Significant figures
A ruler, a scale or a stopwatch shows a number. Which of its digits did the tool really vouch for, and how many digits does an answer worked out from it deserve?
Where you are 14 in What measurement is
Why this exists
Rukmini measures the screen of her phone with a ruler marked in millimeters: 14.6\text{ cm} high and 6.8\text{ cm} wide. Her calculator multiplies them and shows 99.28. The arithmetic is right, but 99.28\text{ cm}^2 claims the area is known to the nearest hundredth of a square centimeter, and the ruler never knew either side that well.
Measurement error wrote the uncertainty of a reading with a \pm. Significant figures do the same job with no \pm: the digits a measurement is written with say how fine the tool was. This lesson covers which digits count, how to write a measurement so nobody has to guess, and two rules of thumb (not laws) for how many digits a calculated answer deserves.
Definition
The significant figures of a recorded measurement are the digits the measuring tool actually supported. They start at the first nonzero digit and end at the digit in the place of the tool’s precision (the space between its marks, or the last digit of its display), and every digit between those two counts, zeros included.
Usually only the written number is in front of you, so the definition turns into rules for reading it:
- Nonzero digits always count. 47.3 has three significant figures.
- Zeros between nonzero digits count. A kitchen scale reading of 207\text{ g} has three: the scale showed whole grams, and the tens digit it showed was 0.
- Leading zeros never count. In 0.0480\text{ kg} the zeros before the 4 only place the decimal point. In grams the same mass is 48.0\text{ g} and those zeros have vanished: a digit that disappears when the unit changes was never measured.
- Trailing zeros after a decimal point count. A lab thermometer reading of 20.0^\circ\text{C} has three. The thermometer showed tenths, and the tenth it showed was zero.
- Trailing zeros in a number with no decimal point are ambiguous. A hedge written down as 1200\text{ cm} might have been measured to the nearest centimeter, ten centimeters or meter. The 1 and the 2 certainly count; the zeros might be measured digits or only placeholders.
Scientific notation removes the doubt. In a \times 10^{n}, with 1 \le a < 10, the power of ten only places the decimal point, so every digit written in a is significant and nothing else is. The hedge becomes 1.2 \times 10^{3}, 1.20 \times 10^{3} or 1.200 \times 10^{3}\text{ cm}: two, three or four significant figures. Putting a number into the form is writing in scientific notation; the new point here is that zeros at the end of a are kept on purpose.
A number that was not measured is exact. Counts (6 jars), unit definitions (1\text{ kg} = 1000\text{ g}) and numbers that are part of a formula (the 2 in twice a length) carry no error, so they have as many significant figures as you care to write.
Intuition
Figure 1 measures the same phone screen with two rulers.
The top ruler is marked only in whole centimeters, and the edge sits nearer 15 than 14, so it reads 15\text{ cm}: two significant figures. The bottom ruler has millimeter marks, so the same edge reads 14.6\text{ cm}: three. The finer tool earned one more digit. The last significant digit is good to within half a unit of its place; nothing to the right of it was measured, and a calculator that prints digits further right is making them up.
The count also depends on the size of the thing: the width, 6.8\text{ cm}, came from the same ruler but has only two figures. So the count tracks how many of the tool’s units fit into the reading, which is the error compared with the reading itself, the relative error. A two-figure reading is 10 to 99 units, so its half-unit error is between about 0.5\% and 5\% of it; with three figures it is ten times smaller. That is why counts, not decimal places, govern multiplying and dividing.
Figure 2 shows the counting rules on written numbers.
Results
Rounding to n significant figures means rounding at the place of the nth significant figure, as in rounding to any place, and filling dropped whole-number places with zeros; if those zeros would be ambiguous, use scientific notation. To two figures and to three:
Two rules of thumb for reporting a calculation honestly follow. They are conventions that usually match the real error, not theorems.
Multiplying and dividing. Keep as many significant figures as the measurement with the fewest.
