Uncertainty in Arithmetic Calculations

You already know how to count significant figures and how to round a result to the right number of digits. But there is a deeper question: when you multiply, divide, add, or subtract measured values, how does the uncertainty in each measurement travel into the final answer? Knowing the rules for this “error propagation” tells you not just how many digits to keep, but how confident you can actually be in your result.

How Errors Combine in Products and Quotients

When you multiply two measured quantities, each one brings its own uncertainty into the result. The rule for combining these uncertainties is surprisingly straightforward: convert each absolute uncertainty into a percentage, then add the percentages.

Think of it this way. If one measurement is uncertain by 0.6% and another by 1%, then the product of the two could be off by as much as 1.6% in the worst case. The percentage errors simply stack up.

Worked Example: Area of a Rectangular Sheet

Problem: The length and breadth of a thin rectangular sheet are measured with a metre scale as 16.2 cm and 10.1 cm. Each measurement has three significant figures. Find the area and its uncertainty.

Solution:

Start by expressing the uncertainty in each measurement. Since the metre scale reads to the nearest 0.1 cm, each value has an absolute uncertainty of $\pm 0.1$ cm.

Length:

$l = 16.2 \pm 0.1 \text{ cm}$

Convert the absolute uncertainty to a percentage:

$\text{Percentage uncertainty in } l = \frac{0.1}{16.2} \times 100 \approx 0.6\%$

So you can write:

$l = 16.2 \text{ cm} \pm 0.6\%$

Breadth:

$b = 10.1 \pm 0.1 \text{ cm}$

$\text{Percentage uncertainty in } b = \frac{0.1}{10.1} \times 100 \approx 1\%$

$b = 10.1 \text{ cm} \pm 1\%$

Combining the errors in the product:

The raw product is:

$l \times b = 16.2 \times 10.1 = 163.62 \text{ cm}^2$

Since this is multiplication, the percentage uncertainties add:

$\text{Percentage uncertainty in area} = 0.6\% + 1\% = 1.6\%$

Convert the combined percentage back to an absolute uncertainty:

$\text{Absolute uncertainty} = 163.62 \times \frac{1.6}{100} \approx 2.6 \text{ cm}^2$

Round the uncertainty to one significant figure ($\approx 3$ cm$^2$) and then round the central value to match:

$l \times b = 164 \pm 3 \text{ cm}^2$

Here, $3$ cm$^2$ is the uncertainty (or error) in the estimated area of the rectangular sheet.

Why Subtraction Can Destroy Significant Figures

For products, a neat rule holds: if each input value is specified to $n$ significant figures, the result is generally valid to $n$ significant figures as well. But this comforting rule breaks down badly for subtraction.

When you subtract two numbers that are close in value, the leading digits cancel each other out. What remains is built almost entirely from the uncertain trailing digits of the original measurements. The result therefore has far fewer reliable figures than either input had.

Worked Example: Mass Difference

Consider $12.9 \text{ g} - 7.06 \text{ g}$. Both values are specified to three significant figures, so you might expect a three-significant-figure answer.

The arithmetic gives $5.84$ g. But wait. The rule for subtraction says the result keeps the fewest number of decimal places, not significant figures. The value 12.9 has only one decimal place, while 7.06 has two. So the result is valid to just one decimal place:

$12.9 - 7.06 = 5.8 \text{ g}$

That is only two significant figures, even though both inputs had three. The cancellation of the leading digits ate up one figure’s worth of reliable information.

This is an important practical lesson: subtraction of close-valued measurements can be a precision killer. Whenever you see a small difference between two large numbers, be extra cautious about the reliability of the result.

Relative Error Depends on the Number Itself

You might assume that two measurements with the same number of significant figures carry the same level of reliability. That is not always true. The relative error (percentage error) of a measurement depends not only on how many significant figures it has, but also on the magnitude of the number.

Seeing This with an Example

Consider two mass measurements, both accurate to $\pm 0.01$ g (the same absolute error):

Measurement 1: $m = 1.02$ g

$\text{Relative error} = \frac{0.01}{1.02} \times 100 \approx \pm 1\%$

Measurement 2: $m = 9.89$ g

$\text{Relative error} = \frac{0.01}{9.89} \times 100 \approx \pm 0.1\%$

Both values have three significant figures. Both have the same absolute error. Yet the relative error differs by a factor of ten. The smaller number (1.02 g) is proportionally ten times less precise than the larger one (9.89 g), even though both came from the same measuring instrument with the same resolution.

This happens because relative error is a ratio: the same absolute uncertainty forms a larger fraction of a smaller number. It is worth keeping this in mind when planning experiments. Measuring a quantity that is very small on the scale of your instrument’s resolution will always give a poor relative error, no matter how many decimal places you write down.

Keeping Extra Digits in Multi-Step Calculations

Here is a practical trap that catches many students. Suppose a calculation involves several steps, and you round to the correct number of significant figures at each intermediate stage. Sounds reasonable, but each rounding introduces a tiny error, and over multiple steps these errors accumulate. By the end, the final answer can drift noticeably from the true value.

The Reciprocal Example

This classic example shows just how quickly things go wrong.

Start with $9.58$ (three significant figures). Calculate its reciprocal and round to three significant figures:

$\frac{1}{9.58} = 0.10438... \approx 0.104$

Now take the reciprocal of that rounded result, again to three significant figures:

$\frac{1}{0.104} = 9.6153... \approx 9.62$

You started with 9.58 and ended with 9.62. The original value has been lost because of the rounding you did at the intermediate step.

Now repeat the calculation, but this time keep one extra digit at the intermediate stage:

$\frac{1}{9.58} = 0.10438... \approx 0.1044 \quad \text{(four digits instead of three)}$

$\frac{1}{0.1044} = 9.5785... \approx 9.58$

The original value comes back perfectly. That single extra digit preserved enough information to prevent the rounding drift.

The Practical Rule

In any multi-step calculation, carry one more significant figure in every intermediate result than the number of digits you need in the final answer. Only round to the proper number of significant figures at the very last step. This small habit costs almost no effort but saves you from the slow accumulation of rounding errors that can silently corrupt your answer.