Rounding Off the Uncertain Digits

When you multiply, divide, or combine measured values, your calculator spits out a long trail of digits. Most of those digits are meaningless because they go far beyond the precision of the original measurements. Rounding trims the result back to the number of digits that are actually justified, so the answer honestly reflects what the measurements can support.

You already know how to count significant figures and how to decide how many figures a result should have after an arithmetic operation. This topic covers the next practical step: how to actually perform the rounding, especially in the tricky borderline cases.

The Two Standard Rules

Rounding boils down to looking at the first digit you are about to drop and making a simple decision.

Rule 1: Dropped digit is greater than 5

If the first insignificant digit (the one you need to remove) is greater than 5, raise the last digit you are keeping by 1.

Example: Round 2.746 to three significant figures.

• The digits you keep are 2, 7, and 4.
• The next digit is 6, which is greater than 5.
• Raise the 4 to 5.
• Result: 2.75

Rule 2: Dropped digit is less than 5

If the first insignificant digit is less than 5, leave the last kept digit unchanged and simply drop everything after it.

Example: Round 1.743 to three significant figures.

• The digits you keep are 1, 7, and 4.
• The next digit is 3, which is less than 5.
• The 4 stays as it is.
• Result: 1.74

These two rules handle most cases without any confusion. The interesting situation arises when the dropped digit is exactly 5.

The Even-Odd Convention for the Digit 5

What should you do when the digit you are dropping is precisely 5? Rounding up every time would introduce a systematic upward bias across many calculations. Rounding down every time would push things the other way. The convention adopted by physicists and engineers splits the difference based on whether the preceding digit is even or odd.

The rule:

• If the preceding digit is even, simply drop the 5. The preceding digit stays as it is.
• If the preceding digit is odd, raise it by 1 (making it even) and then drop the 5.

In either case, the result lands on an even digit. Over a large number of roundings, half will round up and half will round down, so the bias cancels out. This is sometimes called banker’s rounding or round half to even.

Walking Through the Examples

Example 1: Round 2.745 to three significant figures.

• Digits to keep: 2, 7, and 4.
• The digit being dropped is 5.
• The preceding digit is 4, which is even.
• Drop the 5 without changing anything.
• Result: 2.74

Example 2: Round 2.735 to three significant figures.

• Digits to keep: 2, 7, and 3.
• The digit being dropped is 5.
• The preceding digit is 3, which is odd.
• Raise 3 to 4.
• Result: 2.74

Notice that both numbers land on the same rounded value (2.74), but through different paths. That is the even-odd rule doing its job: it always steers toward the nearest even digit.

Multi-Step Calculations: Keep One Extra Digit

In a calculation that involves several steps, rounding at every intermediate stage introduces a small error each time. These errors pile up and can shift the final answer by a noticeable amount.

The practical guideline is straightforward: during intermediate steps, keep one digit more than the number of significant figures you need in the final answer. Do all the algebra and arithmetic with that extra digit retained, and only round to the proper number of significant figures at the very end.

This single habit prevents the slow drift of accumulated rounding errors without requiring you to carry all 10 or 12 digits your calculator shows.

Rounding Well-Known Constants

Some quantities are known to a very high number of significant figures. The speed of light in vacuum, for instance, is known as $2.99792458 \times 10^8$ m/s (nine significant figures). For most classroom problems, carrying all nine digits is unnecessary, so it is commonly rounded to $3 \times 10^8$ m/s. How far you round depends on the precision your problem demands.

Similarly, the value of $\pi = 3.1415926...$ is known to billions of digits. In practice, you pick $3.14$ or $3.142$ depending on how many significant figures the other quantities in your calculation have. The key point is that $\pi$ is an exact mathematical constant, not a measured value, so it never limits the precision of a result. The same goes for exact factors in formulas, like the $2\pi$ in $T = 2\pi\sqrt{\dfrac{L}{g}}$. Both $2$ and $\pi$ carry infinite precision and will never be the weakest link in a calculation.

Worked Example 1: Surface Area and Volume of a Cube

Problem: Each side of a cube is measured to be 7.203 m. Find the total surface area and the volume, reported to the appropriate number of significant figures.

Solution:

The measured length 7.203 m has four significant figures. Since the surface area and volume are calculated from this single measurement, the results must also be rounded to four significant figures.

Surface area:

$\text{Surface area} = 6 \times (7.203)^2 \text{ m}^2$

First, compute the square:

$(7.203)^2 = 51.883209 \text{ m}^2$

Then multiply by 6:

$6 \times 51.883209 = 311.299254 \text{ m}^2$

Round to four significant figures. The first four digits are 3, 1, 1, and 2. The next digit is 9, which is greater than 5, so raise the 2 to 3:

$\text{Surface area} = 311.3 \text{ m}^2$

Volume:

$\text{Volume} = (7.203)^3 \text{ m}^3$

$= 7.203 \times 7.203 \times 7.203 = 373.714754... \text{ m}^3$

Round to four significant figures. The first four digits are 3, 7, 3, and 7. The next digit is 1, which is less than 5, so the 7 stays:

$\text{Volume} = 373.7 \text{ m}^3$

Worked Example 2: Expressing Density with Correct Significant Figures

Problem: 5.74 g of a substance occupies $1.2 \text{ cm}^3$. Express the density with the correct number of significant figures.

Solution:

Count the significant figures in each measurement:

• Mass = 5.74 g has three significant figures
• Volume = $1.2 \text{ cm}^3$ has two significant figures

Density is mass divided by volume (a division), so the result takes the smaller significant-figure count, which is two.

$\text{Density} = \frac{5.74}{1.2} = 4.7833... \text{ g cm}^{-3}$

Round to two significant figures. The first two digits are 4 and 7. The next digit is 8, which is greater than 5, so raise 7 to 8:

$\text{Density} = 4.8 \text{ g cm}^{-3}$