Adding and subtracting. Keep decimal places only down to the coarsest last place among the measurements: the last place that every one of them reached.
Exact numbers never limit. Counts, unit definitions and numbers in a formula are left out when finding the fewest figures or the coarsest place.
Round once, at the end. Carry every digit through each step and round only the final answer; each rounding along the way adds a fresh error (see Boundaries).
Why it is true
Adding and subtracting: errors in the same unit add. Read one length to the nearest tenth of a meter and another to the nearest hundredth. Greatest possible errors add, so their sum could be off by 0.05 + 0.005 = 0.055\text{ m}, more than five hundredths. The hundredths digit of the sum is not supported; the tenths digit is the last one with real information in it. The coarser place decides, not the number of digits, because errors in a sum are compared in the unit itself. Subtraction is the same.
Multiplying and dividing: errors as fractions add. If reading a is off by a fraction p of itself and reading b by a fraction q, then
When p and q are a few hundredths, pq is a few ten-thousandths, so the product is off by a fraction of about p + q, dominated by the rougher measurement. A quotient behaves the same way to the same accuracy. Since the count of significant figures gauges the fractional error, the product earns about as many figures as the input with the fewest.
Why only rules of thumb. Two significant figures can mean a fractional error near 0.5\% or near 5\%, and the worst cases of two measurements seldom line up, so a rule can be off by a digit either way. When the answer matters, work out the smallest and largest values the result could have, as Examples 3, 5 and 6 do.
Worked examples
Counting significant figures
How many significant figures does each recorded measurement have?
(a) 207\text{ g} on a kitchen scale; (b) 0.0480\text{ kg}; (c) 20.0^\circ\text{C} on a lab thermometer; (d) 0.007\text{ L}; (e) 1200\text{ cm} for a hedge.
Solution(a) Three. The 0 sits between the nonzero 2 and 7, so the scale showed it.
(b) Three. The zeros before the 4 only place the decimal point (in grams the reading is 48.0\text{ g}), and the final zero comes after the point, so the tool showed it.
(c) Three. The trailing zero after the point says the thermometer read tenths.
(d) One. Only the 7 counts; as 7\text{ mL} there are no zeros at all.
(e) Two, three or four, and the number alone cannot say which: the two zeros are trailing zeros in a number with no decimal point.
Showing the significant figures
(a) Olwen’s hedge from Example 1 was written down as 1200\text{ cm}. Write the length to show its significant figures if she used a tape marked in centimeters, if she read it to the nearest 10\text{ cm}, and if she paced it to the nearest meter. (b) Write the mass 0.0480\text{ kg} in grams and in milligrams, keeping its significant figures.
Solution(a) The last significant digit sits in the place of the tool’s unit. To the nearest centimeter all four digits count: 1.200 \times 10^{3}\text{ cm}, or 12.00\text{ m}, where the zeros follow a decimal point and show themselves. To the nearest 10\text{ cm}, three count: 1.20 \times 10^{3}\text{ cm}, or 12.0\text{ m}. To the nearest meter, 100\text{ cm}, only the 1 and the 2 count: 1.2 \times 10^{3}\text{ cm}, or 12\text{ m}. One plain number covered three different measurements.
(b) 0.0480\text{ kg} = 48.0\text{ g}, still three figures. In milligrams it is 48{,}000\text{ mg}, and the trailing zeros are ambiguous again; 4.80 \times 10^{4}\text{ mg} shows exactly three. Changing the unit moves the point; it cannot add a measured digit.
The area of a phone screen
Rukmini’s phone screen measures 14.6\text{ cm} by 6.8\text{ cm}, both read to the nearest millimeter. Find its area to an honest number of significant figures, and check the answer against the smallest and largest areas the readings allow.
SolutionThe calculator gives
14.6 has three significant figures and 6.8 has two. The product keeps two, so round at the ones place:
Check against the range. Each reading is good to within 0.05\text{ cm}. The smallest area uses the smallest sides and the largest uses the largest:
The true area is somewhere from about 98.2 to 100.4\text{ cm}^2. Even the ones digit is not certain, so 99 is the honest report and the .28 was noise.
