Significant figures and rounding

How many digits should we write?

You measure a pencil with a ruler marked in millimetres: $8.4\ \text{cm}$. A calculator then divides its length by $3$ and shows $2.8$ followed by a row of threes. Are all those digits true? No. The ruler did not know the length to a millionth of a centimetre, so the answer cannot either.

The significant figures of a number are the digits we trust, plus the first one we are unsure of. Writing more claims a precision we do not have. Writing fewer throws information away.

Counting significant figures

1. Every non-zero digit counts: $284$ has three.
2. Zeros between non-zero digits count: $2.04$ has three.
3. Zeros at the start never count; they only place the point: $0.0036$ has two.
4. Zeros at the end, after a decimal point, count: $3.60$ has three. They were written on purpose.
5. Zeros at the end of a whole number are unclear: $4500$ could have two, three or four. Scientific notation settles it: $4.5 \times 10^3$ (two) or $4.500 \times 10^3$ (four).

Watch out. Changing units never changes the significant figures: $3.60\ \text{m} = 360\ \text{cm} = 3.60 \times 10^{2}\ \text{cm}$, still three.

Rounding

To keep $n$ significant figures, look at the next digit. If it is $5$ or more, round up; if less, leave the last digit as it is: $2.764 \to 2.76$ (three), $2.766 \to 2.77$.

When the dropped part is exactly $5$, NCERT keeps the last digit if it is even and raises it if it is odd: $2.745 \to 2.74$, $2.735 \to 2.74$. Over many results this avoids always rounding up.

Significant figures in calculations

Multiplying or dividing: the answer keeps as many significant figures as the factor with the fewest. See why: a factor with two figures is uncertain in its second digit, and that uncertainty spreads into the product.

You should find that only the first digit, and roughly the second, of the product are certain. That is why it gets two significant figures, like its least precise factor.

Adding or subtracting: the answer keeps as many decimal places as the term with the fewest. $13.54 + 2.7 = 16.24 \to 16.2$, because $2.7$ is unsure in its first decimal place.

Round only at the end. Keep one or two extra digits in the middle of a long calculation, or the rounding errors pile up.

A worked example

A rectangular label is $3.4\ \text{cm}$ by $1.27\ \text{cm}$. Find its area, and its perimeter.

1. Area. $3.4\ \text{cm} \times 1.27\ \text{cm} = 4.318\ \text{cm}^2$. The fewest significant figures is two (in $3.4\ \text{cm}$), so the area is $4.3\ \text{cm}^2$.
2. Perimeter. $2 \times (3.4\ \text{cm} + 1.27\ \text{cm}) = 9.34\ \text{cm}$. The sum is limited by $3.4\ \text{cm}$ to one decimal place, so the perimeter is $9.3\ \text{cm}$.

Think first. Exact numbers, like the $2$ in a perimeter or the $\frac{1}{2}$ in $\frac{1}{2}mv^2$, have unlimited significant figures. They never limit an answer.

Practise with new numbers

Try it in Python

The program rounds a product and a sum the right way.

Remember these

• Significant figures: the sure digits plus the first unsure one.
• Leading zeros never count; trailing zeros after a decimal point do.
• Multiply or divide: fewest significant figures. Add or subtract: fewest decimal places.
• Exact numbers do not limit the answer; round only at the end.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How many significant figures are in $0.00507$, in $5.070$ and in $5.07 \times 10^{4}$?

2. Round $0.08465$ to two significant figures.

3. A car covers $125.4\ \text{m}$ in $5.6\ \text{s}$. Write its average speed with the right number of significant figures.

4. Why should you not round in the middle of a calculation?