# Why are significant figure rules in Multiplication/Division different than in Addition/Subtraction?

I've never understood specifically why this is.

Here's what I mean.

In Addition/Subtraction, what matters are the digits after the decimal point. So for example:

1.689 + 4.3 =

  1.629
+ 4.3XX
---------
5.929
---------
5.9


This makes sense to me. I filled in uncertain values with X, and it makes sense why I can't use the 0.029 in the answer - because I added it to an uncertain value.

However, I don't understand the rules when it comes to Multiplication/Division. The same little trick with X's doesn't help me here.

I know that what matters in Multiplication/Division are the significant figures. So for example:

12.3 * 4.6 =

  12.3
*  4.6
-------
738
492X
-------
56.58
-------
57


The answer is 57 according to significant figure rules of Multiplication/Division, but I just can't make sense of those rules like the way I did with Addition/Subtraction.

Does anyone have an intuitive explanation for the significant figure rules of Multiplication/Division?

• I am reasonably certain that this has been asked before. – Kyle Kanos Jan 6 '16 at 21:04
• More on significant figures. – Qmechanic Jan 6 '16 at 21:09
• I just looked at the posts you linked. None of them address my question. – b0yfriend Jan 6 '16 at 21:12
• You're not going to get a better answer, because significant figures have no theoretical basis. They're a cheap approximation to actual error analysis. – knzhou Jan 6 '16 at 22:01
• @knzhou That (" They're a cheap approximation to actual error analysis.") is the basis of a good answer, if you'd like to expand it. – dmckee --- ex-moderator kitten Jan 6 '16 at 23:23

The number of significant figures is a representation of the uncertainty of a number. $123.4$ has an uncertainty of $0.1$ since the first uncertain digit is usually included. So, your multiplication of $$12.3 \times 4.6$$ is better represented as $$(12.3 \pm 0.1) (4.6 \pm 0.1).$$ I'm going to use @barrycarter's trick of using scientific notation to better represent the values $$(1.23 \pm 0.01) \cdot 10^1 \times (4.6 \pm 0.1) \cdot 10^0.$$ $$(1.23 \pm 0.01) (4.6 \pm 0.1) \times 10^1.$$ From this we can see that the number with more significant digits has less relative uncertainty. Relative uncertainty is approximately the measured uncertainty (also called absolute uncertainty) divided by the value.

Multiplying out: $$((1.23)(4.6) + (1.23)(\pm 0.1) + (4.6)(\pm 0.01) + (\pm 0.01)(\pm 0.1)) \times 10^1.$$ $$(5.658 + (\pm 0.123) + (\pm 0.046) + (\pm 0.001)) \times 10^1.$$ The number of digits we can write down is determined by the total uncertainty, which here is dominated by $\pm 0.123$. So, our result is $$(5.658 \pm 0.123) \times 10^1.$$ $$(5.7 \pm 0.1) \times 10^1.$$ $$57 \pm 1$$ which results in $57$.

To summarize, in multiplication, the largest relative uncertainty dominates the uncertainty of the final answer. This translates into the the number with fewer significant digits determining the significant digits of the final answer. This contrasts with addition, where the largest absolute uncertainty determines the uncertainty of the answer.

• This is great. Thanks for the explanation. One question: If we were to be more precise about the uncertainty, wouldn't we $$(\pm 0.123) + (\pm 0.046) + (\pm 0.001) = \pm 0.17$$ To get the total uncertainty? But we don't do that because significance arithmetic "is just a cheap approximation of actual error analysis?" – b0yfriend Jan 7 '16 at 17:15
• Uncertainties don't add that way because errors in measurements aren't always in the same direction. The actual errors in a given measurement could be $+0.123$, $-0.046$, and $0$, yielding $0.077$. Because of this, errors add in quadrature $\sqrt{err_1^2 + err_2^2 + \cdots}$. In this case, the total error is closer to $0.13$, which is close to $0.123$. See @DavidZ's answer below for a much more precise explanation. – Mark H Jan 7 '16 at 19:17
• What if we try with a different example like $(1.0 \pm 0.1)(1.0 \pm 0.1)$? In this case the computation would give an error $\pm 0.2$ and if we keep multiplying to obtain $(1.0 \pm 0.1)^n$ we get an error $n\cdot 0.1$ that grows eventually reducing the number of significant figures, this seems that would make the product rule false. Wouldn't it? – Marco Disce Dec 23 '16 at 20:48
• @MarcoDisce That's true, which is why the rule is just that: a heuristic rule to estimate the total error. In equations with lots of moving parts, you need more sophisticated error analysis, like in DavidZ's answer. – Mark H Dec 23 '16 at 22:53

It makes more sense if you convert to scientific notation first, so 12.3 * 4.6 becomes: (1.23*10^1 * 4.6*10^1). In this case, you can't expect an answer more accurate than one place after the decimal point, since you don't know what comes after the 6 in 4.6.

In general, in scientific notation, you will essentially be multiplying numbers between 1.000 and 9.999... (and adding the powers of 10), so the accuracy you'd expect is the last decimal position of the least precise number.

Actually, I have two explanations, depending on what you consider more intuitive.

# Interval arithmetic

This one is best explained by example. Remember that a value with a certain number of significant figures is supposed to represent that exact value $\pm 5$ in the first insignificant digit. For example, $12.3$ with three significant figures represents anything in the range $[12.25,12.35]$, and $4.6$ with two significant figures represents $[4.55,4.65]$. If you multiply these two ranges, you can get anything between a minimum of $4.55\times 12.25 = 55.7375$ and a maximum of $4.65\times 12.35 = 57.4275$. So strictly speaking, the result is $[55.7375,57.4275]$.

