Numbers and measurement · Eqora guide
Absolute and relative error: choose the reference
The same numerical gap can describe very different accuracy. Keep the reference, the units and the rounding assumptions visible.
A difference of 0.6 can be small or large, depending on what you are comparing. Reporting that number without its unit or reference hides the information needed to interpret it. Absolute and relative error answer two different questions: how far an approximation lies from a chosen reference, and how large that gap is compared with the reference's magnitude. Neither formula decides whether the result is acceptable until you know the task's requirements.
We will compare an approximate length of 23.4 cm with a reference of 24 cm, then change the scale, units and reference. All numerical values are constructed teaching examples, not laboratory measurements. You can work through the method on paper without any app. Check the original task, notation and assumptions before calculating, and verify the final result independently. Eqora publishes this learning guide, but no tool can guarantee a correct answer or a particular grade.
Absolute and relative error start with a named reference
Write the reference as r and the approximation as a. In our example, r = 24 cm and a = 23.4 cm. The signed difference a − r is −0.6 cm, telling us that the approximation is below the reference. We will call E = |a − r| the absolute error magnitude, so E = 0.6 cm. Taking the absolute value removes direction, not the unit. An approximation above the reference by the same amount has the same E.
Terminology varies: some measurement texts call the signed difference an absolute error, while numerical-analysis courses commonly use its magnitude. State your convention rather than changing signs midway. The University of Illinois reference on absolute and relative error uses the magnitude of the difference and divides by the magnitude of the reference for the relative form. That is the convention followed here. If your assignment explicitly requests a signed deviation, retain its sign and label it separately instead of silently taking an absolute value.
Divide by the reference magnitude before writing a percentage
For a nonzero reference, define e = E/|r|. Here e = 0.6/24 = 0.025. The centimeter units cancel because numerator and denominator describe the same kind of quantity in the same unit. Thus e is dimensionless. Multiply this ratio by 100 to express it as a percentage: 2.5%. The decimal 0.025 and the percentage 2.5% represent the same relative size. Writing 0.025% would describe an error a hundred times smaller.
The denominator is the reference 24, not the approximation 23.4. Using 23.4 instead gives a different ratio, approximately 2.564%, and answers a different question. When a task specifies a reference, preserve it through the calculation. If two experimental values are being compared without a designated reference, first determine which comparison definition the exercise requires. A symmetric percentage difference, for example, is another convention; do not substitute its denominator into a reference-based error formula merely because both expressions contain a percentage sign.
Compare scale before calling an error small
Suppose a second approximation is 239.4 cm against a reference of 240 cm. Its absolute error is again 0.6 cm, but its relative error is 0.6/240 = 0.0025, or 0.25%. The equal absolute gaps occupy different proportions of the references. The second is ten times smaller in relative terms. This does not mean the physical gap vanished; both lengths are still short by the same 0.6 cm. It means that the comparison has changed scale.
Conversely, equal relative errors can correspond to different absolute gaps. A 2.5% error against 240 cm has magnitude 0.025 × 240 = 6 cm. Either 234 cm or 246 cm is that far from the reference. The word “either” matters because a magnitude alone does not specify direction. Before comparing two results, decide whether the task cares about an absolute tolerance, a proportional tolerance or both. A component that must fit a fixed gap may be judged differently from a scale comparison in a mathematics exercise.
Keep unit conversion separate from changing the reference
Convert the first example to millimeters: r = 240 mm, a = 234 mm and E = 6 mm. The numerical absolute error becomes ten times larger, because each centimeter contains ten millimeters. Yet e = 6/240 = 0.025 remains unchanged. This is a useful independent check. A consistent multiplicative unit conversion changes the numerator and denominator by the same factor, which cancels. Converting only one value before dividing would instead produce a meaningless ratio with mixed units.
The guide to unit conversion distinguishes length, area and volume factors. Apply that discipline to both values before subtraction, then carry the resulting unit into the absolute error. The invariance statement here assumes multiplication by a common nonzero scale factor. It does not justify applying a percentage comparison unchanged after an arbitrary shift of the numerical zero. Relative comparisons depend on the meaning of zero in the quantity being studied. Keep this elementary length example separate from scales whose conversion includes an added offset.
