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The rules of sigfigs say that when adding/subtracting, the sum/difference is rounded to the least precise measurement.

$$32.56 + 2.0592 = 34.6192$$ Rounded answer: $34.62$

The rules also say that when multiplying/dividing, the product/quotient is rounded to the lowest number of sigfigs.

$$25.694 \times 1.85 = 47.5339$$ Rounded answer: $47.5$

What seems a little confusing to me is how one rule might relate to the other in a way that is consistent with the relationship between addition and multiplication.

Take the following expression: $$16.09 \times 5$$

As the definition of multiplication (by natural numbers; the rest are logical extensions of that) as an operation is based on repeated addition, the expression mentioned is equivalent to $16.09 + 16.09 + 16.09 + 16.09 + 16.09$. Since all of these numbers share the same precision, the sum will share that precision: $80.45$.

On the other hand, the sigfig multiplication rule would have me round to one sigfig, $80$.

Why exactly is that?

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    If you know that that 5 is an integer exactly, then it's not 5, it's 5.00000000…. On the other hand if you don't know it's exactly.an integer then your $16.09+…+16.09$ idea doesn't work. – MJD Dec 08 '24 at 23:42
  • @MJD I suppose, that works in the case of 5.000000... but what about something like $16.09 \times 9.0000...$? If that's the case then it is $16.09 + ... + 16.09$ nine times, which is $144.81$. Going by the addition rules, that doesn't need to be changed, but by the multiplication rules, it should be $144.8$. – Luis Green Dec 08 '24 at 23:56

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Sig figs are an informal way to keep track of precision. Imagine that we have two values $v_1\pm 10^{e_1}$ and $v_2\pm 10^{e_2}$. Then $$(v_1\pm 10^{e_1})+(v_2\pm 10^{e_2})\\ =v_1+v_2+(10^{e_1}+10^{e_2})\\\approx v_1+v_2\pm 10^{\max(e_1,e_2)}$$ If $e_1\neq e_2$, or the error is small, then this is an acceptable approximation. However, if $e_1=e_2$ and we repeat this addition 5 times, the approximation can be significantly off. For multiplication, it makes more sense to express the values as $v_1(1\pm 10^{-s_1})$ and $v_2(1\pm 10^{-s_2})$. In this case, $$ v_1(1\pm 10^{-s_1})\cdot v_2(1\pm 10^{-s_2})\\=v_1\cdot v_2(1\pm 10^{-s_1}\pm 10^{-s_2}\pm 10^{-s_1-s_2})\\\approx v_1\cdot v_2(1\pm10^{\max(-s_1,-s_2)}) $$ Each of these approximations is better in its own regime. For science, this convention is good enough. Also note that in the case of addition, the errors often balance out rather than accumulating (unless you have a persistent bias in the measurement).

Logan Post
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Take a quick look/scan over this page about error propagation. It's not the simplest thing in the world. To calculate how uncertainty propagates, you need an understanding of some multivariable calculus tools, as well as understanding of probability distributions. That can be beyond the current abilities of a person who simply needs to add together two measurements that each came with uncertainty. Or even if it's not beyond their current abilities, it's time-consuming to do this.

So those sigfig rules are a crude replacement for "doing it right". We don't expect the sigfig rules to propagate error in a way that matches what true error propagation modeling would do. And so we don't expect it to produce logically consistent results if there are calculations like $16.08\times9.\overline{0}$ viewed as addition versus multiplication. But using those sigfig rules is arguably close enough in practice, and the rules are simple to follow, even for people without the calculus and probability/statistics background.

By the way, with something like $16.08\times9$ (where the $9$ is a solid $9$, not $9\pm0.5$), it would matter if this was one single measurement $16.08$ added to itself nine times, or nine separate measurements that all happen to be $16.08$ with their own uncertainties. The first case would result in something with large uncertainty, since the original uncertainty would be multiplied by $9$. The second case would have some positive errors canceling some negative errors, and an overall lower uncertainty in the total compared with the first scenario. The first scenario is appropriate to view as multiplication, the second as addition.

2'5 9'2
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