Significant figures — rules for operations: Worked examples

This is the "throw everything at it" page for Significant figures — rules for operations. Before we solve, we map out every kind of situation these rules must survive. Then each worked example nails a specific box in that map so you never meet a case you haven't practised.

The scenario matrix

Every problem in this topic is really one of these cells. The last column tells you which example below covers it.

Cell	What makes it tricky	Rule that governs it	Example
A Plain multiply	baseline, fewest sig figs wins	×/÷ → sig figs	Ex 1
B Plain add	fewest decimal places wins	+/− → decimals	Ex 2
C Subtraction "digit loss"	subtracting near-equal numbers kills sig figs	+/− → decimals	Ex 3
D Exact / counting number	infinite sig figs — must NOT limit answer	×/÷	Ex 4
E Mixed chain (× then +)	switch rules mid-calculation; guard digits	both	Ex 5
F Leading zeros / scientific notation	leading zeros aren't significant	counting + ×/÷	Ex 6
G Banker's-rounding edge (exactly 5)	tie-breaking to even	rounding	Ex 7
H Real-world word problem	you decide which rule, with units	both	Ex 8
I Zero / degenerate input	what happens at $0$ and at limits	edge case	Ex 9

Read the matrix once. Now we clear it, one cell at a time.

Ex 1 — Cell A: plain multiplication

Worked example

$3.24 \times 5.1 = ?$ Forecast: Guess the number of sig figs in the answer before reading on. (Which factor is weakest?)

Step 1 — Count sig figs of each factor. $3.24$ → all three digits non-zero → 3 sig figs, so $N_{1} = 3$ . $5.1$ → 2 sig figs, so $N_{2} = 2$ . (Recall $N$ = "number of sig figs of that factor", defined at the top.) Why this step? The ×/÷ rule cares only about relative precision, and sig-fig count is our proxy for that. So the very first move is always: how many sig figs does each factor carry?

Step 2 — Multiply raw, carry all digits. $3.24 \times 5.1 = 16.524$ . Why this step? Never round yet — we round once at the end. Right now we keep every digit so no information leaks out early.

Step 3 — Apply $min (N_{1}, N_{2})$ . $min (N_{1}, N_{2}) = min (3, 2) = 2$ sig figs. Round $16.524$ to 2 sig figs → the first two significant digits are $1, 6$ ; the next digit is $5$ … but $524$ after the $6$ is clearly $> 5$ , so round up: $17$ . Why this step? The answer cannot be more precise (relatively) than its weakest factor.

Verify: $17/5.1 = 3.33$ , close to $3.24$ ✓ (order of magnitude and leading digit match). Units carry through as whatever units the factors had.

Ex 2 — Cell B: plain addition

Ex 3 — Cell C: subtraction that destroys sig figs

This is the case people fumble most, so it gets a figure.

Worked example

$7.556 - 7.42 = ?$ Forecast: Both inputs look like 3–4 sig-fig numbers. Guess how many sig figs survive. (Trap!)

Step 1 — Count decimal places. $7.556$ → 3. $7.42$ → 2. The rule is decimals, so $min = 2$ . Why this step? Still an additive operation → decimal places, never sig figs.

Step 2 — Subtract raw. $7.556 - 7.42 = 0.136$ . Why this step? We subtract at full precision first and postpone rounding to the end, exactly as in the addition case — rounding early would corrupt the tiny difference we care about.

Step 3 — Round to 2 decimals. $0.136 \to 0.14$ (digit $6 > 5$ , round up). Why this step? Decimal rule wins.

The shocking part: the inputs had 4 and 3 sig figs, but the answer $0.14$ has only 2 sig figs. Subtracting two nearly-equal big numbers cancels the leading digits and leaves only the noisy tail — this is called catastrophic cancellation. Look at the figure below: the two long bars (lavender $7.556$ and mint $7.42$ ) almost overlap; only the tiny coral sliver on the right is your answer, and it inherits very few reliable digits.

Figure: horizontal bar chart. A lavender bar of length $7.556$ and a mint bar of length $7.42$ reach almost the same point, showing they are nearly equal. A short coral bar drawn from $7.42$ to $7.556$ marks the difference $0.14$ — the only part that survives the subtraction, with just 2 reliable sig figs.

Verify: $7.42 + 0.14 = 7.56 \approx 7.556$ ✓ (to 2 decimals).

Ex 4 — Cell D: an exact / counting number

Worked example A circle's radius is

$r = 2.5 cm$ (2 sig figs). Its diameter is $d = 2 r$ . Report $d$ . Forecast: Does the " $2$ " drag the answer down to… 1 sig fig? Or leave it alone?

