 0.008 472.

In 0.008 472, the zeros in front only show where the decimal point is. The digits 8, 4, 7 and 2 carry the information, so the number has 4 significant figures.

The zeros at the front are needed, because without them the 8 would be in the wrong place. But they say nothing about the measurement itself. In standard form, the number is $8.472 \times 1 0^{- 3}$ , and the zeros disappear completely. The four digits that remain are the significant ones.

That gives the clearest rule of all: write the number in standard form, and count the digits in the number in front. Every digit there is significant. The everyday rules below all follow from it.

Non-zero digits are always significant. 472 has 3 significant figures.
Zeros between non-zero digits are significant. 4003 has 4: in standard form it is $4.003 \times 1 0^{3}$ , and the zeros stay.
Zeros at the front are not significant. 0.0042 has 2: it is $4.2 \times 1 0^{- 3}$ .
Zeros at the end, after the decimal point, are significant. 3.50 has 3. A writer who adds that zero is telling you the measurement was made to the nearest hundredth. Otherwise they would have written 3.5.
Zeros at the end of a whole number are unclear. Does 4500 have 2, 3 or 4 significant figures? From the number alone, you cannot tell. Standard form removes the doubt: $4.5 \times 1 0^{3}$ (2), $4.50 \times 1 0^{3}$ (3) or $4.500 \times 1 0^{3}$ (4).

Now the zeros in $3.00 \times 1 0^{8}$ make sense. They say that the speed was given to 3 significant figures, and that the next two digits are known to be close to 0. Writing just $3 \times 1 0^{8}$ would claim only 1 significant figure. The abbreviation for significant figures is s.f., so we write " $3.00 \times 1 0^{8}$ m/s (3 s.f.)".

Rounding to a number of significant figures

To round to $n$ significant figures:

Start at the first non-zero digit, and count $n$ digits. These are the digits you keep.
Look at the next digit, the deciding digit. If it is 5 or more, add 1 to the last digit you keep. If it is 4 or less, leave the last digit as it is.
Keep the number the right size. Digits you drop before the decimal point become zeros. Digits you drop after the decimal point are simply removed.

For example, 47 638 to 3 s.f.: the digits kept are 4, 7, 6. The deciding digit is 3, so the 6 stays. The 3 and 8 are before the decimal point, so they become zeros: 47 600. Writing 476 would be a disaster, because it is a hundred times too small. Step 3 is there to stop that.

And 0.008 472 to 2 s.f.: the digits kept are 8 and 4 (the zeros in front are not counted). The deciding digit is 7, so the 4 becomes 5: 0.0085.

Explore the rule below with any of the numbers.

Explore — rounding to significant figures

Choose a number, then choose how many significant figures to keep. The kept digits are shaded; the deciding digit has a dark outline.

Significant figures: 3

Each number rounded to 1, 2 and 3 significant figures:

Number	1 s.f.	2 s.f.	3 s.f.
47 638	50 000	48 000	47 600
0.008 472	0.008	0.0085	0.008 47
0.073 651	0.07	0.074	0.0737

The speed of light, 299 792 458 m/s, is 300 000 000 to 1, 2 or 3 s.f.: it looks the same as an ordinary number every time. Only standard form shows the difference: 3 × 10⁸, 3.0 × 10⁸ and 3.00 × 10⁸.

Did you watch the speed of light? Rounded to 1, 2 or 3 s.f., it is 300 000 000 every time, as an ordinary number. Only standard form shows how many figures were kept: $3 \times 1 0^{8}$ , $3.0 \times 1 0^{8}$ or $3.00 \times 1 0^{8}$ . This is one of the best reasons to use standard form in science.

Stop and think

A student rounds 0.0396 to 2 s.f. and writes 0.04. Is that right?

