Finding number of significant figures

[vcsharp2003]

Homework Statement:

What is the number of significant figures for a measurement of 8400.00 cm?

Relevant Equations:

None

I am using these rules to determine number of significant digits.
(1) All trailing zeros following decimal are significant
(2) All trailing zeros in a number without decimal point are not significant
(3) All non-zero digits are significant
It seems rule 2 will not apply here and therefore number of significant digits is 6 in this measurement.

But what confuses me is if the measured quantity was 8400, then due to rule 2 the number of significant digits would be just 2.

So, the same number has different significant figures when expressed with or without decimal, and this is confusing to me.

 

[malawi_glenn]

Think like this, if the number is 8400.00 cm. How big is the error? In other words what is the range of numbers that has been rounded down or up to 8400.00 ?

 

[Orodruin]

The thing with (2) is that it is not really well stated. It should be "Trailing zeros in a number without a decimal point may or may not be significant" you simply do not know without further information because it could be that you are measuring with two, three, or four digits precision - you just don't know from 8400. It could be any of the following:

• Measurement precision with two significant digits, anywhere between 8350 and 8449.999...9
• Measurement precision with three significant digits, anywhere between 8395 and 8404.999...9
• Measurement precision with four significant digits, anywhere between 8399.5 and 8400.4999...9

You simply do not know from 8400. This is where scientific notation is better because you only have significant digits. The three cases would be $8.4\cdot 10^3$, $8.40\cdot 10^3$, and $8.400\cdot 10^3$, respectively. This makes the number of significant digits clearer.

Of course, if you can estimate the error you can always use a notation such that the error is clear, which is even better. One common notation is to quote the error in the last digits in parentheses, ie, 8400 with an error of 15 would be written 8400(15).

 

[vcsharp2003]

Think like this, if the number is 8400.00 cm. How big is the error? In other words what is the range of numbers that has been rounded down or up to 8400.00 ?

Would it be $8400.00 \pm 0.01$? I am thinking about the least measurement possible, which would be $0.01$ in this case.

 

[vcsharp2003]

Measurement precision with two significant digits, anywhere between 8350 and 8449.999...9

Are you using $\text {measured value} \pm (\text {least count}/2)$ to determine the range? In first case you took measured value as $8.4 \times 10^3$, so least count is $0.1$

 

[apostolosdt]

In your post, you state "a measurement of 8400.00 cm." That fixes everything. When a numerical value is given and described as "measurement," you cannot modify it without some well-based reason. It expresses the precision of the instrument involved in the process. Assuming the numerical value stated is correct, it cannot be rounded off unless it is involved in some arithmetic. Then, the other numbers' significant figures will settle the issue.

Regarding your comments on what confuses you, a numerical value as a result of a measurement can indeed be expressed in a variety of significant figures, but that has nothing to do with rules---it is what the precision of the measuring device is. A 'cheap' one would provide less significant figures; an expensive one many more. In experimental physics, $1\ne 1.0$, right?

Hope that helps.

 

[Orodruin]

Are you using $\text {measured value} \pm (\text {least count}/2)$ to determine the range? In first case you took measured value as $8.4 \times 10^3$, so least count is $0.1$

Yes. Significant digits are essentially the digits that you are confident in so you are effectively rounding to that number of digits. It is sort of a poor-man’s-error-analysis. Quoting the appropriate error and doing proper error propagation is usually a better option.

 

[vcsharp2003]

Significant digits are essentially the digits that you are confident in so you are effectively rounding to that number of digits.

Ok, that makes sense. It seems we must include or exclude the end points of the range so that they round off to the measured value and not just a plus minus half of least count. That's why you took the highest point as $8.4499999 \text {...}$ rather than $8.45$. Is that right? So, using $8.4 \pm 0.05$ would appear as $8.35 \text { to } 8.45$, which would not be correct.

 

[Orodruin]

Ok, that makes sense. It seems we must include or exclude the end points of the range so that they round off to the measured value and not just a plus minus of least count. That's why you took the highest point as $8.4499999 \text {...}$ rather than $8.45$. Is that right? So, using $8.4 \pm 0.05$ would appear as $8.35 \text { to } 8.45$, which would not be correct.

No, that is not the reason. The probability of getting exactly 8.45 is negligible so it really does not matter. The reason that I don't want to use $\pm$ is that it gives the impression of being an actual estimated error. All you are saying with significant digits is that that is an upper bound on the error.

 

[vcsharp2003]

In your post, you state "a measurement of 8400.00 cm." That fixes everything. When a numerical value is given and described as "measurement," you cannot modify it without some well-based reason. It expresses the precision of the instrument involved in the process. Assuming the numerical value stated is correct, it cannot be rounded off unless it is involved in some arithmetic. Then, the other numbers' significant figures will settle the issue.

Regarding your comments on what confuses you, a numerical value as a result of a measurement can indeed be expressed in a variety of significant figures, but that has nothing to do with rules---it is what the precision of the measuring device is. A 'cheap' one would provide less significant figures; an expensive one many more. In experimental physics, $1\ne 1.0$, right?

Hope that helps.

But, isn't it assumed in Physics that any measurement device will have an error range for a given reading.

Which measuring device is always precise in a Physics lab?

 

[malawi_glenn]

Which measuring device is always precise in a Physics lab?

None. But let's use my subpar foodscale as an example. It can only measure weights in whole grams.

So if I now put my shaker with proteinpowder on it, it reads 38 g (I pressed TARE before I inserted the protein powder into the shaker btw).

If I were to put my shaker on the scales I have at the physics lab in my school which can measure weights in hundreds of a gram, I know it would show a value between 38.49 g and 37.51 g. And if I were to use the chemistry labs scales, which can measure thousands of a gram, it would show a value between 38.499 g and 37.501 g.

At least naively. In reality i have to know how the scales are manufactured and calibrated and so on

 

[Orodruin]

In reality i have to know how the scales are manufactured and calibrated and so on

This is important. Particularly for scales that use the local gravitational field to infer mass and claim to measure masses with five significant digits.

But let's use my subpar foodscale as an example. It can only measure weights in whole grams.

It does however likely work with more digits internally. It is just prudent on the part of the manufacturer to do the significant digits job for you and only display the significant digits.

This will be the case for most devices with a display, but there are cases where you would still need to deduce the number of significant digits yourself. For example, measuring a length with a ruler or if your bathroom scale has an analogue scale or if your electric bathroom scale displays tenths of a kg but the displays varies between showing 91.1 and 91.4.