Rounding to one significant figure
A newspaper reports a crowd of 2347 people as about 2000. The number has been rounded to one significant figure: only its first digit, the 2 in the thousands, is kept, and every place after it is filled with a zero.
Whether the 2 stays or rounds up depends on the digit just after it, the 3 in the hundreds. 3 is less than 5, so the 2 stays and 2347 rounds down to 2000. A crowd of 2647 would round up to 3000, because its second digit, 6, is 5 or more.
2347 is below the halfway mark, 2500, so it is nearer to 2000 than to 3000.
More significant figures
To round to two significant figures, keep the first two digits, 2 and 3, and let the next digit decide. It is 4, which is less than 5, so 2347 rounds to 2300. To three significant figures, keep 2, 3 and 4 and look at the 7, which is 5 or more: the 4 rounds up and 2347 becomes 2350.
The zeros at the end of 2300 are not significant figures. They hold the 2 in the thousands place and the 3 in the hundreds, so the rounded number stays about the same size as 2347. Writing 2347 to two significant figures as 23 would turn a crowd of thousands into 23 people.
Between 2300 and 2400 the halfway mark is 2350. 2347 is just below it, so to two significant figures it is 2300.
Where the counting starts
Significant figures are counted from the first digit that is not zero. In 0.0052 the zeros at the front only show where the decimal point is, so they are not significant. Counting starts at the 5, and 0.0052 has two significant figures, the 5 and the 2. Written in grams, 0.0052 kg is 5.2 g, and the same two figures are all that is left.
A zero between two nonzero digits does count. 4.06 has three significant figures, because its 0 sits between the 4 and the 6.
Rounding works the same way once you know where to start. To round 0.0052 to one significant figure, keep the 5 and look at the digit after it, 2, which is less than 5: the answer is 0.005.
The first significant figure of 0.0052 is the 5, in the thousandths, so to one significant figure it rounds to a whole number of thousandths. 0.0052 is below the halfway mark, 0.0055, so it rounds to 0.005.
each zero holds a place: × 10 moves every figure one place, and a zero fills in or drops out, but the figures and their count stay the same
Show three zeros at once, and keep significant figures = 2.
The handle moves the decimal point while the figures 5 and 2 stay the same. The zeros that appear only hold places, so there are always 2 significant figures, while the number of decimal places changes.
Significant figures are not decimal places
Decimal places are counted from the decimal point. Significant figures are counted from the first nonzero digit. So the two instructions give different answers: 0.052 to one decimal place is 0.1, but 0.052 to one significant figure is 0.05.
Significant figures are the right choice when you do not know in advance how big a number will be. Two significant figures keep the same amount of detail in a million as in a thousandth.
Worked example: A Mass with Leading Zeros and a Count That Rounds to 35 000
Question A museum label gives the mass of a gold coin as 0.004 06 kg and says that 34 962 people visited the museum last year. (a) State the number of significant figures in 0.004 06, and write the mass correct to 2 significant figures. (b) Write the number of visitors correct to 2 significant figures and correct to 3 significant figures, and say how the two answers differ.
1.The zeros at the front of 0.004 06 only show where the decimal point is, so they are not significant. Counting starts at the first non-zero digit, which is the 4.
The zeros in front of the 4 only place the decimal point. Counting starts at the 4. 2.The digits from the 4 onwards are 4, 0 and 6. The zero between the 4 and the 6 is significant, because it lies between two significant digits. The mass has 3 significant figures.
The digits 4, 0 and 6 are significant, so the mass has 3 significant figures. 3.To round to 2 significant figures, keep the 4 and the 0 and look at the next digit, which is 6. Since 6 is 5 or more, the 0 rounds up to 1. (a) The mass is 0.0041 kg, correct to 2 significant figures.
(a) The next digit, 6, rounds the 0 up to 1: the mass is 0.0041 kg, correct to 2 significant figures. 4.In 34 962 the first two significant figures are 3 and 4. The next digit is 9, so the 4 rounds up to 5. Correct to 2 significant figures the number of visitors is 35 000. The three zeros only keep the 3 and the 5 in their places.
Keep 3 and 4. The next digit is 9, so the 4 rounds up: 35 000, correct to 2 significant figures. 5.Correct to 3 significant figures, keep 3, 4 and 9 and look at the next digit, which is 6. The 9 rounds up, so 349 hundreds become 350 hundreds and the number is again 35 000.
Keep 3, 4 and 9. The next digit is 6, so 349 hundreds become 350 hundreds: 35 000 again. 6.(b) Both answers are written 35 000, but correct to 3 significant figures the first zero is significant: it says that the number is nearer to 35 000 than to 34 900 or 35 100. Check: 34 962 is 38 away from 35 000 and 62 away from 34 900.
(b) Both answers are written 35 000, but correct to 3 significant figures the first zero is significant.
Answer: (a) 3 significant figures; 0.0041 kg; (b) 35 000 both times, but correct to 3 significant figures the first zero is significant
Common mistakes
- Counting the zeros at the front and saying that 0.004 06 has 5 or 6 significant figures. Those zeros would disappear if the mass were written as 4.06 g, so they carry no information about the precision.
- Writing 34 962 correct to 2 significant figures as 35. Rounding must not change the size of the number. The zeros in 35 000 are needed to keep the 3 in the ten-thousands place.
More significant figures and estimation problems, worked step by step →
Round at the end, not in the middle
When a calculation has several steps, keep the exact values until the last step and round only the final answer. A value rounded early carries its small error into every step after it, as the next problem shows.
Worked example: A Cable Mass Worked Out with an Early Rounding and Without
Question A 6 m length of steel cable has a mass of 1.3 kg. A bridge needs 2400 m of the same cable. (a) Ravi finds the mass of 1 m of cable, rounds it to 2 significant figures and then multiplies by 2400. Find his answer. (b) Find the mass of the cable without rounding the mass of 1 m, and find the difference between the two answers.
1.The mass of 1 m of cable is 1.3 ÷ 6 = 0.21666… kg. The digit 6 repeats without end.
The mass of 1 m of cable is 1.3 ÷ 6 = 0.21666… kg. 2.Correct to 2 significant figures this is 0.22 kg, because the third significant figure is 6 and the 1 rounds up to 2.
Correct to 2 significant figures, 0.21666… is 0.22. 3.(a) Ravi's answer is 0.22 × 2400 = 528 kg.
(a) Ravi's answer is 0.22 × 2400 = 528 kg. 4.Without rounding, keep the mass of 1 m as the fraction 1.36. The mass of the cable is 1.36 × 2400 = 1.3 × 400 = 520 kg, because 2400 ÷ 6 = 400.
Kept exact, the mass is 1.36 × 2400 = 1.3 × 400 = 520 kg. 5.(b) The mass is 520 kg, and Ravi's answer is 528 − 520 = 8 kg too large. Check: the rounding added 0.22 − 0.21666… = 0.00333… kg to each meter, and 0.00333… × 2400 = 8 kg.
(b) Rounding early makes the answer 528 − 520 = 8 kg too large.
Answer: (a) 528 kg; (b) 520 kg, so rounding early makes the answer 8 kg too large
Common mistakes
- Thinking that a rounding error of about 0.003 kg is too small to matter. It is an error in every meter, and there are 2400 meters, so it grows to 8 kg.
- Rounding 0.21666… to 0.21 by cutting off the later digits. The third significant figure is 6, so correct to 2 significant figures the value is 0.22.
More significant figures and estimation problems, worked step by step →