1. Sep 2, 2004

### Imparcticle

I really am having a rather difficult time trying to understand how to add and subtract, multiply and divide significant figures. I know how to round them, how to identify how many significant figures there are in a given measurement, but not add/substract/multiply/divide them.

I was using the following online quiz, (I have my score on it, which is pretty bad, I know but I'm trying to understand WHY I made the errors I did.) to practice for my quiz tomorrow. http://www.syvum.com/cgi/online/serve.cgi/members/gillilandd/quizzes/sigfigs.tdf

If my score does not show, here are the problems I had trouble with:

1.2 cm + 3.78 cm + 45.38 cm

my answer: 50. this is incorrect, so the computer says.

34.8 cm x 98 cm
my answer: 3400 this too is incorrect.

Can anyone please give me an explanation as to what my errors were?

thanks.

2. Sep 2, 2004

### Chi Meson

When adding or subtracting, line up the numbers vertically (you can often do this mentally), making sure that the decimal points are all on top of each other. Notice the number that has the last decimal furthest to the left (In your example it is 1.2). THis position is the position of the last significant digit of the answer (50.4 cm)

In other words, the answer to a sum can only be precise to the same place value as the least precise number being added.

For multiplication and division, you count the sig figs of each number. The answer will have the same number of sig figs as that of the input number with the least sig figs.

3. Sep 2, 2004

### needhelpperson

ya why does it say its wrong?

Last edited: Sep 2, 2004
4. Sep 2, 2004

### Werdatothewise

You added wrong - your first answer should be 50.36 cm, and the second you multiplied wrong - should be 3410.4cm*2

~Werdas' @2004

Last edited: Sep 2, 2004
5. Sep 2, 2004

### Werdatothewise

Originally Posted by Imparcticle
I really am having a rather difficult time trying to understand how to add and subtract, multiply and divide significant figures. I know how to round them, how to identify how many significant figures there are in a given measurement, but not add/substract/multiply/divide them.

I was using the following online quiz, (I have my score on it, which is pretty bad, I know but I'm trying to understand WHY I made the errors I did.) to practice for my quiz tomorrow. http://www.syvum.com/cgi/online/ser...zes/sigfigs.tdf

If my score does not show, here are the problems I had trouble with:

1.2 cm + 3.78 cm + 45.38 cm

my answer: 50. this is incorrect, so the computer says.

34.8 cm x 98 cm
my answer: 3400 this too is incorrect.

Can anyone please give me an explanation as to what my errors were?

thanks.

~Werdas' @2004

6. Sep 3, 2004

### HallsofIvy

Staff Emeritus
"Werdatothewise", in addition to having a problem with spelling, isn't giving you a lot of help here. Chi Meson's remark, that your answer is correct, is ALMOST on target. It's not really a matter of "adding, multiplying", etc. Just go ahead and do the arithmetic the way you normally would (with a calculator!?) and then remember that you answer must have no more significant figures that the LEAST ACCURATE data.

In this case, 1.3 cm has only two significant figures so, no matter how accurate the other numbers are, you answer should have two significant figures.

So why isn't 50 correct? Because you got caught by a "loophole". We tend to think of integers as being exact (unlike measuring, counting is exact) so to make it clear that this is a measured quantity and is NOT exact, it is better to write it as 50.0. I suspect that is the desired answer.

It would be even better to use "scientific notation" and write 5.0x 101 which clearly shows that it has two significant figures.

(The exact integer "50" would not be written in scientific notation- just as 50.)

We run into the same problem with large numbers. Suppose I have measured something to be 5000 miles to the nearest 10 miles (in other words, there are 3 significant figures). How do I show which of the "0"s are "significant and which are not? Again, the answer is scientific notation: 5.00 x103.

7. Sep 3, 2004

### Chi Meson

HallsofIvy, I would usually conceed to your better knowledge, but here I truly know what I'm talking about.

Unless you can point to something that says otherwise, I am going to stand by my statement. OR have the rules changed in the last year?
When adding, the rules are different. If you measure out a distance with a ruler and get a measurement of 115.37 cm and then you measure an additional distance with a caliper and get 0.1478 cm, do not tell me that, when adding them together, the correct answer is 115.5 cm

The least precise measurement here is 115.37 cm . The extra precision of the caliper will not make the first measurement more precise, nor will it make the first measurement less precise. Rather the result will be limited to the decimal place value of the least precise number.

After rounding, the above sum would be 115.52 cm.
I guarantee you, the answer to the original sum is 50.4 cm

I do agree about the multiplication sample though.

Last edited: Sep 3, 2004
8. Sep 5, 2004

### Imparcticle

This is what I do not understand. I understand the laws which allow me to determine the nature of significant figures in a given number, but I fail to understand why they're used, and how you know that there are 3 significant figures in your aforementioned example.

BTW, I got a B+ on my test (8/10). The ones I missed had to do with addition. My error common to both problems missed concerned the significant digits in my final answer whose place values did not match that of the number with the least value of significant figures. This is so according to my teacher (if what I wrote is incorrect, I hope u know what I mean). I'm not sure I understand because I come across this error only on a few problems but not all. :uhh:

THANK YOU!!