Show the answer

The value is right, but it shows only 1 significant figure. The digits kept are 3 and 9. The deciding digit is 6, so the 9 goes up, which carries: 39 becomes 40. To show that 2 figures were kept, the zero must stay: 0.040. That final zero is significant, because it comes after the decimal point. In standard form, the answer is $4.0 \times 1 0^{- 2}$ .

How many figures should an answer have?

A calculator gives as many digits as its screen can hold. That does not mean they are all worth writing down. An answer cannot be more precise than the numbers it came from.

Suppose the sides of a triangle are measured to 2 significant figures, and the calculator gives the area as 12.7431 cm². The last four digits are not real information, because the measurements were not that precise. A sensible answer is 13 cm² (2 s.f.). Writing 12.7431 cm² would claim a precision the measurements do not have. Scientists call this false precision.

A common guide is to give an answer to the same number of significant figures as the least precise number you used. It is a guide rather than an exact law, but it stops answers from claiming too much.

There is one more rule, and it matters just as much: round only at the end. In the last lecture, the number of protons with the mass of the Earth was found by first rounding $1 \div (1.67 \times 1 0^{- 27})$ to $5.99 \times 1 0^{26}$ , then multiplying by $5.97 \times 1 0^{24}$ . That gave $3.58 \times 1 0^{51}$ . If you keep the full calculator value until the very end, the answer is $3.5749 \dots \times 1 0^{51}$ , which is $3.57 \times 1 0^{51}$ to 3 s.f. Rounding in the middle changed the last figure. Keep every digit while you work, and round once, at the end.

That is it. That is the whole idea of this lecture. Significant figures are a way of being honest about how much you know. Keep the digits that carry information, round only at the end, and do not let a calculator make your answer look more certain than your data.

Test your counting below.

Sorter — how many significant figures?

Choose a card, then choose the box. Each card explains itself.

Each number and its number of significant figures:

Number	s.f.	Reason
0.050	2	front zeros do not count; the final zero, after the point, does
4.5 × 10³	2	count the digits in the number in front
0.0042	2	front zeros do not count
306	3	a zero between non-zero digits counts
3.50	3	a final zero after the point counts
0.007 01	3	front zeros do not count; the middle zero does
4003	4	both middle zeros count
7.000 × 10⁶	4	the zeros in front of × 10⁶ are written on purpose
2.050	4	the middle zero and the final zero both count

Worked examples

Worked Example 1 — One number, four roundings

Round 82 547 to 1, 2, 3 and 4 significant figures.

1 s.f.: keep 8; the deciding digit is 2, so it stays: 80 000.

2 s.f.: keep 8, 2; the deciding digit is 5, so the 2 goes up: 83 000.

3 s.f.: keep 8, 2, 5; the deciding digit is 4: 82 500.

4 s.f.: keep 8, 2, 5, 4; the deciding digit is 7, so the 4 goes up: 82 550.

Check: every answer is still about eighty thousand. Each one is the right size.

Worked Example 2 — A small decimal

Round 0.073 651 to 1, 2 and 3 significant figures.

The first significant figure is the 7. To 1 s.f.: the deciding digit is 3, so 0.07. To 2 s.f.: keep 7, 3; the deciding digit is 6, so 0.074. To 3 s.f.: keep 7, 3, 6; the deciding digit is 5, so the 6 goes up: 0.0737.

Worked Example 3 — Real constants

Round the Earth's mass, $5.9722 \times 1 0^{24}$ kg, and the average Earth–Sun distance, $149.598 \times 1 0^{6}$ km, to 3 significant figures, in standard form.

Mass: keep 5, 9, 7; the deciding digit is 2: $5.97 \times 1 0^{24}$ kg.

Distance: $149.598 \times 1 0^{6} = 1.49598 \times 1 0^{8}$ . Keep 1, 4, 9; the deciding digit is 5, so the 9 goes up, which carries: 149 becomes 150. The answer is $1.50 \times 1 0^{8}$ km. The final zero must be written, because it is the third significant figure.