Speed along a corridor
Margot measures 60.0\text{ m} along a straight school corridor with a tape, and times Lior over it with a stopwatch: 8.00\text{ s}. Find Lior’s average speed in meters per second, then in kilometers per hour.
SolutionSpeed is distance divided by time:
Both measurements have three significant figures, so the quotient keeps three, and the third is a zero that must be written: 7.50\text{ m/s}. Writing 7.5 would claim less than the tape and the stopwatch gave.
One meter per second is 3600 meters an hour, or 3.6\text{ km/h}. That 3.6 comes from unit definitions, so it is exact and does not limit the answer:
The answer keeps three figures: 27.0\text{ km/h}.
Adding lengths and subtracting temperatures
(a) Lior measures a hedge in two parts: 12.4\text{ m} with a tape marked in tenths of a meter, and 3.27\text{ m} with a tape marked in centimeters. How long is the hedge?
(b) Naveen reads a lab thermometer marked in tenths of a degree. A beaker of water is at 18.6^\circ\text{C}, and ten minutes later at 19.1^\circ\text{C}. How much did the temperature rise?
Solution(a) 12.4 + 3.27 = 15.67. The first part reaches only tenths, so the sum keeps tenths:
The range shows why. The parts could be as short as 12.35 and 3.265, or as long as 12.45 and 3.275, so the hedge is from 15.615\text{ m} to 15.725\text{ m}. The hundredths digit could be almost anything. Both inputs have three significant figures, yet counting decides nothing here: in a sum, the place is what matters.
(b) 19.1 - 18.6 = 0.5. Both readings reach tenths, so the rise is 0.5^\circ\text{C}: three significant figures in each reading, one in the answer. Subtracting nearly equal measurements cancels the digits they share. Each reading could be off by 0.05^\circ\text{C}, so the rise is from 0.4 to 0.6^\circ\text{C}, which is all one significant figure claims.
A rule that is too cautious
Olwen’s decking is a long rectangle. Its length, from a tape marked in centimeters, is 14.25\text{ m}; its width, from a folding rule marked only in tenths of a meter, is 0.9\text{ m}. Find its area by the rule for products, then check against the range.
SolutionThe calculator gives 14.25 \times 0.9 = 12.825. The width has one significant figure, so the product keeps one, at the tens place: 10. A plain 10 would be ambiguous, so write
The smallest possible area is 14.245 \times 0.85 = 12.10825 and the largest is 14.255 \times 0.95 = 13.54225. The true area is between about 12.1 and 13.5\text{ m}^2, close to 13 and nowhere near 10. One significant figure covers readings from 1 unit, where the half-unit error is 50\%, to 9 units, where it is about 5.6\%, and 0.9 sits at the good end. Reporting “about 13\text{ m}^2”, or the range itself, is more honest than 12.825 or 1 \times 10^{1}. The rule is blunt, but it does point at the fix: the width is the weak measurement, and measuring it to the centimeter would help the area most.
Boundaries
Only measured numbers are limited. In pure arithmetic, \tfrac{1}{3} = 0.333\ldots is exact; rounding it is a choice about how much to write, not a report on a tool.
Significant figures describe precision, not accuracy. A tape measure with a bent end hook can show 2.364\text{ m}, four significant figures, and still be a centimeter off. The digits say how finely the tool reads, not whether it reads true: the difference between precision and accuracy.
Ambiguous zeros stay ambiguous. Some books mark measured zeros in a whole number with a bar over the last one, or with a decimal point at the end as in 1200., but these marks are not used everywhere. Scientific notation is the form every reader understands.
