# Significant Figures

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?

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 $7 s$.

• Common convention for division is to let the number with the least number of sig figs dictate how many are in the answer. So yeah, 22/7 would yield 3 in this case. Jul 12 '17 at 15:25
• This would mean that the true value lies between $2$ and $4$, would it not? If yes, then that's a very big uncertainty.
– R004
Jul 12 '17 at 15:36
• That is correct. You're going to get large uncertainties when you have a measurement with a single significant digit, there's no way around that other than get better ways of measuring the quantities in question. Sorry! Jul 12 '17 at 15:45
• Here's what's bothering me. Let's take the two numbers to talk about magnitudes of some physical quantities. The uncertainty in 22 m( say ) is 1 m, and the uncertainty in 7 s( say ) is 1 s. Here, 22 m/7 s gives us the average speed. Now, the maximum possible average speed is given by 22+1/7-1 = 3.83...< 3.9( our max. is 4 ). And the minimum possible average speed is given by, 22-1/7+1 = 2.6...>2.5 ( our minimum is 2 ). So you do see that the uncertainties derived from studying the significant digits are a little different. I hope I could convey the problem.
– R004
Jul 12 '17 at 16:00
• It sounds to me like you're more concerned about error propagation. Read around about it a little and see if it clears some things up. Jul 12 '17 at 16:09

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.

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.

• @R004 The frameworks I'm working with assumes that measurement errors are introduced "multiplicatively." That is, say there's some physical number with true value $x$. We measure it through some process, and get a measured value $\hat{x}$. On purely mathematical grounds, so long as $x \neq 0$, $\hat{x} = (1+\epsilon) x$ for some unknown $\epsilon$. The "multiplicative" framework assumes that what we know about our measurement process/equipment is some particular upper bound for $|\epsilon|$. There are other possible frameworks for how to model measurement error, but this one is common.(cont'd)
– user231101
Jul 12 '17 at 17:10
• @R004 Why I used, e.g., $\frac{22}{1+\epsilon_1}$ is because, in your setup, 22, is the measured value. That is, 22 plays the role of $\hat{x}$. To write $x$ in terms of $\hat{x}$, we divide both sides by $1 + \epsilon$.
– user231101
Jul 12 '17 at 17:12
• @R004 Also of note: sometimes $\epsilon$ is called the "relative error" of the measurement, whereas $\hat{x} - x$ is called the "absolute error." We're assuming we know a bound on the relative error.
– user231101
Jul 12 '17 at 17:14
• @R004 Yes(ish), assuming we are using $\frac{1}{10}$ as our bound on the relative error of the measurement. The "ish" is because I'd write $|\epsilon| < \frac{1}{10}$, or even $-\frac{1}{10} < \epsilon < \frac{1}{10}$, instead of "$\epsilon = \pm\frac{1}{10}$", which, if we're being precise, might mislead someone to think $\epsilon$ is exactly equal to one of those two numbers ($\pm\frac{1}{10}$). The true value of $\epsilon$ is unknown.
– user231101
Jul 12 '17 at 17:21
• @R004 Wait, I think I may have misunderstood your last comment. If you intend $22.2$ to indicate that we know three significant digits, that would mean our bound on $|\epsilon|$ is $\approx \frac{1}{1000}$, not $\frac{1}{10}$.
– user231101
Jul 12 '17 at 17:26

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.

• Let us say that I am calculating the average speed of an object over a distance of $22 m$ in a time interval of $7 s$( I could have taken other integers, but since I've begun with two above, I shall proceed with them ). Now, what would be my average speed, $3 m/s$ or $3.1 m/s$?
– R004
Jul 12 '17 at 15:41
• Shouldn't you use significant figures for Avogadro's number in calculations, though? Lack of digits for that number past a certain point represents real experimental uncertainty, IIRC. (Although in practice we probably know more digits for Avogadro's number than for the other numbers in a typical calculation.)
– user231101
Jul 12 '17 at 15:42
• A better example might be the speed of light, which is defined to have an exact value.
– user231101
Jul 12 '17 at 15:44
• @R004 If you counted "one mississippi, two mississippi, ..." in order to measure the $7 s$, then the avg speed is $3 m/s$. However, if the stopwatch on your smart phone gives you tenths of a second, and you measured $7.0 s$, then you have two significant figures and you can say the average speed is $3.1 m/s$. It all depends on the accuracy of your measurements. Jul 12 '17 at 17:30