A rounded measurement represents an interval of possible values. If a strip is recorded as 8.4 cm to the nearest tenth of a centimetre, its actual length need not be exactly 8.4 cm. Rounding and error bounds questions ask you to recover that interval and then decide how the uncertainty affects a calculation.
Begin with the stated accuracy, not with the number of digits that happen to be printed. The number 300 could be exact, rounded to the nearest unit or rounded to the nearest hundred. Those descriptions lead to different bounds.
First identify the rounding unit
Decimal places count positions after the decimal point. Significant figures count from the first nonzero digit. Consequently, rounding 0.006284 to three decimal places gives 0.006, while rounding it to three significant figures gives 0.00628.
The rounding unit is the value of the final retained position. For two decimal places it is 0.01, so the half-unit is 0.005. For a value recorded to the nearest ten, the rounding unit is 10 and the half-unit is 5.
For positive measurements in this lesson, use the school convention that a halfway value rounds up. If r is the rounded measurement and h is the rounding unit, the original value x satisfies:
r − h/2 ≤ x < r + h/2.
The lower endpoint is included and the upper endpoint is excluded. Other tie-breaking conventions exist, so follow any rounding rule stated in a question. Do not apply this interval to a value obtained by truncation, which has a different error pattern.
Worked example 1: recover a measurement interval
Question: A length L is recorded as 8.4 cm to the nearest 0.1 cm. Find its bounds and maximum absolute rounding error.
- The rounding unit is 0.1 cm.
- Half of that unit is 0.05 cm.
- The lower bound is 8.4 − 0.05 = 8.35 cm.
- The upper bound is 8.4 + 0.05 = 8.45 cm.
Therefore 8.35 ≤ L < 8.45 cm. The maximum absolute rounding error is 0.05 cm. The lower endpoint 8.35 rounds to 8.4; the upper endpoint 8.45 rounds to 8.5, so it cannot be included.
Writing 8.3 ≤ L < 8.5 uses a whole rounding unit on each side and makes the interval twice as wide as it should be. Writing L = 8.4 loses the measurement uncertainty altogether.
Choose extremes from the operation
Before substituting any endpoints, ask what makes the result larger. For positive independent quantities, increasing either factor increases a product. Increasing a denominator decreases a quotient. These observations determine the correct pairing of bounds.
| Expression | Lower-bound calculation | Upper-bound calculation |
|---|---|---|
| a + b | Lower a + lower b | Upper a + upper b |
| a − b | Lower a − upper b | Upper a − lower b |
| ab | Lower a × lower b | Upper a × upper b |
| a/b | Lower a ÷ upper b | Upper a ÷ lower b |
The positivity and independence conditions matter. If a variable can be negative, or the same variable appears more than once, blindly choosing endpoints separately can give an incorrect result. Establish the allowed interval and how the expression changes across it.
Worked example 2: bounds for a rectangular area
Question: A rectangle has recorded length 8.4 cm and width 3.2 cm, each rounded to the nearest 0.1 cm. Find bounds for its area A.
We already know that 8.35 ≤ L < 8.45. Similarly, 3.15 ≤ W < 3.25. Both lengths are positive, so the smallest area uses both lower bounds:
Alower = 8.35 × 3.15 = 26.3025 cm2.
The upper-bound calculation uses both upper bounds:
Aupper = 8.45 × 3.25 = 27.4625 cm2.
Hence 26.3025 ≤ A < 27.4625 cm2. The recorded-value calculation gives 8.4 × 3.2 = 26.88 cm2, which lies inside the interval.
Do not take 26.88 and add or subtract 0.05. The 0.05 uncertainty belongs to each length in centimetres, not directly to the area in square centimetres. Calculate the input intervals first, then pass their endpoints through the area formula.
Worked example 3: a quotient needs opposite extremes
Question: A distance is recorded as 240 m to the nearest 10 m, and a time as 12 s to the nearest second. Find bounds for average speed.
The distance d satisfies 235 ≤ d < 245. The time t satisfies 11.5 ≤ t < 12.5. Since speed is d/t, the smallest possible limiting value uses the least distance and greatest time:
vlower = 235/12.5 = 18.8 m/s.
The upper-bound calculation uses the greatest distance and least time:
vupper = 245/11.5 = 490/23 m/s, approximately 21.30435 m/s.
Here 18.8 < v < 490/23 m/s. Neither limiting endpoint is reached: the lower calculation requires the excluded time 12.5, and the upper calculation requires the excluded distance 245. An upper or lower bound need not itself be a possible value.
The recorded speed is 240/12 = 20 m/s, which is inside these bounds. Dividing the two lower input bounds would not find the minimum speed because it would also make the denominator smaller.
Worked example 4: state the percentage-error denominator
Question: A positive measurement is recorded as 25.0 cm to the nearest 0.1 cm. Estimate the maximum percentage rounding error using the recorded value as the reference.
The absolute rounding error is at most 0.05 cm. With the denominator explicitly defined as the recorded value, the estimate is:
(0.05/25.0) × 100% = 0.2%.
This is different from defining relative error using the actual value. Under that definition, the largest percentage occurs at the lowest permitted length, 24.95 cm, and is (0.05/24.95) × 100%, approximately 0.2004%. State the convention and follow the question’s definition rather than silently switching denominators.
Practice with solved checks
Practice 1: a trailing zero still carries information
A mass m is 6.20 kg correct to two decimal places. Give its interval.
Solution: The unit is 0.01 kg and the half-unit is 0.005 kg. Therefore 6.195 ≤ m < 6.205 kg. Treating 6.20 as a one-decimal-place measurement would discard the stated accuracy.
Practice 2: a difference of measured lengths
Two lengths are 15.2 cm and 9.6 cm, each to the nearest 0.1 cm. Find bounds for their difference D.
Solution: The intervals are [15.15, 15.25) and [9.55, 9.65). The lower calculation is 15.15 − 9.65 = 5.50; the upper is 15.25 − 9.55 = 5.70. Both limiting endpoints require an excluded input, so 5.50 < D < 5.70 cm.
Practice 3: significant figures determine the unit
A positive quantity is recorded as 7,300 to two significant figures. Find its bounds.
Solution: The second significant digit occupies the hundreds position. The unit is 100 and the half-unit is 50. Thus 7,250 ≤ x < 7,350. The number of printed zeros alone does not determine the accuracy.
Frequently asked questions about bounds
Is an upper bound the largest possible measurement?
Not necessarily. With the rounding convention used here, the upper endpoint is excluded. Values may approach it as closely as desired without reaching it.
Why is the interval width one rounding unit?
It extends half a unit below and half a unit above the recorded value. Adding those two half-units gives one whole unit.
Are decimal places and significant figures interchangeable?
No. Decimal places start after the decimal point; significant figures start at the first nonzero digit. They may occasionally produce the same answer.
Do I round the calculated bounds straight away?
Keep exact endpoints or enough digits during the working. Rounding a lower bound upwards or an upper bound downwards can accidentally exclude valid values.
What if a number is explicitly exact?
Do not add rounding uncertainty. An exact count or defined value does not acquire an error interval merely because it appears beside rounded measurements.
Why do division bounds use opposite endpoints?
For a positive quotient, increasing the numerator increases the result, whereas increasing the denominator decreases it. The two inputs therefore work in opposite directions.
Does this calculate every source of measurement error?
No. It tracks the uncertainty implied by the stated rounding. Instrument calibration and other measurement effects need separate information.
Continue with related Mathematics resources
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