Significant figures and magnitude

Ask a student to measure the length of a desk with a metre scale and a caliper, and you get two different-looking numbers even though both are "correct" — say $75.4\ \text{cm}$ from the scale and $75.42\ \text{cm}$ from the caliper. Neither is wrong; they simply carry different amounts of reliable information. Significant figures are the bookkeeping system physics uses to say, honestly, how much a number can be trusted, and a companion idea, order of magnitude, lets you compare wildly different sizes — an atom and a galaxy — using nothing but a power of ten. Together these two ideas run through every measurement, every calculation, and every reported answer in the "Physical World and Measurement" unit: get them wrong and a perfectly correct calculation can still lose marks for over-claiming precision.

The core insight

Significant figures record exactly how precisely a quantity is known (reliable digits plus one estimated digit, with fixed rules for zeros and for rounding sums versus products), while order of magnitude rounds that same quantity to the nearest power of ten for quick comparison of scale.

The core idea

Significant figures in a measured number are all the digits known with certainty, plus one final digit that is estimated (uncertain). A length written as $12.3\ \text{cm}$ carries 3 significant figures; written as $12.30\ \text{cm}$ it carries 4 — the extra trailing zero is a genuine claim that the instrument resolved the hundredths place, not decoration.

Counting rules. Non-zero digits always count. Zeros sandwiched between non-zero digits always count ( $1005$ has 4 s.f.). Leading zeros never count — they only fix the decimal point's position ( $0.0025$ has 2 s.f.). Trailing zeros after a decimal point always count ( $1.500$ has 4 s.f.). Trailing zeros with no decimal point are ambiguous in plain notation ( $1500$ could be 2, 3, or 4 s.f.); scientific notation removes the ambiguity, since $1.500\times10^{3}$ is unambiguously 4 s.f. Multiplying or dividing by an exact counting number (like "2" in "twice the radius") never reduces significant figures.

Rounding after arithmetic. The two arithmetic operations follow different rules, and mixing them up is the single most common exam slip:

Addition and subtraction: round the result to the same number of decimal places as the least precise term.
Multiplication and division: round the result to the same number of significant figures as the least precise factor.

Standard convention for the boundary case: if the digit being dropped is $>5$ , round up; if $<5$ , round down; if exactly $5$ with nothing after it, round to the nearest even digit, which prevents a systematic upward bias when many such roundings accumulate.

Order of magnitude expresses the approximate size of a quantity as a single power of ten, obtained from its scientific-notation form:

x = a \times 10^{n}, \qquad 1 \le a < 10.

where: $x$ is the quantity, $a$ is the coefficient, and $n$ is an integer exponent. The order of magnitude is $10^{n}$ if $a < \sqrt{10} \approx 3.16$ , and $10^{n+1}$ if $a \ge \sqrt{10}$ — the coefficient is rounded to whichever power of ten it sits closer to on a logarithmic scale, not on the ordinary number line. This single-power-of-ten shorthand is what lets a physicist compare an atomic radius ( $10^{-10}\ \text{m}$ ) with a galactic distance ( $10^{21}\ \text{m}$ ) at a glance, without carrying full decimal precision.

Figure

Two rulers comparing a length recorded as 12.3 centimetres with three significant figures against the same length recorded as 12.30 centimetres with four significant figures, showing the finer instrument resolves one more digit

Derivation AWhy the order-of-magnitude threshold is the square root of ten

The goal is to show, from the definition of order of magnitude as "the power of ten a quantity is logarithmically closest to," that the coefficient in scientific notation should be rounded up to the next power exactly when it reaches $\sqrt{10} \approx 3.162$ , and not at the halfway point $5$ that ordinary linear rounding would suggest — the reason the electron-mass coefficient $9.1$ rounds up to $10$ while a coefficient like $2$ or $3$ stays down, and the rule this article's core idea uses without proof.

Step 1. Write the quantity in scientific notation, $x = a \times 10^{n}$ with $1 \le a < 10$ . On an ordinary (linear) number line, rounding $a$ to the nearer of $1$ or $10$ would flip at the midpoint $a = 5.5$ . But order of magnitude is a logarithmic idea — it asks which power of ten, $10^{n}$ or $10^{n+1}$ , is closer to $x$ when sizes are compared multiplicatively rather than by subtraction.

Step 2. Take logarithms (base 10) of the quantity: $\log_{10} x = n + \log_{10} a$ . Since $1 \le a < 10$ , the fractional part $\log_{10} a$ lies between $0$ and $1$ . Rounding $\log_{10} x$ to the nearest integer is precisely what "nearest power of ten" means, so the order of magnitude is $10^{n}$ if $\log_{10} a < 0.5$ , and $10^{n+1}$ if $\log_{10} a \ge 0.5$ .

