# B Why we take the least # of significant digits when multiplying?

#### HuuChi1778

This question just flew right into my head when I was in the mid of a shower :), and I cannot resist finding out the answer. Question: Why do we take the least number of significant digits when multiplying/ dividing, and the least number of decimal places when adding/subtracting? Thank you!

Related Other Physics Topics News on Phys.org

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
Have you tried figuring the answer out for yourself? If so, what were your thoughts?

#### HuuChi1778

I do have a few ideas on this topic. I think that significant digits are there to determine the amount of precision of the given data. I believe that the reason why we take the least number of decimal places when adding and subtracting is because we cannot have a result that is more precise than the least precise number(s). For example, when adding 5.1 + 5.20, the answer should be 10.3. It is because the precision (+- 0.1) of the number 5.1 is less precise than the precision of the number 5.20 which is +- 0.01. However, I cannot think of a good reason why we apply the rule of rounding to the least number of significant digits instead round the number to the least decimal places. I would like an answer from an expert like you. Thank you!

Last edited:

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
What happens to the errors when multiplying?

#### HuuChi1778

That is a very tough question from you. I just did a quick check and realized that when we multiply the errors, in this case, 1x10^-1 and 1x10^-2. The result is 1x10^-3, which makes it even smaller. That means the result is more precise than it can actually be. Is that the reason why?

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
I suggest you think of it this way in terms of errors. A number is given by $a \pm \delta a$. For multiplication purposes, it is more convenient to write this as $a (1 \pm \delta a/a)$, here $\delta a/a$ is the relative error and essentially gives you an estimate on how many significant digits you should use. If $\delta a/a$ is (roughly) of order $10^{-n}$, you use $n$ significant digits.

$$(a \pm \delta a) + (b \pm \delta b) = (a+b) \pm \delta a \pm \delta b = (a+b)\left( 1 + \frac{\pm \delta a \pm \delta b}{a+b}\right).$$
Clearly, the larger of $\delta a$ and $\delta b$ here dominates the error and you should use the larger error (the number with the smaller number of decimals has the larger error) to determine how many decimals you can reasonably use.

Multiplying two numbers gives you
$$(a \pm \delta a) (b \pm \delta b) = ab (1\pm\delta a/a)(1\pm\delta b/b) \simeq ab (1 \pm \delta a/a \pm \delta b/b),$$
where we have assumed that the errors are relatively small so that the product of the errors is negligible. Clearly, the number with the larger relative error determines the error in the product.

#### David Lewis

Why do we take the least number of significant digits when multiplying/dividing?
Actually you round to the least number of significant figures after multiplying/dividing. During the process itself I believe you keep one or two more significant figures (if you have them) than the least precise multiplicand/dividend/divisor contained in your product/quotient.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
Actually you round to the least number of significant figures after multiplying/dividing. During the process itself I believe you keep one or two more significant figures (if you have them) than the least precise multiplicand/dividend/divisor contained in your product/quotient.
You never round your answer until all computations are done, regardless of the operation.

#### jbriggs444

Homework Helper
You never round your answer until all computations are done, regardless of the operation.
*gets up from his front porch chair and shakes his cane at the young whippersnappers walking by*. In my day...

If you are computing with longhand arithmetic, you will want to round off. Each multiplication would otherwise double the number of figures you are computing with and each division would likely result in an infinite string. If you are using a log or trig table in your handy CRC, those results are already rounded.

Edit: And if you are a whippersnapper, using a calculator or computer, you are almost certainly rounding off anyway. Which can bite you if you are unaware of it.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
*gets up from his front porch chair and shakes his cane at the young whippersnappers walking by*. In my day...

If you are computing with longhand arithmetic, you will want to round off. Each multiplication would otherwise double the number of figures you are computing with and each division would likely result in an infinite string. If you are using a log or trig table in your handy CRC, those results are already rounded.
I think "never" is accurate enough for a B-level thread in an age where people are using calculators and computing numbers with at most three significant digits ...

#### Khashishi

A number like 5.5 can be interpreted as 5.5 +- 0.05. So, if we add two numbers,
3.24 + 5.5
(3.24 +- 0.005) + (5.5 +- 0.05)
Now, if we don't have any additional information about the distributions of the errors, then we just use the rule of thumb where we "absorb" the smaller error into the larger one, and we get
8.74 +- 0.05
which is just 8.7

Now, as an exercise, try the same thing with multiplication.

By the way, if you do have more information about the distribution of errors, then you probably shouldn't use the rules of significant figures and instead use explicit propagation of errors.

"Why we take the least # of significant digits when multiplying?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving