Why are zeros after a decimal point significant?

[curiousstudent]

I understand the rules of significant figures. One of those rules says that zeros to the right of a decimal point are counted as significant figures. I don't understand why that is. If you have the number 8 the zeros following are understood. In 8.00 the two zeros don't need to be there in regular problems, so why does the rule change so that you keep them there and count them as significant if they don't need to be there?

 

[philosofeem]

It signifies the accuracy of the measurement. In that specific case it would mean that the measurement was accurate enough to know that it was 8.00 and not 8.07 or 8.04. If it were 8.0, the number in the hundredths place would be uncertain because whatever measuring instrument you were using couldn't measure to that level of certainty. Also if you multiplied 8.00 times 8.0 the zero in the hundredths place would be made uncertain and therefore, insignificant. Someone please correct me if I'm wrong

 

[curiousstudent]

So if you had something that could measure exactly 8, do we add an infinite number of zeros

 

[DrewD]

So if you had something that could measure exactly 8, do we add an infinite number of zeros

Well... yes, but nothing ever could, so no. This is not math, it is a notation convention to simplify how one writes a measured quantity. Mathematically it is obvious that 8=8.0, but if it is a measured quantity, then a different convention is used.

 

[curiousstudent]

What I don't understand is why this is the first time I have done this. in science we have never before used unnecessary zeros. What if something can only be measured to a whole number such as 8? In that situation there are understood zeros but they aren't measured. What would you right, and I still don't understand why they are significant figures if they are simply place holders.

 

[philosofeem]

What I don't understand is why this is the first time I have done this. in science we have never before used unnecessary zeros. What if something can only be measured to a whole number such as 8? In that situation there are understood zeros but they aren't measured. What would you right, and I still don't understand why they are significant figures if they are simply place holders.

If the measurement is just 8, then it is not 8.0 or 8.00 or 8.0000-to infinity. If the measurement is plain, whole 8 then all numbers after the decimal point are insignificant because they are uncertain. With a better measuring tool it could actually be 8.18 or 8.0000001. With measurements, 8. Is very different from 8.0000000000.

 

[Nugatory]

What if something can only be measured to a whole number such as 8?

There's no such thing, because whether you get a whole number or not depends on your choice of units. Suppose you tell me that you had measured the mass of something to be exactly 500 grams, and 500 is a nice whole number. OK, what's its mass in ounces?

However, you are right to be uncomfortable about the this custom of extending zeroes to the right to show the limits of a measurement. The modern style, much preferred, is to show the error limits explicitly by writing something like $7.53\pm.02$ or $8.00\pm.02$; in that form the trailing zeroes in the second example don't look so much like a weird special case.

 

[Curious3141]

I understand the rules of significant figures. One of those rules says that zeros to the right of a decimal point are counted as significant figures.

Just to clarify a misconception, when the leading number before the decimal point is zero, the zeros to the right of the point but preceding a non-zero digit are considered non-significant. The reason is that all those zeros are simply "placeholders" to tell you the magnitude of the number. If you have a number like 0.0000000052123 and you wanted 2 sig figs, you would truncate to 0.0000000052 and not 0.0 or even 0.00.

Often, an easy way to correctly round a number like this is to express it in scientific notation $a \times 10^b$, and apply the rounding only to $a$.

 

[curiousstudent]

With measurements, 8. Is very different from 8.0000000000.

If you add those measurements it is exactly the same though, correct? If you add 8+8.000 the answer would be 16 because the other zeroes are "uncertain". So the significance there is null. And not to mention but there are also people who want me to accept this but then say that zeroes to the left of a decimal point and to the right of significant figures are not significant (ex. 8200 has 2 significant figures) This does not make sense since they are also very important to the accuracy. 8200 is a lot different than 82, and what if it were measured more accurately like 8200.0, then are the zeroes to the left of the decimal significant? I apologize for my long windedness but I am very confused on the subject.

 

[Nugatory]

8200 is a lot different than 82

Not so.

The difference between 8200 and 82 is just the difference between measuring the exact same volume with the exact same measuring apparatus accurate to one part in one hundred, but with the scale labeled in centiliters instead of liters.