Several operations go one at a time, with one rounding. For the edging round Olwen’s decking, 2(14.25 + 0.9), the sum is judged by places (tenths) and the doubling by counts, with the 2 exact. Carrying every digit gives 30.30, which rounds to 30.3\text{ m}. Rounding the sum 15.15 to 15.2 first and then doubling gives 30.4\text{ m}: the early rounding added an error of 0.05, and doubling doubled it.
Rounding a final 5. This lesson rounds a 5 up. Some fields round it to the even neighbor instead, so that many roundings do not drift upward. The number of significant figures kept is the same either way.
Where this goes next
Choosing how accurately to report a multi-step answer, with units carried through every step, is reasoning with units, and these rules are its tools. Every answer in multiplying and dividing in scientific notation ends with a leading number a calculator gives to many places; significant figures say where to round it. When a measured value is compared with an accepted one, the gap as a fraction of the accepted value is percent error, this lesson’s relative error written as a percent. Friendly numbers for estimating round for speed; here the digits dropped are the ones the tool never supported.
Exercises
Counting significant figures
How many significant figures does each recorded measurement have? If the written number cannot say, give every possibility.
- 47.3\text{ g} on a kitchen scale
- 0.0062\text{ L}
- 5.080\text{ cm}
- 9000\text{ m}
- 30.20\text{ s} on a stopwatch
- 0.100\text{ kg}
Writing measurements to show their significant figures
- Write 0.00052\text{ kg} in scientific notation. How many significant figures does it have?
- A hedge is written down as 2400\text{ cm}, measured to the nearest 10\text{ cm}. Write the length in scientific notation so that it shows its significant figures, and say how many there are.
- A kitchen scale that reads to the nearest gram shows 3000\text{ g}. Write the reading in scientific notation, and in kilograms, so that both show its significant figures.
- Round 0.04617 to two significant figures and to three.
- Round 72{,}851 to two significant figures and to three, writing each in scientific notation.
Multiplying and dividing measurements
- A phone screen measures 12.6\text{ cm} by 5.9\text{ cm}. Find its area to an honest number of significant figures.
- Pietro-Sal runs 100.0\text{ m}, measured with a tape, and a stopwatch reads 13.7\text{ s}. Find the average speed.
- Six identical jars of jam together weigh 1.38\text{ kg} on a kitchen scale. Find the mass of one jar.
- Rukmini pours the oil from a bottle into a measuring cylinder marked every 10\text{ mL} and reads 500\text{ mL}. The oil’s mass is 452\text{ g}. Find the mass of one milliliter of the oil, in grams per milliliter.
- Concrete for a driveway is laid 21.5\text{ m} long, 1.60\text{ m} wide and 0.08\text{ m} deep. Find the volume of concrete.
Adding and subtracting measurements
- A hedge is measured in three parts: 8.5\text{ m}, 11.36\text{ m} and 4.08\text{ m}. Find its total length.
- Margot pours 128.4\text{ mL} of water from one measuring cylinder, and 47\text{ mL} from another read to the nearest milliliter, into a bottle. How much water is in the bottle?
- A digital lab thermometer shows a beaker of water at 36.82^\circ\text{C}. Later a glass thermometer marked in tenths of a degree reads 37.1^\circ\text{C}. Find the rise in temperature, and say how many significant figures it has.
- An empty bowl weighs 386\text{ g} on a kitchen scale. With flour in it, a second scale reads 1.250\text{ kg}. Find the mass of the flour in grams.
Reasoning
- Explain why the zeros in 0.0062 are not significant figures but the final zero in 5.080 is.
- Naveen says that Rukmini’s 14.6\text{ cm} and 6.8\text{ cm} came from the same ruler, so they should have the same number of significant figures. Explain what the two readings do share, and why their counts still differ.
- Explain why the rule for sums looks at decimal places, while the rule for products counts significant figures.
- A label says a kitchen worktop is 4500\text{ mm} long. List the possible numbers of significant figures. For each, write the length in scientific notation and give the range of true lengths it allows.
- 9.9 and 1.1 both have two significant figures. For each, find its greatest possible error as a fraction of the reading, and compare the two. What does this say about the rule for products?