Three classic mistakes

Trap 1 — Losing the size of the number

47 638 to 2 s.f. is 48 000, not 48. Digits dropped before the decimal point must become zeros, or the answer is a thousand times too small.

Trap 2 — Counting the zeros in front

In 0.0042, the zeros in front are place holders. The first significant figure is the first non-zero digit, the 4. Starting the count at the first zero gives the wrong digits.

Trap 3 — Rounding too early

Rounding in the middle of a calculation can change the final answer, as the proton count showed (3.58 against 3.57 × 10⁵¹). Keep the full value on your calculator, and round once at the end.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

How many significant figures do these have? (a) 0.0603 (b) 5020 written as $5.020 \times 1 0^{3}$ (c) 0.90

Hint 1
Write each in standard form, then count the digits in the number in front.

Hint 2
(a) is $6.03 \times 1 0^{- 2}$ . (c) is $9.0 \times 1 0^{- 1}$ .

Full answer
(a) 3. (b) 4. (c) 2: the final zero comes after the decimal point, so it counts.
Round (a) 6.0974 to 2 s.f., (b) 2504 to 2 s.f., (c) 0.000 384 6 to 3 s.f.

Hint 1
Find the first non-zero digit, count the digits to keep, and look at the next one.

Hint 2
In (b), the dropped digits are before the decimal point, so they become zeros.

Full answer
(a) Keep 6, 0; the deciding digit is 9, so 6.1. (b) Keep 2, 5; the deciding digit is 0, so 2500. (c) Keep 3, 8, 4; the deciding digit is 6, so 0.000 385.
A rectangle is measured as 4.6 cm by 3.2 cm. A student writes its area as 14.72 cm². Explain why this is not a sensible answer, and give a better one.

Hint 1
How many significant figures do the measurements have?

Hint 2
The answer should not claim more figures than the least precise measurement.

Full answer
Both measurements have 2 s.f., so the area can be trusted to only about 2 s.f. 14.72 claims four figures of precision that the measurements do not have. A better answer is 15 cm² (2 s.f.).
A runner covers 100 m (measured to the nearest metre) in 9.58 s. A calculator gives the average speed as 10.438413… m/s. Give a sensible answer, and explain why "10.438413 m/s" claims too much. Then explain why 100 m here has 3 significant figures, even though it ends in zeros.

Hint 1
"To the nearest metre" tells you that the last zero of 100 is a real measured digit.

Hint 2
Both numbers then have 3 s.f.

Full answer
Measured to the nearest metre, 100 m means 100, not 90 or 110 or 101: all three digits are known, so it has 3 s.f. (in standard form, $1.00 \times 1 0^{2}$ m). The time, 9.58 s, also has 3 s.f. So a sensible answer is 10.4 m/s (3 s.f.). The calculator's 10.438413 claims precision to a millionth of a metre per second, which neither measurement could support. This is why the trailing zeros of a whole number need a note about how it was measured, or standard form, before you can count them.

The bigger idea

Every measured number is a little uncertain, and significant figures are how mathematics and science write that honestly. A number with three significant figures says "these three digits I trust". Adding more digits does not make a result more accurate. It only makes it look more accurate, which is worse.

Standard form and significant figures work as a pair: the power gives the size, and the digits in front give the precision. In the next lecture, you will round on purpose, to 1 significant figure, to estimate an answer quickly and check that a calculator result makes sense.

Key concepts in this lesson

A significant figure is a digit that carries information about the size of a number, rather than only holding a place.
In standard form, every digit in the number in front is significant, so standard form shows the number of significant figures clearly.
Zeros in front are never significant, zeros between digits always are, and zeros at the end are significant when they come after the decimal point.
To round to n significant figures, keep n digits from the first non-zero digit, use the next digit to decide, and keep the number the right size.
An answer should not claim more precision than its data, and rounding should be done once, at the end of a calculation.