This will be more clear if you use the modern style, where we'd report the measurement as $(8.2\pm.1)\times 10^1$ liters or $(8.2\pm.1)\times10^3$ centiliters.

 

[curiousstudent]

But if you were o convert 8200 liters to centiliters it would not be the same, your point does not apply. The context of that quote is in reference to significant figures and why zeroes have such strange and seemingly unnecessary rules.

 

[Nugatory]

If you add those measurements it is exactly the same though, correct? If you add 8+8.000 the answer would be 16 because the other zeroes are "uncertain".

No, they're still certain. We don't carry them through to the sum because the uncertainty in the 8 on the left is so large. Let's try writing $8+8.000$ out carefully: It's $(8\pm1)+(8\pm.0001)=16\pm1\pm.0001\approx16\pm1$; the accuracy in the second addend is swamped by the inaccuracy in the first.

 

[curiousstudent]

I think I'm starting to understand. Now the big question here is why don't we use significant figures elsewhere like in math.

 

[DrewD]

I think I'm starting to understand. Now the big question here is why don't we use significant figures elsewhere like in math.

Because there are no measurements. Everything is exact.

 

[curiousstudent]

but what about when finding the perimeter or something like that. 3 feet should be more accurately portrayed right

 

[Nugatory]

But if you were to convert 8200 liters to centiliters it would not be the same, your point does not apply.

How so? If the measurement is accurate to one part in one hundred, then it's accurate to one part in one hundred no matter how I convert the units - the error just multiplies or divides along with the base measurement.

The context of that quote is in reference to significant figures and why zeroes have such strange and seemingly unnecessary rules.

Those "strange and seemingly unnecessary" rules are the rules necessary to make significant figure calculations work as a reasonable approximation to the the more precise $\pm$ error bounds method.

There's some history here. Once upon a time, back in the old days when aspiring young science and math students had to ride dinosaurs to school, calculations were done with an archaic device called a "slide rule" (and I swear I'm not making this up when I say that some of the best ones were made of bamboo). Slide rules can do multiplication and division but not addition and subtraction, which is a real problem for the $\pm$ way of describing error bounds. Thus, significant digits, along with the funny zero-digit rules, were invented as a way of representing the error bounds in a purely multiplicative way, and they're much less useful now that we have electronic calculators that can handle addition and subtraction as easily as multiplication and division.

 

[DrewD]

I haven't measured anything in a math class since middle school. Sure, we've talked about geometry, but it was purely theoretical so there was no measurement.

I'm going to tell you the truth. The only place I've ever used sigfigs is in Chem lab. Everyone else just reports the precision of their measuring device. I would assume that any math class considering measurement/sampling errors would do the same thing.

 

[philosofeem]

but what about when finding the perimeter or something like that. 3 feet should be more accurately portrayed right

Do you mean like "find the area of a 3.078292737297271ft by 8.0002727627227ft room?" Math is a tool to learn the mathematical relationships between objects, such large numbers for such simple calculations would be unnecessary

 

[voko]

There's no such thing, because whether you get a whole number or not depends on your choice of units.

There are such things. The number of electrons in an atom does not depend on any units, and it is always whole.

Both instances of number "2" in $mv^2/2$ are exact and whole.

Let's face it, the significant figures notation is not universally applicable.

 

[CompuChip]

I posted this answer in another thread a while back.
Maybe it will help you.

Before answering your question, maybe a slightly more practical example will be useful. Suppose that I want to know how tall you are. By visual inspection - as it's often formally called - I can easily determine that your height is about 2 m. But of course, it can be 1.90, or 1.84. In fact, just by looking at you it would be crazy to pretend that I know your height to even within 10 cm. Now I could specify the result of my "measurement" as 2.000 m. But that would imply that I know that it is 2 m to great accuracy, while really all I can say is that it is "somewhere between 1.5 m and 2.5 m - well, 2.5 is a bit high but it could be even 2.20 - in any case it will round to 2 m." This is what I express by saying "2" instead of "2.000".