However, it's a lot more convenient to represent this range as a single number, using the same significant figure convention as before, where the uncertainty is $\pm 5$ in the first insignificant digit. We can't actually do this, though; the size of the range is $1.69$, which is not $5\times 10^n$ for any $n$. The sensible options are either to write $60$, which represents the range $[55,65]$, or $57$, which represents the range $[56.5,57.5]$. But there isn't a straightforward rule that justifies the first choice, without doing interval arithmetic every time. The second choice, on the other hand, can be justified by keeping the lower number of significant figures from either of the original operands. So that's what we do.

# General error propagation

For a better justification, significant figure rules are chosen as they are because they're a decent match to the general formula for propagation of uncertainties. That general formula is as follows: if you have some quantity $y$ which is a function of several other variables, $y = f(x_1, x_2, \ldots)$, then the uncertainty in $y$ is given by $$\sigma_y = \sqrt{\biggl(\frac{\partial y}{\partial x_1}\biggr)^2\sigma_{x_1}^2 + \biggl(\frac{\partial y}{\partial x_2}\biggr)^2\sigma_{x_2}^2 + \cdots}$$ Actually, this formula only applies under certain conditions, but they're very common conditions, and it would take a whole separate answer (or even more) to explain all the subtleties. (See here and here for examples.) So let's just take that formula as a starting point.

If a value $x$ has $n$ significant figures, that means its relative uncertainty $r = \frac{\sigma_x}{\lvert x\rvert}$ is between $(5\times 10^{-n})$, if $x$ is a power of $10$, and $(5\times 10^{-(n+1)})$, if $x$ is just less than a power of $10$ (like $99.99999$). Or within an order of magnitude, anyway. You could argue that maybe the $5$ should be a $3$ or something like that, but that gets into the conditions I already decided to skip discussing.

## Multiplication and division

I do this case first because it's easier. If the function is $f(x_1, x_2) = x_1x_2$, the general error propagation formula gives $$\sigma_y = \sqrt{x_2^2\sigma_{x_1}^2 + x_1^2\sigma_{x_2}^2}$$ Or substituting in $rx$ for the uncertainty, all the $x_1$s and $x_2$s cancel out and we get $$r_y = \sqrt{r_1^2 + r_2^2}$$ Going through the argument for division, the algebra is a little different but you still get to this same formula for relative uncertainty.

There are three cases to consider here:

• $r_1 \ll r_2$: $r_1$ is basically insignificant and you can write $r_y \approx r_2$
• $r_1 \gg r_2$: $r_2$ is basically insignificant and you can write $r_y \approx r_1$
• $r_1 \approx r_2$: in this case you can approximate them as roughly equal and write $r_y \approx \sqrt{2} r_1$. Depending on the exact relative magnitudes, this factor might be a little different, but both $r_1$ and $2 r_1$ are likely to be around the same order of magnitude, which is all that matters for an uncertainty.

One can make this argument a little more precise, but the point is, the resulting uncertainty $r_y$ is more or less the same as the larger of $r_1$ and $r_2$. But remember, $r$ is related to the number of significant figures: $r \approx 10^{-n}$, meaning that a larger relative uncertainty $r$ corresponds to a smaller number of significant figures. So the final quantity $y$ will typically have the same number of significant figures as the less precise (fewer significant figures) of the two operands $x_1$ and $x_2$.

It's also worth checking that this works for addition and subtraction.

If the function is $f(x_1, x_2) = x_1 \pm x_2$, then the partial derivatives are both equal to $1$. So the error propagation formula gives $$\sigma_y = \sqrt{\sigma_{x_1}^2 + \sigma_{x_2}^2}$$ Here substituting in the relative uncertainty gives $$r_y(x_1 \pm x_2) = \sqrt{r_1^2 x_1^2 + r_2^2 x_2^2}$$ but this isn't so useful because things don't cancel out. So let's go back to the previous formula, with the $\sigma$s. This time, the three cases break down as follows:

• $\sigma_1 \ll \sigma_2$: $\sigma_1$ is negligible and you can set $\sigma_y \approx \sigma_2$
• $\sigma_1 \gg \sigma_2$: $\sigma_2$ is negligible, so $\sigma_y \approx \sigma_1$
• $\sigma_1 \approx \sigma_2$: following more or less the same reasoning as before, $\sigma_y \approx \sigma_1$ to within an order of magnitude or so

Now, the actual uncertainty $\sigma$ is related, not to the number of significant figures, but to their position. The larger uncertainty corresponds to the leftmost insignificant figure among the two operands, which is exactly what you use.

• Could you please explain as to how "depending on the exact relative magnitudes, you might have as much as $r_{y}\approx 2r_{1}$". Thanks. – Procyon Apr 30 '16 at 0:50
• @Procyon Actually, I don't remember exactly what I meant by that. I think I'll take it out. – David Z Apr 30 '16 at 7:56
• If $r_1 \approx r_2$ then you can actually have an error that is greater than both $r_1$ and $r_2$ and this could result in a smaller number of significant figures. Doesn't it make the general rule of the product false? – Marco Disce Dec 23 '16 at 20:46
• @MarcoDisce The significant figure rules are only approximate anyway. – David Z Dec 23 '16 at 21:12