Step 3. Convert the threshold back out of logarithms. Setting $\log_{10} a = 0.5$ and solving gives $a = 10^{0.5} = \sqrt{10} \approx 3.162$ . So the switch-over coefficient is $\sqrt{10}$ , not $5$ — a direct consequence of comparing sizes on a log scale rather than a linear one.

Step 4. Check against the worked case already used in the core idea: the electron's mass is $9.1\times10^{-31}\ \text{kg}$ , so $a = 9.1 \ge 3.162$ , and the order of magnitude rounds up to $10^{-30}\ \text{kg}$ . A coefficient like $a = 2.0$ stays below $3.162$ , so a quantity such as $2.0\times10^{6}$ keeps the order of magnitude $10^{6}$ , not $10^{7}$ .

Worked example

Problem: A rectangular plate has length $l = 5.62\ \text{cm}$ (3 s.f.) and breadth $b = 2.3\ \text{cm}$ (2 s.f.). Find the area to the correct number of significant figures, state its order of magnitude, and then find the plate's perimeter to the correct number of decimal places.

Step 1. Multiply the length and breadth directly, keeping every digit the calculator gives: $A = l \times b = 5.62 \times 2.3 = 12.926\ \text{cm}^{2}$ .

Step 2. Apply the multiplication rule: round to match the least number of significant figures among the factors. Here $b = 2.3$ has only 2 s.f., so the area must be reported to 2 s.f.: $A \to 13\ \text{cm}^{2}$ .

Step 3. Find the order of magnitude of the area. Write $13 = 1.3 \times 10^{1}$ . Since $1.3 < \sqrt{10} \approx 3.16$ , the exponent is not rounded up, giving order of magnitude $10^{1}\ \text{cm}^{2}$ .

Step 4. Now the perimeter, which is addition, not multiplication: $P = 2(l+b) = 2(5.62 + 2.3)$ . Inside the bracket, the addition rule applies first — $5.62$ has 2 decimal places, $2.3$ has 1, so the sum $7.92$ must be rounded to 1 decimal place: $5.62 + 2.3 \to 7.9\ \text{cm}$ . Doubling an exact integer does not change significant figures or decimal places, so $P = 2 \times 7.9 = 15.8\ \text{cm}$ .

The area is $13\ \text{cm}^{2}$ (order of magnitude $10^{1}\ \text{cm}^{2}$ ) and the perimeter is $15.8\ \text{cm}$ — two different rounding rules applied correctly within the same problem.

Common traps

Reporting every digit a calculator displays, forgetting that the least precise input caps the precision of the final answer.
Swapping the two rounding rules — using "match significant figures" for a sum, or "match decimal places" for a product — the single most common exam error on this topic.
Treating leading zeros as significant: in $0.0025$ , only the digits $2$ and $5$ count (2 s.f.); the zeros merely place the decimal point.
Confusing order of magnitude with number of significant figures: order of magnitude is about overall size (the power of ten), significant figures are about precision; a number can have many significant figures and still sit at a small order of magnitude, or vice versa.

Ask the tutor: unsure whether your final answer should keep 2 or 3 significant figures, or want to practise finding the order of magnitude of real physical quantities — ask the AI tutor to check your rounding step by step.

Why does multiplying by an exact number like 2 not reduce significant figures, but multiplying by a measured 2.0 does?

An exact (counting) number such as "2" in "twice the radius" has no uncertainty at all — it is not a measurement, so it never limits precision, and the answer's significant figures come only from the measured quantity. But $2.0$ written as a measured value carries exactly 2 significant figures of its own uncertainty, so if it is the least precise factor in a multiplication, it does cap the result to 2 s.f. The distinction is whether the number came from counting/definition or from an instrument reading.

Why is the rounding rule different for addition than for multiplication?

Because decimal places and significant figures measure different things. In addition, the absolute uncertainty of each term (fixed by its last reliable decimal place) simply adds up, so the sum can only be as precise, in decimal places, as its least precise term. In multiplication, it is the relative (percentage) uncertainty that combines, and relative uncertainty tracks the number of significant figures, not decimal places — so the product is capped by the factor with the fewest significant figures.

Is order of magnitude the same as rounding to one significant figure?

Not quite, though they look similar. Rounding to 1 significant figure keeps a single leading digit unchanged (so $9.1\times10^{-31}$ rounds to $9\times10^{-31}$ ). Order of magnitude discards even that digit and reports only the power of ten, using the $\sqrt{10} \approx 3.16$ threshold to decide whether to round the exponent up — so $9.1\times10^{-31}$ has order of magnitude $10^{-30}$ , not $9\times10^{-31}$ . Order of magnitude is a coarser, purely exponential comparison of scale.

Significant figures and magnitude

The core idea

Worked example

Common traps

Before this

Where this leads