Derive exact density bounds and conditional identification

A density result contains a model

Density is mass divided by volume for the same sample and conditions. Our synthetic measurements are 20 g and 10 mL, with adjustable absolute bounds. The nominal ratio is 2 g/mL. A long decimal expansion of that ratio would not provide more information about the actual density. The task is to propagate the specified feasible inputs before comparing hypotheses.

Derive endpoint bounds

For positive volume, increasing mass increases m/V and increasing volume decreases it. Therefore the minimum ratio pairs the smallest mass with the largest volume; the maximum pairs the largest mass with the smallest volume. This is an exact monotonicity argument for a rectangular feasible region. It is not an approximate “same number of significant figures” rule.

Work the original packet

With mass bound 0.20 g and volume bound 0.30 mL, the interval is [19.8/10.3, 20.2/9.7], approximately [1.9223, 2.0825] g/mL. The exact hypothetical candidate 1.95 g/mL is compatible; 2.10 g/mL is not. This conclusion depends on matched temperature, correct sample pairing, valid bounds and the declared density model.

Compare measurement improvements

Reduce only the mass bound, then restore it and reduce only the volume bound. Compare the resulting interval widths rather than assuming the visually smaller uncertainty is automatically less important. A gram and a milliliter are different dimensions. Relative sensitivities depend on the nominal values: an absolute 0.1 mL error is 1% of 10 mL, while 0.1 g is 0.5% of 20 g.

Avoid false identification

An interval containing one listed candidate excludes the other listed candidate under the model. It does not prove that every possible material has been considered, that the sample is pure, or that a mixture cannot have the same density. Request another observable whose prediction differs across remaining explanations. Density by itself is often an underdetermined identification problem.

Challenge the feasible region

If mass and volume share a calibration variable, not every corner of an independently drawn rectangle may be possible. The conservative rectangle may then be too broad, or the assumed bounds may be wrong. Describe a joint-error model before narrowing the interval. For a fresh case, 12±0.2 g and 4±0.1 mL give [11.8/4.1,12.2/3.9], approximately [2.8780,3.1282] g/mL. Retain units at every step.

Source and evidence boundary

These explanations, calculations and diagrams are original. Conceptual references checked 30 October 2026: NIST uncertainty framework, OpenStax dimensional calculations. The linked textbook is a reference, not copied lesson text or imagery. All displayed datasets are synthetic teaching packets. Saved numerical agreement or prose is not a performed practical, independent reasoning review or laboratory certification. No handling, heating or chemical preparation procedure is requested.