Significant Figures

Question

The number $22$ has two significant figures while the number $7$ has one significant figure. Should $\frac{22}{7}$ have one significant figure, giving us an answer $3$, or should it have two significant figures, thereby giving us an answer $3.1$?

From what I have read, the result of division should have one significant figure, yielding the number $3$. This number has an uncertainty $1$ which is huge!

Am I going wrong somewhere?

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Edit: Let us consider the numerator to represent the physical quantity distance and the denominator to represent the physical quantity time. The fraction would then give us the average speed of an object over a distance of $22 m$ in an interval of $7 s$.

Answer 1

You're doing everything right; the uncertainty really is that big. Think about it: the real value might be as low as $\frac{21.5}{7.5} \approx 2.87$ or as high as $\frac{22.5}{6.5} \approx 3.46$, which is a pretty big swing.

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For the details of why it works out that way, what follows is more than you ever wanted to know about it.

When we say "22 has two significant figures" and "7 has one significant figure", what we really mean is that the genuine value that 22 approximates is $\frac{22}{1+\epsilon_1}$ and the genuine value that 7 approximates is $\frac{7}{1+\epsilon_2}$, where $|\epsilon_1| \approx \frac{1}{100}$ and $|\epsilon_2| \approx \frac{1}{10}$. Note that $\epsilon_1$ and $\epsilon_2$ might be positive or negative, because our approximations might be greater than or less than the genuine value.

With that in mind, the genuine value that $\frac{22}{7}$ approximates is \begin{align*} \frac{\frac{22}{1+\epsilon_1}}{\frac{7}{1+\epsilon_2}} &= \frac{22}{7}\frac{1+\epsilon_2}{1 + \epsilon_1} \\ &= \frac{22}{7}\frac{(1-\epsilon_2)(1+\epsilon_2)}{(1-\epsilon_2)(1+\epsilon_1)} \\ &= \frac{22}{7}\frac{1 - \epsilon_2^2}{1 + \epsilon_1 - \epsilon_2 - \epsilon_1\epsilon_2} \\ &\approx \frac{22}{7}\frac{1}{1 + (\epsilon_1 - \epsilon_2)} \end{align*} (In the final "$\approx"$ line, we're discarding $\epsilon_2^2$ and $\epsilon_1\epsilon_2$ because, as the product of small numbers, they'll generally be smaller than everything else. There are sometimes flaws in that assumption, but that assumption is baked in to the "standard advice" about significant figures.)

Given the above algebra, the error in our approximation is approximately \begin{equation*}|\epsilon_1 - \epsilon_2| \approx \frac{1}{100} + \frac{1}{10} \approx \frac{1}{10}\end{equation*} so one significant figure is appropriate.

Answer 2

Significant figures (or significant digits, when I was in school) are only relevant when taking measurements. How much does this sample weigh? How long is this item? How far away is the sun? If our measurements are accurate, we have more significant digits. You are correct that a result has only as many significant digits as the least amount of digits that goes into the calculation.

However, your choice of $22$ and $7$ for your example implies you are approximating $\pi$. Since $\pi$ is a constant, it is not measured in laboratory or real-world contexts. Significant figures don't apply to constants. Do you calculate significant digits based on the significant figures in Avogadro's number?

Further, rational numbers are defined as the division of integers. Integers have infinite precision, so $\frac{22}{7}=\frac{22.000...}{7.000...}$. Therefore, you can use as many significant digits as you want. However, since $\frac{22}{7}$ is merely an approximation of $\pi$, I wouldn't recommend using more than 3 significant digits.