Significant figures and rounding rules: Visual walkthrough

Every measured number is really a little cloud, not a sharp point. We are going to see that cloud, watch it grow when we do arithmetic, and read the significant-figure rules straight off the geometry.

Step 1 — A measurement is a fuzzy interval, not a point

WHAT. Take a ruler marked every millimetre and measure a pencil. You can read confidently to the mm and guess one digit past it. So " $4.7$ cm" does not mean the exact real number $4.7000 \dots$ ; it means "somewhere near $4.7$ ."

WHY. Before we can talk about "how many digits are honest," we must picture what a digit claims. The last written digit is the uncertain one — it marks the edge of what you can honestly see.

PICTURE. Look at the red band below. The pencil tip is not a knife-edge; it sits inside a band of width roughly $\pm 0.05$ cm. The number $4.7$ is the centre of that band; the band itself is the honesty.

Figure — Significant figures and rounding rules

$4.7$ — the reported value (centre of the band).
$\pm 0.05$ — the half-width $Δ (4.7)$ ; the last digit ( $7$ ) is where this fuzz lives.

Related idea in the vault: Accuracy vs precision — precision is exactly how narrow this band is.

Step 2 — Writing more digits lies about the band

WHAT. Suppose we rewrite the same pencil as " $4.700$ cm." That symbol claims the band is now only $\pm 0.0005$ cm — a band 100 times narrower.

WHY. We must see why extra digits are forbidden: each digit you append shrinks the promised band by a factor of ten. If your ruler cannot support that shrinkage, the extra digit is an invention.

PICTURE. The wide black band is what the ruler truly gives. The tiny red band is what " $4.700$ " pretends. The red band does not fit inside the truth — it claims knowledge you do not have.

So: the number of significant figures = the number of digits whose band the instrument can honestly support. That single sentence is the whole topic; everything below is what happens when we do arithmetic with these bands.

Step 3 — Multiplying: bands grow by relative fuzz

WHAT. Let a rectangle have side $A = 6.38$ cm and side $B = 2.0$ cm. The area is $A \times B$ . What band does the area carry?

WHY. We introduce relative uncertainty — the fuzz as a fraction of the value — because multiplication scales things proportionally, and proportions are exactly what fractions track.

$relative fuzz of the answer \frac{Δ ( area )}{area} \approx relative fuzz of A \frac{Δ A}{A} + relative fuzz of B \frac{Δ B}{B}$

$Δ A, Δ B$ — the band half-widths (absolute fuzz) of each side, exactly the $Δ$ from Step 1.
$\frac{Δ A}{A}$ — that fuzz as a fraction of the side, e.g. $\frac{0.005}{6.38} \approx 0.08%$ .
The sum tells us the answer's relative fuzz is dominated by the term with the bigger fraction.

PICTURE. The rectangle's fuzzy edges make a fuzzy border (red). Side $B = 2.0$ has the fatter relative border ( $\pm 0.05/2.0 = 2.5%$ ) versus $A$ 's $0.08%$ . The red border is thick on the short side — that side controls the total.

Step 4 — Adding: bands grow by absolute fuzz

WHAT. Now lay two rods end to end: $12.11$ cm and $0.3$ cm. The total length is a sum, not a scaling.

WHY. When you stack lengths, the fuzz you care about is the absolute half-width in the same units — because bands add end-to-end, physically. Fractions no longer matter; the raw width does.

$Δ (sum) \approx \pm 0.005 Δ (12.11) + \pm 0.05 Δ (0.3)$

$Δ (12.11), Δ (0.3)$ — again the band half-widths from Step 1, one per rod.
The $\approx$ is here for the same reason as Step 3: worst-case bands add almost exactly, and the tiny correction is ignorable.
$Δ (12.11) = \pm 0.005$ cm — known to the hundredths place.
$Δ (0.3) = \pm 0.05$ cm — known only to the tenths place; a fatter absolute band.
The sum's fuzz is dominated by the widest absolute band, i.e. the term with the fewest decimal places.

