# Assuming significant figures

## Main Question or Discussion Point

I have a question which states that a car's speed is 110km/h. Should I assume this speed to be 3 significant figures or 4 significant figures? why??

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the safest thing to do is stick to the number of figures given in the question, which is 3. If you feel that you must use 4 because calculations comu up like that then work with that but give 3 in your final answer.
Whatever you do...do not go beyond 4 !!!
The figures you do not know are, by definition, NOT SIGNIFICANT

Ok, I figured it would be 3 in this case, but I was taught that the last non-zero digit is considered the first uncertain digit, which would mean that 110km would have 2 significant figures not 3.

This is where I'm confused. Could you explain?

You have hit on a common problem !!!
It is difficult to know wheter the zero is significant or not.
If the speed was 115km/hr then there is no problem
I would say that in practice it is safe totake it to be 3 sig figs.
The problem is that these discussions about significant figures is not an exact science since they are used to convey accuracy of quantities. If greater accuracy is required then more significant figures are needed.
I am fairly certain you will get some other replies giving a variety of interpretations.

Hope this helps.
ps
if you think about it, giving a number to 3 sig figs implies that you know the value to better than 1%
when you write 115 you are saying that it is NOT 114 or 116, so you know the number to within +/- 1....about 1%
This is what makes it tricky deciding whether 110 is 3 or 2 sig figs. If it is only 2 then it means the value could be between 100 and 120 which is probably not intended !!! because it implies an accuracy of only +/-10%

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I'd say that if the speed is 110 +/- 5 then it should be written 1.1x10^2 km/h, then it's far simpler to derive implicit errors from stated significant figures.

Borek
Mentor
I was always under impression that in most physics courses nobody cares about sig figs, they are treated seriously only by chemists (and even then not by all).

I was always under impression that in most physics courses nobody cares about sig figs, they are treated seriously only by chemists (and even then not by all).
They are treated as a means of assessing and conveying accuracy but it would be laborious if every example had to have this error analysis applied.
They are very important in practical situations and in practical exams students are penalised if they cannot show an awareness of the significance of significant figures.
This usually shows up when pupils write down the number from a calculator without any regard as to how many figures should be shown.

Borek
Mentor
They are treated as a means of assessing and conveying accuracy but it would be laborious if every example had to have this error analysis applied.
I know what their application is, but I have seen on many occasions people being told to report the result with a "reasonable number of digits" - so obviously 1.23678923467854 doesn't make sense, but 1.24 or 1.237 is acceptable. Sig figs are about as wrong as easy to use, and treating them too religiously is a bad idea, as it gives students an unjustified feeling that they are doing something important.

Let me use an example to see opinion: ½(23.1 cm3 − 20.32 cm3 + 19.0 cm3)

Would you equate that to 3 significant digits or 1? I am suggesting 1 because the numerator & denominator of the fraction 1/2 are both composed of 1 significant digit.

I'd say common sense needs to be used, and the person doing the analysis needs to be aware of what terms are experimental data and what terms are exact. Without any further information (which we'd have if we did the experiment), I'd guess that 1/2 is an exact term and the other terms are to 3/4 s.f., so I'd quote the result to 3.

I agree there is some vagueness on what is acceptable, but there should definitely be clear pointers on what is unacceptable. I see all the time students measure results, round them, then average them (re-introducing false significant figures) and quoting the average to the same precision as the data. Unless you go through all the maths you'd never realise their result is ~10 standard deviations from the true mean. IMO that's very very bad practice. Rounding mid-calculation is obviously wrong, but a second error is quoting too many significant figures.

Borek
Mentor
Would you equate that to 3 significant digits or 1? I am suggesting 1 because the numerator & denominator of the fraction 1/2 are both composed of 1 significant digit.
Are you suggesting that kinetic energy has never more that 1 sigfig, as in $\frac{mv^2}2$ 2 has only one sigfig?

No, I think it should be more, but just wanted to verify the ambiguity of sig figs.

jbriggs444
Homework Helper
2019 Award

1.4h pretty clearly.

The 130 km has either two or three significant figures depending on your guess as to whether the trailing zero is significant. The 95 has two significant figures. A quotient has as many significant figures as the less accurate of the dividend and the divisor. That's two either way.

However... That's a leading one on the 137 in the result. And that's a leading nine on the 95 in the input. That means that the relative error bounds on the input are tighter than "two significant figures" would normally suggest. And the relative error bounds on an output of 1.4 would be looser than "two significant figures" would normally suggest. So I would be quite willing to report 1.37 h if I were sure that the 130 km were accurate to +/- 0.5 km and that the 95 km/h were accurate to +/- 0.5 km/h. That way I wouldn't be throwing away accuracy in the name of significant figures.

When you say there is a leading 1 and a leading 9, what significance does that have on deciding on the number of significant figures??

jtbell
Mentor
To take a somewhat contrived example, consider two numbers, each with two significant figures:

(a) 11 - changing the last figure by ±1 is a change of about ±9%.

(b) 99 - changing the last figure by ±1 is a change of about ±1%.

Clearly and (a) and (b) have significantly different levels of precision, even though both have two sig figs.