Step 1 — Classify the "2". The $2$ in $d = 2 r$ is a definition, not a measurement. It has infinite sig figs. Why this step? Only measured numbers carry uncertainty. A defined multiplier is perfectly exact, so it can never be the "weakest" factor.

Step 2 — Multiply. $d = 2 \times 2.5 = 5.0 cm$ . Why this step? We compute the raw product first — at full precision, before any sig-fig rounding — so the rounding decision in Step 3 acts on the true value rather than a prematurely-trimmed one.

Step 3 — Sig figs = fewest among measured factors. Only $r$ is measured → 2 sig figs → $5.0 cm$ (keep the trailing zero after the decimal!). Why this step? Exact numbers are skipped in the $min$ . So the answer inherits $r$ 's 2 sig figs, not 1.

Verify: $5.0/2 = 2.5 = r$ ✓, and unit is cm (length) ✓.

Ex 5 — Cell E: mixed chain, guard digits

Worked example Compute

$(4.2 \times 3.15) + 0.087$ . Forecast: The multiply and the add obey different rules. Where do you apply each?

Step 1 — Do the multiply, keeping guard digits. $4.2 \times 3.15 = 13.23$ . By the ×-rule this "wants" 2 sig figs, but we do not round yet — we carry $13.23$ as a guard value. Why this step? Rounding mid-chain throws away information and accumulates error. Keep 1–2 extra digits until the very end.

Step 2 — Note the decimal precision the multiply earned. $4.2 \times 3.15$ : weakest factor $4.2$ has 2 sig figs, so the product is trustworthy to only 2 sig figs. The product $13.23$ has its first sig fig in the tens place (the " $1$ ") and its second in the units place (the " $3$ "). "2 sig figs" therefore means the last reliable digit sits in the units place — so the first uncertain digit is already the tenths, giving 0 reliable decimal places. Why this step? For any number just above $10$ , counting off 2 significant digits lands you on the units digit; everything to the right of the decimal point is beyond the trustworthy range. That is exactly why "2 sig figs on $13.23$ " translates into "reliable to the units place = 0 decimals" — we need this decimal precision to add correctly next.

Step 3 — Add, then apply the decimal-place rule. $13.23 + 0.087 = 13.317$ . Decimal places to keep = $min (0, 3) = 0$ → round to whole number → $13$ . Why this step? The sum can't be more precise than its worst term, which is the multiply's units-level result.

Verify: the tiny $0.087$ barely nudges $13.23$ ; final $13$ agrees with " $\approx 4.2 \times 3.15 \approx 13$ " ✓.

Ex 6 — Cell F: leading zeros & scientific notation

Worked example

$0.00680 \div 0.4 = ?$ Forecast: How many sig figs does $0.00680$ actually have? (Leading zeros are decoys.)

Step 1 — Count sig figs carefully. $0.00680$ : leading zeros ( $0.00$ ) are placeholders → not significant. The digits $6, 8, 0$ are — trailing zero after a decimal counts → 3 sig figs ( $N_{1} = 3$ ). $0.4$ → 1 sig fig ( $N_{2} = 1$ ). Why this step? Leading zeros only mark the decimal point's position; they carry no measured information. The final zero, however, is a deliberate claim of precision.

Step 2 — Divide raw. $0.00680/0.4 = 0.017$ . Why this step? We carry out the division at full precision before rounding, so the sig-fig cut in Step 3 is applied to the exact quotient rather than a pre-rounded stand-in.

Step 3 — Apply $min = 1$ sig fig. $min (N_{1}, N_{2}) = min (3, 1) = 1$ . Round $0.017$ to 1 sig fig → $0.02$ , better written $2 \times 1 0^{- 2}$ . Why this step? The single-sig-fig $0.4$ is the weak link; Scientific notation makes the lone sig fig unambiguous.

Verify: $0.02 \times 0.4 = 0.008 \approx 0.0068$ ✓ (same order of magnitude, 1-sig-fig level).

Ex 7 — Cell G: the exactly-5 tie (banker's rounding)

Worked example Round each to the stated precision: (a)

$2.45$ to 1 decimal, (b) $8.750$ to 2 sig figs. Forecast: With a lone $5$ at the cutoff, do you always round up? (No.)

Step 1 — Spot the tie. (a) cutting $2.45$ after the tenths: the dropped part is exactly " $5$ " with nothing after. (b) cutting $8.750$ after two sig figs: dropped part exactly " $50$ " = exactly half. Why this step? A trailing exact-5 is the only case where "round up" would create a systematic upward bias over many measurements.