PICTURE. Two rods on an axis. The first has a hair-thin red fuzz; the second a fat red fuzz. When laid nose-to-tail the total fuzz is basically the fat one — you cannot claim the sum to the hundredths place when one rod is only known to the tenths.

Step 5 — Rounding is trimming the answer back to its band

WHAT. The raw product $6.38 \times 2.0 = 12.76$ carries fake sharpness. To round to 2 sig figs we look at the first dropped digit (the decider) — here the $7$ .

WHY. Rounding is not decoration; it erases digits that live outside the honest band. The decider tells us which way the true value most likely leans.

PICTURE. A number line zoomed on $12.76$ . The two candidate keepers are $12$ and $13$ . The value sits past the halfway mark (red midpoint), so it snaps up to $13$ .

Round $12.76 \to 13$ because the decider $7 > 5$ .

Step 6 — The exact-halfway case and the fair coin

WHAT. What if the decider is exactly $5$ with nothing after, e.g. round $2.5$ and $3.5$ to whole numbers? Neither candidate is nearer.

WHY. If we always pushed $5$ upward, then across thousands of calculations we would drift the average upward — a systematic bias. To stay fair, we flip toward the even neighbour ("round half to even").

PICTURE. $2.5$ sits dead-centre between $2$ and $3$ ; the red arrow points to the even side → $2$ . $3.5$ sits centred between $3$ and $4$ ; the even side is → $4$ . One went down, one went up — the bias cancels.

Step 7 — Degenerate cases: exact numbers and zeros

WHAT. Some inputs have no band at all.

WHY. A counted or defined number (" $12$ eggs", " $1000$ m per km", the $2$ in $2 π r$ ) is exact — its band half-width $Δ$ is zero. A zero-width band added to any other band changes nothing, so exact numbers never limit the answer.

PICTURE. Three bands on one axis: a measured value with a real red band; an exact number drawn as a red point (zero width); and their combination — the point does not widen the band at all.

Also the placeholder zeros. In $0.0042$ , the leading zeros give the value location, not width — they are outside the band-defining digits, so they are not significant. Rewriting as $4.2 \times 1 0^{- 3}$ makes the band-carrying digits ( $4, 2$ ) obvious. Contrast a sandwiched zero ( $205$ ): you cannot reach the $5$ without crossing that $0$ , so it is inside the measured band and is significant.

The one-picture summary

Everything collapses into one image: a measurement is a band; multiplication adds bands as fractions (→ fewest sig figs), addition adds bands as widths (→ fewest decimal places), rounding trims to the band edge, and a fair coin handles the dead-centre $5$ .

Recall Feynman retelling of the whole walkthrough

Every number you measure is really a little smudge, not a dot. Its width is set by your ruler, and the last digit you write sits right at the smudge's edge — writing extra digits is claiming a skinnier smudge than you really have, which is lying. Now do maths. When you multiply, the smudges combine as percentages, so whichever number is the sloppiest percentage-wise rules the answer — and "percentage sloppiness" is just significant figures, so you keep the fewest sig figs. When you add, the smudges stack end to end in real units, so whichever number is fuzzy at the biggest place value rules — that's decimal places, so you keep the fewest decimals. To shrink an answer back to its honest smudge you round: look at the first digit you're throwing away; past halfway you go up, before halfway you go down. And if you land exactly on the halfway 5 with nothing after, you flip toward the even neighbour so you don't secretly cheat upward every single time. Counted things like "12 eggs" have zero smudge, so they never make your answer fuzzier. That's the entire topic — in smudges.

Recall Quick self-test

Multiplication keeps which count? ::: The fewest significant figures. Addition keeps which count? ::: The fewest decimal places. Why two different rules? ::: Multiplying adds relative fuzz (sig figs); adding stacks absolute fuzz (decimal places). Round $2.5$ to a whole number, and why? ::: $2$ — round half to even to avoid upward bias. Do exact numbers limit precision? ::: No — their uncertainty band has zero width. Sig figs in $0.0042$ ? ::: $2$ — leading zeros are placeholders, outside the band.