Now consider the following number with three sig figs:

(c) 101 - changing the last figure by ±1 is a change of about ±1%.

Clearly (b) and (c) have about the same level of precision, even though one has two sig figs and the other has three.

AlephZero
Homework Helper
If you want to estimate the error in the answer, then do it properly.

"Counting significant figures" is such a crude method as to be no real practical use. If the answer happens to be 99, then to 2 significant figures you have an error of about 1%. If it happens to be 101, to 2 s.f. you have an error of about 10%. If you think that makes any sense, then carry on counting significant figures....

Edit: I started typing this before jtbell's post appeared.

I'm still a little confused. How about 65 * 1.96

What would you state that answer as?

If you want to estimate the error in the answer, then do it properly.

"Counting significant figures" is such a crude method as to be no real practical use. If the answer happens to be 99, then to 2 significant figures you have an error of about 1%. If it happens to be 101, to 2 s.f. you have an error of about 10%. If you think that makes any sense, then carry on counting significant figures....

Edit: I started typing this before jtbell's post appeared.
101 is not given to 2 significant figures, it is 3. 101 means you know it is not 100 and it is not 102.
The problem arises when a value of 100 is quoted I would say

I'm still a little confused. How about 65 * 1.96

What would you state that answer as?
You should give this to 2 significant figures, 130.
You should not use more significant figures than in the least precise number.

AlephZero
Homework Helper

101 is not given to 2 significant figures, it is 3. 101 means you know it is not 100 and it is not 102.
The problem arises when a value of 100 is quoted I would say
Maybe the words were not quite clear, so I''l try again.

Suppose you measure an acceleration of 9.9m/s^2. The value is given to to 2 s.f, and the error is about 1%. Agreed?

Now suppose we convert the value into g. The standard value of earth gravity is 9.80665 m/s2, exactly, by definition.

My calculator says 9.9 m/s^2 = 1.0095190508481489601443918157577g. Too much information!

But to 2 sf, that number is 1.0g, which would be interpreted as having an error of about 10%.

So if you believe significant figures mean something, just changing the units for a quantity can make it 10 times less accurate!

Of course the common sense thing to do is call it 1.01g, and not worry that the number of significant figures is "wrong".

Suppose you measure an acceleration of 9.9m/s^2. The value is given to to 2 s.f, and the error is about 1%. Agreed?

Now suppose we convert the value into g. The standard value of earth gravity is 9.80665 m/s2, exactly, by definition.

The DIFFERENCE between these values is 0.1.....1% The nice bit of slight of hand does not change this.
If 'g' is defined then you can just leave it as a symbol (like c for speed of light) it is not a measured quantity. The 'measured' g of 9.9 implies it is between 9.8 and 10.0 Use these values and you will get the measured value = 9.9/9.8 = 1.0g or 10/9.8 = 1.0g...the .0 is significant.

The whole idea of significant figures is to do with reporting measurements which must have some degree of uncertainty about them. If you make measurements in an experiment and report values it is a big mistake to quote answers that can be shown to be 'too accurate' so you are obliged to use the number with the least number of sig figs as a 'worst case' example of the calculated value.
We can all dream up examples around 99, 100, 101 to show how easy it is to get confused and lost in the detail.

If we use e-zeros example of 65 x 1.96 this could represent the length of a steel bar and the width of the steel bar. If you are required to calculate the area then only 2 figures should be used because the 65 could be a value anywhere between 64 and 66 whereas the 1.96 could be any value between 1.95 and 1.97.
If these were important measurements this tells you that you need a better way to measure the length (or a less precise way to measure the width)

jtbell
Mentor
65 x 1.96 this could represent the length of a steel bar and the width of the steel bar. If you are required to calculate the area then only 2 figures should be used because the 65 could be a value anywhere between 64 and 66 whereas the 1.96 could be any value between 1.95 and 1.97.
Let's check this using a more accurate method of propagating the uncertainties:

(65 ± 1)(1.96 ± 0.01)
= (65 ± 1.54%)(1.96 ± 0.51%)
= 127.4 ± √(1.54%2 + 0.51%2)
= 127.4 ± 1.62%
= 127 ± 2 (after rounding off)

So the answer in this case is more precise than 130 ± 10, and not as precise as 127 ± 1. Nevertheless, I would say the third figure is still significant.

Borek
Mentor
The whole idea of significant figures is to do with reporting measurements which must have some degree of uncertainty about them.
Sorry, but that's not the case. Significant figs are a poor mans way of dealing with uncertainties. Problem is, many people do believe it is THE way of dealing with them, while they are not. Sometimes they are better than nothing, but people that treat their results seriously don't use significant figs, but statistical approach and error propagation.

According to NIST electron mass is 9.10938291x10-31 kg with a standard uncertainty of 0.00000040x10-31 kg (which is sometimes written as 9.10938291(40)x10-31). I wonder how you are going to tell that with significant figures.

'Sometimes they are better than nothing'......exactly....that is the case......they do have some value and poor men (and women) making a start in physics with the hope of developing into 'people that will be able to treat their results seriously' need to be guided in the development of techniques.