9. Sep 5, 2004

### Moonbear

Staff Emeritus
The reason why significant figures are used is to indicate the uncertainty in a number. So, in Chi Meson's example of using a ruler vs a caliper, a ruler has larger distances between graduations (lines) on it than a caliper, so you cannot be as precise using a ruler to obtain a measurement as you can be with a caliper. The last digit of your measurement is the least precise, because of the limits of your instrument. For example, if you have a ruler that has marks every .1 cm, you can't record the length of something as .125 cm because you don't have that degree of precision to your instrument (ruler).

In the example of 5.00 X 10^3 miles, there are three significant digits because you count the zeros to the right of the number. This tells you your instrument was capable of measuring to the nearest 10 miles. You wouldn't be able to tell this if it was written as "5000". This is the purpose of using scientific notation for numbers. On the other hand, you do NOT count zeros to the left of the first number. They would just be place holders, such as with .00045, then you would only have two significant digits.

Chi Meson is correct about addition rules for significant figures. However, one thing to be careful of is that you should add before rounding your numbers, and then round the answer to match the decimal number of the least precise number.

For the original multiplication question you posed, HallsofIvy's answer about using scientific notation becomes important for getting the correct answer. With an answer of 3400, it's ambiguous if you intend to convey that you have precision of 3400+100, 3400+10, or 3400+1. So, the only way to clarify this is to use scientific notation of 3.4x10^3

Does this help?

10. Sep 5, 2004

### Fredrik

Staff Emeritus
I think the easiest way to understand this is to insert slightly different numbers into the calculations.

1.2 cm + 3.78 cm + 45.38 cm

If we take the numbers to be exact, the correct answer is 50.36 cm. Let's see how much that changes if we increase the last digit in one of the terms by one.

1.3 cm + 3.78 cm + 45.38 cm = 50.46 cm (0.1 cm too large)
1.2 cm + 3.79 cm + 45.38 cm = 50.37 cm (0.01 cm too large)
1.2 cm + 3.78 cm + 45.39 cm = 50.37 cm (0.01 cm too large)

As you can see, the uncertainty in 1.2 causes an uncertainty in the final result of order 0.1 cm. The contribution to the uncertainty in the final result from the other two numbers is only of order 0.01 cm, and is therefore irrelevant.

Since the uncertainty is of order 0.1 cm, the answer should be expressed as 50.4 cm.

Let's try the same trick with the other calculation (34.8 cm x 98 cm). The exact result is 3410.4 cm^2.

34.9 cm x 98 cm = 3420.2 cm^2 (almost 10 cm^2 too large)
34.8 cm x 99 cm = 3445.2 cm^2 (almost 35 cm^2 too large)

As you can see, the uncertainty is of order 10 cm^3, so the answer should be expressed as 3.41*10^3 cm^3.

11. Sep 5, 2004

### Imparcticle

THANK YOU for all you help. this is all very clear now.

12. Sep 6, 2004

### Leong

imparcticle,
so, the answer for the 2nd question is $$3.4X10^3 cm^2$$?

if i write the answer as 3400 (2 significant figures) cm square, is it the same with the above answer because i think that it is clearly written here that the answer is to 2 significant figures only and implies that the last two zeros serve as a place holder.

is your user name supposed to be 'impracticle'?

13. Sep 6, 2004

### Moonbear

Staff Emeritus
The same rules don't apply to multiplication as to addition with significant figures. The answer would have to have only two significant figures, because 98 only has two significant figures, so 3.4x10^3, not 3.41x10^3. If there were supposed to be 3 significant figures, 98 would be written instead as 98.0

14. Sep 6, 2004

### Moonbear

Staff Emeritus
That way of writing it would be unambiguous, but not consistent with standard conventions, and quite cumbersome to use. Scientific notation also gains greater convenience when you work with very large and very small numbers, as you often will do in chemistry.

15. Sep 6, 2004

### Imparcticle

YES.

3.4 x 103 = 3400
If the answer is 3400., then it has 4 significant figures. So yes, the last two zeros (if there is no decimal point after the last digit) would not be considered significant digits.

Is it supposed to be 'impracticle' or did I chose for it to be a variation of the aforementioned root?
I chose my user name after seeing a spelling error with the word "impractical". They spelled it "impracticle", and hence, after a bit of revising of the word, I came up with "imparcticle". My user name has been and will further continue to be "imparcticle". I hope that adequately answers your inquiry.

16. Sep 6, 2004

### Leong

Moonbear,
if you were to grade me, would you give me wrong for the answer i had provided ?
Sometimes, all that matter to a student is 'right or wrong' only, 'why' will always come in the second. at least, it is true for me.

17. Sep 6, 2004

### Fredrik

Staff Emeritus
You're right, the result should be written as 3.4*10^3. This would have been obvious if I had changed the number 98 more. If I e.g. had used 102 instead (to see the effect of an uncertainty of 4 in the last digit) I would have gotten the result 3549.6, which is 239.2 too large. So the uncertainty is definitely in the second digit.