Now let me get my measuring tape which has a cm scale with small ticks between the integer values, and measure it more accurately to be 1.84 m. Now of course, I could again say "You are 2 m tall", but that would give less information than I actually have. On the other hand, I could interpolate between the marks and say "You are 1.843 m tall". But of course my interpolation is a bit uncertain - my eye is not good enough to distinguish between 1.8424 and 1.8437. So stating the number as 1.843 would again fool you into thinking I measured it more accurately than I did. Basically, by reading off the number on a 0.5 cm scale, all I know that it is between 1.835 and 1.845. All these values (possibly with the exception of 1.845 exactly) round to 1.84. So I should say 1.84; in contrast 1.840 would imply that I actually measured it up to mm.

Now when you add or subtract values, you have to take this accuracy into account. Suppose I also measure my own height in the first way - I will also find 2m! Subtracting the two gives 2m - 2m = 0m. This does not mean, of course, that we have the same height. We do have the same height within the accuracy of our measurement - i.e. we are equally tall up to a difference of +/- 1 m. To sketch a "worst-case" scenario: a more accurate measurement may reveal us to be 2.49 m and 1.51 m respectively, giving a difference of 0.98 m. In contrast, if I would have measured us both to be 1.84m in 2 significant digits, the biggest difference we could have would be between 1.835 and (slightly under) 1.845, so 0.01m.
Now of course, when I report this difference, again I have to take this into account. In the first case, where I only have 1 significant digit in the measurement, I cannot report the difference between the measurements more accurately than that, so I should not pretend I can and write 0.0 m.
Even if I would know from my ID that my height is 1.84 m, but I only know yours in 1 significant digit as 2m, I could not say that the difference is 0.16m. After all, this would imply that the actual difference is within 0.005m. This would be true if my more accurate measurement turns out to give 1.84 for you as well, but you could be 2.4m, for all I know - the "2m" does not give me more information. Therefore, even if I would know my height up to a fraction of a centimeter, knowing yours in one significant digit will still force me to give a less accurate result. A little thought shows that the "worst" case here is you turn out to be 2.49m or 1.51, giving a difference of about 0.34m either way. So in any difference I calculate, there is a (big!) uncertainty even in the first decimal, meaning I cannot give the difference more accurately than meters (again giving 2 m - 1.84 m = 0 m, with the right significance).Yes.Yes, otherwise you should either round it correctly - if you know that it is that much bigger than .234670 - or you should give it in fewer significant digits.You're not entirely right, and I think you are worrying too much about the edge case. The "rounding 5 up" rule - whichever one you use - is only relevant if the number ends in ...5. If the actual value were ...50001 you would round it up anyway, and if it were ...499990 you would round it down. So either the error would be .00000049999... (repeating) or 0000005 - and these numbers are the same (by the famous 0.9999... = 1).

 

[Nugatory]

So if you had something that could measure exactly 8, do we add an infinite number of zeros

There's no such thing

There are such things. The number of electrons in an atom does not depend on any units, and it is always whole.
Both instances of number "2" in $mv^2/2$ are exact and whole.

But they are not measurements.

 

[voko]

But they are not measurements.

The number of electrons in an atom is not a measurement? Then what is it?

If I count the number of persons in a room, is that not a measurement? And if I say the headcount is 8, does that mean there might actually be 7.8 or 8.3 men?

 

[Borek]

But they are not measurements.

Where is a difference between counting and measuring?

Tardigrades are eutelic - that means in a given species all organisms has exactly the same number of somatic cells. Say all tardigrades from a given species have 39013 cells. Isn't it an exact number - even if it was "measured" by "counting"?

Edit: voko was faster. Checking details always takes time.

 

[DrewD]

The number of electrons in an atom is not a measurement? Then what is it?

If I count the number of persons in a room, is that not a measurement? And if I say the headcount is 8, does that mean there might actually be 7.8 or 8.3 men?

The 2 in $\frac{mv^2}{2}$ is not a measurement.

Perhaps, as your headcount point shows, it is more appropriate to state that sigfigs are used for continuum measurements, not discrete measurements.

Also, I don't think anyone here thinks they are universally applicable. I think we all agree that they are just a short hand notation used in some instances, but perhaps I shouldn't speak for others.

