# Why are zeros after a decimal point significant?

1. Aug 27, 2013

### 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?

2. Aug 27, 2013

### 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

3. Aug 27, 2013

### curiousstudent

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

4. Aug 27, 2013

### DrewD

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.

5. Aug 27, 2013

### 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.

6. Aug 27, 2013

### philosofeem

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.

7. Aug 27, 2013

### Staff: Mentor

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.

8. Aug 27, 2013

### Curious3141

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 $a$.

9. Aug 27, 2013

### curiousstudent

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 alot 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.

10. Aug 27, 2013

### Staff: Mentor

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.

11. Aug 27, 2013

### 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.

12. Aug 27, 2013

### Staff: Mentor

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.

13. Aug 27, 2013

### 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.

14. Aug 27, 2013

### DrewD

Because there are no measurements. Everything is exact.

15. Aug 27, 2013

### curiousstudent

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

16. Aug 27, 2013

### Staff: Mentor

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.

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.

17. Aug 27, 2013

### 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.

18. Aug 27, 2013

### philosofeem

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

19. Aug 28, 2013

### voko

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.

20. Aug 28, 2013