Step 2 — Round to nearest EVEN. (a) tenths digit is $4$ (even) → keep it → $2.4$ . (b) second sig-fig digit is $7$ (odd) → bump to even $8$ → $8.8$ . Why this step? Banker's rounding forces the kept digit to be even, so upward and downward rounds cancel on average.

Verify: both kept digits ( $4$ and $8$ ) are even ✓ — that's the defining check for a correct banker's round.

Ex 8 — Cell H: real-world word problem

Worked example A runner covers

$d = 400. m$ (3 sig figs, the trailing dot means the zero is real) in $t = 62.3 s$ . Report her average speed with correct sig figs. Forecast: Is speed a multiply/divide (sig figs) or an add/subtract (decimals) situation?

Step 1 — Choose the rule. Speed $= d / t$ → division → use the sig-fig rule. Why this step? Always ask "multiplicative or additive?" first. Division = relative errors add = count sig figs.

Step 2 — Count sig figs. $400.$ → the decimal point makes all three zeros/digits count → 3 sig figs ( $N_{1} = 3$ ). $62.3$ → 3 sig figs ( $N_{2} = 3$ ). Why this step? The written decimal point is the author's signal that those trailing zeros are measured, not placeholders (see Scientific notation).

Step 3 — Compute raw, then round. $400./62.3 = 6.42054 \dots m/s$ . $min (N_{1}, N_{2}) = min (3, 3) = 3$ → $6.42 m/s$ . Why this step? Both factors give 3 sig figs, so the answer keeps 3.

Verify: $6.42 \times 62.3 = 400.0 \approx 400.$ m ✓; units $m / s$ = speed ✓ (Dimensional analysis).

Ex 9 — Cell I: zero and degenerate inputs

Worked example (a)

$0 \times 9.87 = ?$ — how many sig figs? (b) What happens to $d / t$ as $t \to 0$ ? Forecast: Does "0" have sig figs? Does the ×-rule even make sense at the edges?

Step 1 — Handle exact zero. The $0$ here is a pure zero (e.g. "no displacement"), so it is exact — infinite sig figs, like a counting number. Then $0 \times 9.87 = 0$ , and the answer is exactly $0$ , not " $0.00$ ". Why this step? A pure zero isn't a measurement with an uncertain last digit; it's an exact count, so it never limits sig figs (Cell D logic). Had the input instead been a measured $0.0$ , we would keep its decimal-place information as in Ex 2.

Step 2 — Degenerate/limiting case $t \to 0$ . In $v = d / t$ , as $t \to 0^{+}$ with $d > 0$ fixed, $v \to + \infty$ . No sig-fig rule can rescue a division by (near-)zero — the relative error $Δ t / t$ blows up. Why this step? Sig-fig bookkeeping assumes each factor's relative error is small (< a few %). Near a zero denominator that assumption fails, so the rule stops being meaningful — you must report the measurement as unresolved, not fake a precise answer.

Verify: $lim_{t \to 0^{+}} 1/ t = + \infty$ (the ratio grows without bound) ✓; and $0 \times 9.87 = 0$ exactly ✓. This connects to Error propagation — relative vs absolute: dividing by a value near its own uncertainty makes relative error explode.

Recall Recall check — one line each

×/÷ answer keeps which count? ::: The fewest significant figures. +/− answer keeps which count? ::: The fewest decimal places. Why can subtracting near-equal numbers lose sig figs? ::: Leading digits cancel, leaving only the noisy tail (catastrophic cancellation). How many sig figs does the "2" in $d = 2 r$ have? ::: Infinite — it is exact. Round $8.750$ to 2 sig figs (banker's) ::: $8.8$ (nearest even). Round $2.45$ to 1 decimal (banker's) ::: $2.4$ (nearest even). $400./62.3$ to correct sig figs ::: $6.42 m/s$ . Difference between a pure $0$ and a measured $0.0$ ? ::: Pure $0$ is exact (infinite sig figs); measured $0.0$ carries decimal-place uncertainty.

Connections

Significant figures — rules for operations — the parent rules these examples exercise.
Error propagation — relative vs absolute — why cancellation and division-by-small blow up error.
Scientific notation — disambiguates trailing zeros (Ex 6, Ex 8).
Measurement & uncertainty — the physical meaning of the "uncertain last digit".
Orders of magnitude & estimation — sanity-checking answers (used in every "Verify").
Dimensional analysis — units check in Ex 8.