 

[voko]

The 2 in $\frac{mv^2}{2}$ is not a measurement.

No it is not. But in $G \frac {m_1 m_2} {r^2}$ it is, and accurate to more than just one figure. Yet we just write "2" most of the time.

Also, I don't think anyone here thinks they are universally applicable. I think we all agree that they are just a short hand notation used in some instances, but perhaps I shouldn't speak for others.

That was what I wanted to say: it is a convention, very frequently useful, but sometimes confusing or even inapplicable. One should not just treat any number as if it were automatically subject to this convention, some consideration is necessary.

 

[Nugatory]

Perhaps, as your headcount point shows, it is more appropriate to state that sigfigs are used for continuum measurements, not discrete measurements.

That's probably the best way of thinking about it.

 

[CompuChip]

For discrete measurements I would use scientific notation to indicate the accuracy, e.g. I would write
$88 \cdot 10^0$
$9 \cdot 10^1$
$1 \cdot 10^2$
depending on how accurately I counted to 88.

 

[ThomasO]

So if you had something that could measure exactly 8, do we add an infinite number of zeros

I would write 8.0 with a dot over 0, in the same way we represent 1/3=0.3333... as 0.3 with a dot over 3. But I am not sure anyone would understand it.

 

[Algr]

So basically significant zeros are a simple, but somewhat limited way to show measurement accuracy. If I say there are 8.0 people in a room, I am sort of implying that 8.1 might be possible.

Significant zeros also show up where you have integer-like limits to accuracy. For example, US currency is limited to 2 significant digits. A penny is .01, but you can't divide that further using physical coins and bits. You can include or omit the sub-dollar amounts, $2 or $2.00, but $2.0 would look strange and make people think some kind of misprint had happened.

 

[HallsofIvy]

No it is not. But in $G \frac {m_1 m_2} {r^2}$ it is, and accurate to more than just one figure. Yet we just write "2" most of the time.

If you are referring to the "2" in r^2, no it is NOT. That is Newton's law of gravity and the "2" is precisely that- we are squaring the r. I suspect you have read somewhere that, due to Einstein's "General Theory of Relativity" says that this should be "2 point something". (I think I heard that when I was about 14.) That is hogwash promulgated by science "popularizers" who are far too much over simplifying what they themselves do not undstand. The General Theory of Relativity treats gravity through tensor equations that have nothing to do with Newton's equation and are NOT just adding decimal places.

 

[HallsofIvy]

So basically significant zeros are a simple, but somewhat limited way to show measurement accuracy. If I say there are 8.0 people in a room, I am sort of implying that 8.1 might be possible.

No, you are not. Read DrewD's post, above:
"Perhaps, as your headcount point shows, it is more appropriate to state that sigfigs are used for continuum measurements, not discrete measurements."

Significant zeros also show up where you have integer-like limits to accuracy. For example, US currency is limited to 2 significant digits. A penny is .01, but you can't divide that further using physical coins and bits. You can include or omit the sub-dollar amounts, $2 or $2.00, but $2.0 would look strange and make people think some kind of misprint had happened.

I would not consider that a matter of "significant digits", spefically because " A penny is .01, but you can't divide that further using physical coins and bits." It is, again, matter of counting, not measuring.

 

[Borek]

I suspect you have read somewhere that, due to Einstein's "General Theory of Relativity" says that this should be "2 point something".

I believe it was suggested as an explanation when the anomalies of the perihelion precession of Mercury orbit became evident for the first time. These anomalies are now nicely explained by GR and are typically listed between tests showing GR is correct.

 

[voko]

If you are referring to the "2" in r^2, no it is NOT. That is Newton's law of gravity and the "2" is precisely that- we are squaring the r.

If you teach physics axiomatically, perhaps.

 

[pwsnafu]

If you teach physics axiomatically, perhaps.

Huh? Even if it isn't axiomatically taught, that's what Newton's law of universal gravitation says. It's a statement. Whether or not that statement matches observations is irrelevant, as a mathematical expression that 2 is exact.