Answers
- Three. All three digits are nonzero.
- Two. The leading zeros only place the decimal point; the 6 and the 2 count.
- Four. The first zero lies between nonzero digits, and the final zero comes after the decimal point.
- One, two, three or four. The 9 counts; the three zeros are trailing zeros in a number with no decimal point.
- Four. The final zero comes after the decimal point, so the stopwatch showed it.
- Three. The zero before the decimal point is a leading zero; the 1 and the two zeros after it count.
- 5.2 \times 10^{-4}\text{ kg}, two significant figures. The leading zeros disappear into the power of ten.
- 2.40 \times 10^{3}\text{ cm}, three significant figures: to the nearest 10\text{ cm}, the digits down to the tens place count.
- 3.000 \times 10^{3}\text{ g} and 3.000\text{ kg}, four significant figures each. To the nearest gram the ones digit was measured, so every zero counts.
- 0.046 and 0.0462. To two figures the next digit, 1, rounds down; to three, the next digit, 7, rounds the 1 up to 2.
- 7.3 \times 10^{4} and 7.29 \times 10^{4}. 72{,}851 is nearest 73{,}000 and, to three figures, nearest 72{,}900.
- 74\text{ cm}^2. 12.6 \times 5.9 = 74.34, and 5.9 has two significant figures.
- 7.30\text{ m/s}. 100.0 \div 13.7 = 7.299\ldots; the time has three significant figures and the distance four, so keep three, writing the final zero.
- 0.230\text{ kg}. 1.38 \div 6 = 0.23. The 6 is an exact count, so the answer keeps the three figures of 1.38.
- 0.90\text{ g/mL}. 452 \div 500 = 0.904. Read to the nearest 10\text{ mL}, 500 has two significant figures, so keep two: 0.90, with its zero written.
- 3\text{ m}^3. 21.5 \times 1.60 \times 0.08 = 2.752, and the depth 0.08 has one significant figure.
- 23.9\text{ m}. 8.5 + 11.36 + 4.08 = 23.94, and the coarsest last place is the tenths, from 8.5.
- 175\text{ mL}. 128.4 + 47 = 175.4, and 47 reaches only the ones place.
- 0.3^\circ\text{C}, one significant figure. 37.1 - 36.82 = 0.28, and the glass thermometer reached only the tenths.
- 864\text{ g}. 1.250\text{ kg} = 1250\text{ g} read to the nearest gram, and 1250 - 386 = 864; both readings reach the ones place.
- The zeros in 0.0062 only place the decimal point: in milliliters the volume is 6.2\text{ mL} and they are gone. The final zero in 5.080 comes after the decimal point, so the tool showed it: the reading was 5.080, not 5.081 or 5.079.
- They share their last place: both are good to the nearest tenth of a centimeter. But the height is 146 tenths and the width only 68, so the height takes three digits and the width two. The count tracks how many of the tool’s units fit into the reading, which depends on the size of the thing measured.
- In a sum, errors are compared in the unit itself: a half-tenth error in one term passes straight into the total, so the coarsest place decides. In a product, what passes through is each error as a fraction of its reading, and the count of significant figures gauges that fraction.
- Two: 4.5 \times 10^{3}\text{ mm}, from 4450 to 4550\text{ mm}. Three: 4.50 \times 10^{3}\text{ mm}, from 4495 to 4505\text{ mm}. Four: 4.500 \times 10^{3}\text{ mm}, from 4499.5 to 4500.5\text{ mm}. Each range is half a unit either side of 4500, the unit being 100, 10 or 1\text{ mm}.
- Half a unit in the last place is 0.05 for both. 0.05 \div 9.9 \approx 0.0051 and 0.05 \div 1.1 \approx 0.045, nine times as much, since 9.9 \div 1.1 = 9. The rule treats both as two significant figures, so it can misjudge a product’s error by close to a factor of ten, which is why it is a rule of thumb.