Handle negative and zero references explicitly
A negative reference is not a reason to report a negative error magnitude. With r = −8 and a = −7.8, the signed difference is +0.2, the absolute error is 0.2 and e = 0.2/8 = 0.025, or 2.5%. The positive signed difference says the approximation lies above the reference. The positive ratio describes the size of that discrepancy relative to |r|. Omitting the absolute value in the denominator would mix these two purposes and reverse the sign of the size measure.
At r = 0, the relative formula is undefined because it would divide by zero. For a = 0.001, the absolute error is 0.001; there is no reference-based relative percentage under this definition. Do not report zero percent, an invented finite percentage or infinity as a calculated replacement. Near zero, the ratio can also look large: r = 0.002 and a = 0.003 give E = 0.001 but e = 0.5, or 50%. Report the absolute scale as well so the reader can interpret what happened.
Distinguish a computed error from a rounding bound
An actual error can be calculated only relative to the stated reference. For the exact fraction 1/3 and approximation 0.33, E = 1/3 − 33/100 = 1/300. Dividing by 1/3 gives e = 1/100 = 0.01, so the percentage error is exactly 1%. The approximation is below the reference, but E remains nonnegative. Keeping fractions until the final presentation avoids replacing one approximation by another before you have completed the comparison.
A rounding bound answers a different question. Rounding to the nearest hundredth changes a number by at most 0.005, assuming an ordinary nearest-value rule. That is a bound, not a claim that every rounded number has error exactly 0.005. For example, 12.346 rounds to 12.35, with actual absolute error 0.004. Different tie-breaking conventions affect which endpoint rounds up at an exact halfway value, but the at-most bound still holds. Truncating rather than rounding to the nearest value requires its own bound and should not be described as the same procedure.
Do not replace an unknown error with a guessed certainty
In a real measurement, the true value may not be known exactly. A reference may itself be an accepted approximation with uncertainty. The difference between two displayed numbers is then an observed discrepancy relative to that reference, not proof of the measurement's unknown true error. Repeating a measurement consistently does not establish that its common value is correct. A stable bias could remain. State what the reference represents and what information is actually available before attaching an accuracy claim to a calculation.
Likewise, an interval of possible values is not the same object as one known error. If all you know is that a value rounded to 12.35 to the nearest hundredth, you can discuss the interval implied by that rounding rule; you cannot infer the unique original value 12.346. The independent answer-checking guide encourages testing what the data really support. Here that means checking whether your conclusion is an equality, a bound or an estimate, and preserving that distinction in the final sentence rather than displaying extra decimals as evidence of certainty.
Translate a tolerance into a checkable interval
Suppose an exercise allows a relative error of at most 1% against the positive reference 80. Then E/80 ≤ 0.01, so E ≤ 0.8. The acceptable approximations lie in [79.2, 80.8]. Both endpoints meet an “at most” condition. If the instruction instead says “less than 1%,” the endpoints are excluded. The approximation 80.9 has E = 0.9 and e = 0.01125, or 1.125%, so it fails either version of the one-percent requirement.
This interval can be checked independently by substituting its endpoints back into |a − 80|/80. Notice that an error percentage is not automatically an acceptance threshold: the threshold must come from the problem. Keep all working digits until after making the comparison, because rounding 1.125% too coarsely could hide a failure near the boundary. If a result is exactly on a boundary using rounded input data, say what assumption makes that classification possible instead of presenting more certainty than the input permits.
Practise with a changed reference, then reverse the calculation
Try a reference of 50 cm and an approximation of 49.75 cm without looking back at the first example. The signed difference is −0.25 cm, E = 0.25 cm and e = 0.25/50 = 0.005, or 0.5%. In millimeters the comparison is 500 against 497.5 with E = 2.5 mm; the ratio is still 0.005. Now reverse the question: approximations with exactly 0.5% error relative to 50 cm are 49.75 cm and 50.25 cm. Explain why there are two possibilities.
Finally, consider subtracting nearby values. Exact values 12.01 and 12.00 differ by 0.01. If both are first rounded to one decimal place, their displayed difference is 0.0. Relative to the exact difference 0.01, the result has absolute error 0.01 and relative error 100%. Small input changes can therefore have a large relative effect on a small difference. Finish by writing the reference, signed difference, magnitude, ratio, percentage and assumptions yourself. Those labels make the calculation auditable and prevent one correct-looking number from answering the wrong question.
