# Expressing answer to correct degree of accuracy

In Physics when we multiply or add two quantities with different significant figures, why do we express our answer with the lowest significant figures? Consider we measured current to be 0.65A and we wanted to find: $$\frac{1}{0.65}$$ Why would our answer be 1.5? Furthermore, is there some general rule to expressing our answers to the correct degree of accuracy?

• Wikipedia has quite extensive guidlines <en.wikipedia.org/wiki/Significant_figures> (including a discussion of how to handle error estimates). Commented Feb 10, 2023 at 23:26
• If I calculate a time interval 10T to 2 decimal places, to what should I express T? Commented Feb 12, 2023 at 7:31

Always keep in the back of your mind that you are trying to estimate the error.

It helps enormously if information is given as to how a value was obtained.
In your case you quote a value of $$0.65$$. I have ignored the unit in the interests of clarity.
If the value came from a non-digital scale reading, how far apart were the scale readings? With the value given it could be that the scale divisions were separated by $$0.1$$ or $$0.02$$ or $$0.01$$ etc.
With a digital readout was the last significant figure the result of rounding ($$0.646$$ is displayed as $$0.65$$) or truncation ($$0.656$$ is displayed as $$0.65$$)?

Given no further information one might guess that any actual value between $$0.645$$ and $$0.655$$ results in a reading of $$0.65$$.
Given that you want to estimate the error in a reciprocal you need to estimate the fractional/percentage error which in this case is $$0.005/0.65 \times 100 \approx 0.77\%$$.

$$1/0.65 \approx 1.538$$, and this value poses another problem which has significance when a value is around zero.
Compare values $$0.98$$ and $$1.02$$. I would suggest that an estimation of the error in both these value would yield the same result and yet one value is quoted to two significant figures and the other to three significant figures.
So the value you suggested, $$1.5$$, especially without any qualification, is probably better replaced by $$1/0.65=1.5384 \approx 1.54$$ with an error of $$\pm0.01$$.

Consider we measured current to be 0.65A and we wanted to find: 1/0.65 Why would our answer be 1.5?

So $$I=0.65 \mathrm{\ A}$$ means that the current is somewhere between $$0.645 \mathrm{\ A}$$ and $$0.655 \mathrm{\ A}$$. That means that $$1/I$$ is somewhere between $$1.45 \mathrm{\ A^{-1}}$$ and $$1.53 \mathrm{\ A^{-1}}$$. This is roughly what is implied by $$1.5 \mathrm{\ A^{-1}}$$

is there some general rule to expressing our answers to the correct degree of accuracy?

The general rule is called the propagation of errors. Significant figures are a shortcut used by students as “training wheels”. In the professional scientific literature the uncertainty is directly and explicitly reported. It is not left to be inferred from the use of significant figures.

• Perhaps just a dissent about words, but still worth mentioning IMO: Significant figures are also important even when working with explicit error estimates – you still only report the significant figures of your number. E.g. it is usually considered an error the give more than two places of the error (as statistical errors are seldomly more accurately known and more precise numbers don't really help), and then in turn it does not make sense to give more places of the central value than of the errors. Commented Feb 12, 2023 at 11:23