 

[voko]

Huh? Even if it isn't axiomatically taught, that's what Newton's law of universal gravitation says. It's a statement. Whether or not that statement matches observations is irrelevant, as a mathematical expression that 2 is exact.

Axiomatic physics illustrated.

 

[pwsnafu]

Axiomatic physics illustrated.

In that case you are going to have to explain how non-axiomatic science exists at all. Because in all languages (including all mathematical sciences) you have "what the statement says" and "whether the statement is true". The only way I can interpret you is that the former doesn't exist (i.e. the statement is not saying what the statement is claiming to say).

 

[voko]

In that case you are going to have to explain how non-axiomatic science exists at all. Because in all languages (including all mathematical sciences) you have "what the statement says" and "whether the statement is true".

There is a science called "physics". Anything that science says is a model or a consequence from a model, and no model is ever (well, not ever, but for quite a while now) considered "true". It is just considered (or not) to be in some good agreement with experimental and observational data within some specified or implied limits.

As regards gravitation, there is even a model where it is constant. It works fine where it is applicable. Is it "true" or not? There are other models. Newton's model; Einstein's model (and the post-Newtonian spin-off); Yukawa-modified gravity, just to name a few.

 

[pwsnafu]

There is a science called "physics". Anything that science says is a model or a consequence from a model, and no model is ever (well, not ever, but for quite a while now) considered "true". It is just considered (or not) to be in some good agreement with experimental and observational data within some specified or implied limits.

As regards gravitation, there is even a model where it is constant. It works fine where it is applicable. Is it "true" or not? There are other models. Newton's model; Einstein's model (and the post-Newtonian spin-off); Yukawa-modified gravity, just to name a few.

You are right I shouldn't have said "truth". I did mean "agreement with observations" (is there is a word for that? I guess "accurate"?) when I made my post.

Getting back. You claimed that the 2 in r², was not exactly two (i.e. not an integer 2), and when this was pointed out you claimed that was axiomatic. Newton's statement is that it is an integer 2 and not 2 point something. It just happens to fail in certain situations, and (just as you say) we can use other models.

Am I interpreting you correctly?

If that is so, I fail to see how your second paragraph is not also axiomatic physics. We have what the Newton's model says (the 2 is exact) and then afterwards we have whether it it satisfies the observations. You can't compare it to data unless you understand the model first.

Or maybe I have this "axiomatic vs non-axiomatic" division completely wrong?

 

[voko]

Getting back. You claimed that the 2 in r², was not exactly two (i.e. not an integer 2), and when this was pointed out you claimed that was axiomatic. Newton's statement is that it is an integer 2 and not 2 point something. It just happens to fail in certain situations, and (just as you say) we can use other models.

Am I interpreting you correctly?

I said that the law was a result of measurement. Or observation, if you will. As such, it is bound to have some uncertainty - that's why we have this entire "significant figure" business to begin with. There is no a priori reason why the 2 in the law is on a completely different footing than, for example, the G in that same law.

 

[jbriggs444]

I said that the law was a result of measurement. Or observation, if you will. As such, it is bound to have some uncertainty - that's why we have this entire "significant figure" business to begin with. There is no a priori reason why the 2 in the law is on a completely different footing than, for example, the G in that same law.

You are suggesting, I suppose, that if our ability to measure r accurately over a range of radii were impaired and our ability to measure M and m accurately over a range of test objects were improved and if we treated the problem purely as modelling a set of data points by regression to a function of the form F = GmM/rⁿthen we could plausibly know G to a greater accuracy than n rather than the reverse.

Sure, that seems fair.

 

[D H]

There is no a priori reason why the 2 in the law is on a completely different footing than, for example, the G in that same law.

Yes, there is. Gravitation is instantaneous in Newtonian physics and space is isotropic, three dimensional, and distinct from time. That 2 is exact given those assumptions.

Discard those assumptions and you don't get Newtonian gravity. You get general relativity, which doesn't look like Newtonian gravity. In the limit of small masses, small velocities, and large distances, general relativity does simplify to Newton's law of gravitation -- and the 2 is exact.

 

[CompuChip]

Sorry to interrupt this fascinating discussion, but is this still at all relevant to the original question? Or has the original topic starter left us long ago?