working with significant figures for math calculations

M

mes

New member
Joined
Sep 12, 2012
Messages
3
Not sure if this is the right place to ask this. I thought I knew the rules for working with significant figures but I don't know why my answer is wrong. Here is the density problem. 46.9/1.114 = 42.100. I thought I need to look at the anwer as having the same significant figures as the least precise number in my calculation. Seems to me that 46.9 has only 3 significant figures so my answer should have 3 significant figures . Therefore I said my answer should be 42.1 but the prompt says that is the wrong number of significant digits. What is the right answer and why?
 
But it is NOT "silliness" for scientists or engineers, as opposed to mathematicians. Because "significant digits" is not a matter of mathematics, it is a matter of measurement. If you say that a measurement, the length of a rectangle, say, is "49.3 m" then you saying that your measuring apparatus allowed you to measure that length to the nearest tenth of an meter- that the actual value might be anywhere from 49.25 to 49.35 meters. If you measure the width of the rectangle as 3.2 meters, also to "the nearest tenth of an meter", the actual width could be any where from 3.15 to 3.25 meters.

Just multiplying the measured lengths would give an area of (49.3)(3.2)= 157.76 square meters. But, in fact, it could be as little as (49.25)(3.15)= 155.1375 square meters or as large as (49.35)(3.25)= 160.3875 square meters. That is, we could be several square meters off. The simplest way to show that is to give the result as "160 square meters", rounding to two "significant digits" (better would be "scientific notation"- \(\displaystyle 1.6\times 10^2\) clearly showing that last "0" is NOT a significant digit), the number of significant digits in the least accurate measurement, 3.2 meters.

mes, I agree with you. "4.69" has three significant digits so your answer should. I recommend you talk to your teacher about this.
 
Last edited:
I've always struggled with measurement units. 1/10 is quite arbitrary. I've always thought that I could judge 1/4 a lot better than 1/10, but 1/4 adds another significant digit, so it is not allowed. Even if I believe it is exactly 1/4, I have to pick 0.2 or 0.3.
 
tkhunny said:
I've always struggled with measurement units. 1/10 is quite arbitrary. I've always thought that I could judge 1/4 a lot better than 1/10, but 1/4 adds another significant digit, so it is not allowed. Even if I believe it is exactly 1/4, I have to pick 0.2 or 0.3.
Click to expand...
"1/10"is NOT a "measurement unit". A typical meter stick has a marks at each centimeter and then smaller marks at each millimeter. Measuring something with a meter stick you should be able to estimate which of the two millimeter marks it is closer to. That is, measuring something with a meter stick you normally have something like ".432 meters", with 3 significant figures, with the understanding that it could be any where from .4315 meters to .4325 meters.

This is also why we use "significant figures" rather than "number of decimal places" you may remember from elementary school. "0.432 meters" is exactly the same is exactly the same as "43.2 cm" or "432 mm". The number of significant figures is 3 but the number of decimal places varies.
 
For engineers, significant digit is very important - because precision costs money. So in general, we reduce it down to "tolerance" - and that is why sometimes we hear about NASA paying $10 for a bolt (a 2.000" long and 0.5000 dia bolt made of tantalum can even cost $100 dollars).

These days I teach basic engineering in a community college. In my class no more than 3 sig digits allowed (no matter whatever the precision of the input and people get crazy about π ) - no body is going to pay for that on as long peoples lives are not dependent on that structure.

Anyway, back to the original question - the answer should be 4.69, in my opinion.
 
HallsofIvy said:
"1/10"is NOT a "measurement unit".
Click to expand...

Whoa, there. I quite deliberately wrote my comments without reference to any specific units.

Maybe I had a different kind of physics teacher in high school. We were required to estimate 1/10 of the smallest completely marked standard unit (usually inch, ounce, centimeter, gram, or millilitre for the small things we were working on), no matter the other markings on the device. If it was inches and 1/5 inches, we still used 1/10 inches as the last significant digit. That can be pretty confusing when 1/16 inch (0.0625") is the smallest marking. Never confusing with metric units, of course.

I hadn't realized until this moment that there remained variation of thought on the matter.

I always wanted to make measurements of "between a and b" and produce range estimates for calculated final results. Again, my high school physics teacher never liked that, either.

No wonder I didn't go into physics!
 
tkhunny said:
Whoa, there. I quite deliberately wrote my comments without reference to any specific units.

Maybe I had a different kind of physics teacher in high school. We were required to estimate 1/10 of the smallest completely marked standard unit (usually inch, ounce, centimeter, gram, or millilitre for the small things we were working on), no matter the other markings on the device. If it was inches and 1/5 inches, we still used 1/10 inches as the last significant digit. That can be pretty confusing when 1/16 inch (0.0625") is the smallest marking. Never confusing with metric units, of course.
Click to expand...
I agree that that is foolish- which is why I objected to your talking about "1/10" as a "measurement". As I said in my first post, you can generally estimate which of the smallest marking is closer to the actual measurement and so giving the nearest mark gives a \(\displaystyle \pm\).5 .

I hadn't realized until this moment that there remained variation of thought on the matter.

I always wanted to make measurements of "between a and b" and produce range estimates for calculated final results. Again, my high school physics teacher never liked that, either.

No wonder I didn't go into physics!
Click to expand...
 
Last edited:
mes said:
Not sure if this is the right place to ask this. I thought I knew the rules for working with significant figures but I don't know why my answer is wrong. Here is the density problem. 46.9/1.114 = 42.100. I thought I need to look at the anwer as having the same significant figures as the least precise number in my calculation. Seems to me that 46.9 has only 3 significant figures so my answer should have 3 significant figures . Therefore I said my answer should be 42.1 but the prompt says that is the wrong number of significant digits. What is the right answer and why?
Click to expand...

Significant digits are important in "real world" measurements. They are a reflection of the fact that we are limited in the accuracy with which we measure things.

Since your problem only involves multiplication or division, your approach and answer are correct; use 3 sig digs in your answer.

Remember that the rules for addition/subtraction are different. Then you are limited by the number with the least accuracy. For example, "101 + .2" does not equal 101.2. We only know the 101 number to the nearest digit -- not to the nearest tenth of a digit. The true value might be 101.452 or 100.6374. We simply don't know. We only know that the value is closer to 101 than it is to either 100 or 102. Therefore, the answer to "101 + .2" is still "101".
 
I like to remember this simple example:

The Earth + One Grain of Sand = The Earth

It's not a lesson in addition. It's a lesson in significance.
 
You must log in or register to reply here.
Top