What is the proof of the rules of significant figures?

In summary, the rules of significant figures are a simplified version of error analysis and propagation. When multiplying and dividing, the number of significant figures in the result should be equal to the number with the least significant figures. This is because the error in the least precise measurement will dominate the overall error in the calculation. Essentially, significant figures represent the level of precision in a measurement and should not be interpreted as exact values.
  • #1
fxdung
388
23
Please prove the rules of significant figures. I do not know why when multiplying and dividing we have to retain the same number of significant figures as in the number with the least of them.
 
Mathematics news on Phys.org
  • #2
The rules of significant figures is a poor man’s version of error analysis and propagation. It is more rules of thumb about how far you can trust your precision than actual rules.
 
  • Like
Likes Klystron and jedishrfu
  • #3
Let's say for instance you have a measured value which you write as ##x = 2.3##. What that is supposed to mean is that you're confident of the 2 and the 3, but not of any further figures. You don't mean that the real value is 2.3000. You mean it's approximately 2.3. You mean that it could be 2.32. It could be 2.27. You can't distinguish between those possibilities with your measuring equipment. It's something that rounds to 2.3, so it could range from 2.25 to just under 2.35. We could write this as 2.30##\pm##0.05. Let's call this ##x + dx## where ##-0.05 \leq dx \lt 0.05## (technically it can't equal +0.5 exactly as that would round up, not down).

Now let's say we have another measured value ##y = 4.12##. On this one, I'm confident of the three digits, but no more. So actually it's 4.12##\pm##0.005 or ##y + dy## with ##-0.5 \leq dy \lt 0.5##.

What will we say about a calculated value ##z = xy##? Well since the correct values of ##x## and ##y## are actually a range of values, we have a range of possible values for ##z##.
##z + dz = (x + dx)(y + dy) = xy + y\;dx + x\;dy + dx\,dy##
So the error part that we add to ##xy## is ##dz = y dx + x dy + dx dy##.
That means that ##dz/z = (y\;dx)/z + (x\;dy)/z + dx\,dy = (y\;dx)/(xy) + (x\;dy)/(xy) + dx\,dy/(xy)## = ##(dx/x) + (dy/y) + (dx/x)(dy/y)##.
In the example we have ##dx/x = 0.05/2.3 = 0.022## a ##2.2\%## error and ##dy/y = 0.005/4.12 = 0.0012##, a ##0.12\%## error. So the relative error in ##z## is going to be at least ##2.2\% + 0.12\%##, and that last term is even smaller, a tiny fraction of a percent. So we typically ignore it.

We know ##y## to about 0.1%, but we only know ##x## to about ##2\%##, and that causes us to only know ##z## to about ##2\%##. It is the error in your least precise term that dominates in your overall error.
 
Last edited:
  • #4
fxdung said:
I do not know why when multiplying and dividing we have to retain the same number of significant figures as in the number with the least of them.
This isn't a "proof", rather an intuit:

You're trying to measure the area of your rectangular table to make a pattern on it out of beads. You need to go to the bead store with a value for the area to get the right amount of beads.

You use a millimetre tape measure to measure the length of your table and get a value of 1,000 millimetres (1.000 m).
Your tape measure gets busted and now you're left with only a metre stick with all the markings worn off.
You measure the width of your table and get a value of 1m. Without any markings, you have no way of knowing whether the precise width of your table is 0.5m or 1.5m or anything in between.

When you go to the bead store, what can you tell them about the area of your table?

1.000 x 1(rounded up or down) is not 1.000. You simply don't know its area to within a millimetre. All you know is that your table's area is somewhere between 0.5 and 1.5 m2.

In other words the only meaningful thing you can say, without artificially adding digits, is that your table is 1 metre square.
 

1. What are significant figures and why are they important in scientific calculations?

Significant figures are digits in a number that represent the precision of a measurement. They are important because they indicate the level of accuracy in a calculation and help maintain consistency in reporting data.

2. How do I determine the number of significant figures in a given number?

The general rule is that all non-zero digits are significant, as well as any zeros between non-zero digits. Zeros at the beginning of a number are not significant, but zeros at the end of a number after a decimal point are significant.

3. What is the purpose of rounding when dealing with significant figures?

Rounding is used to ensure the correct number of significant figures in a calculation. When multiplying or dividing, the final answer should have the same number of significant figures as the number with the fewest significant figures. When adding or subtracting, the final answer should have the same number of decimal places as the number with the fewest decimal places.

4. Can significant figures be applied to all types of numbers, including whole numbers and fractions?

Yes, significant figures can be applied to all types of numbers. However, for whole numbers, the number of significant figures is determined by the number of digits in the number. For fractions, the number of significant figures is determined by the number of digits in both the numerator and denominator.

5. How do significant figures affect scientific measurements and their uncertainty?

Significant figures play a crucial role in determining the uncertainty of a measurement. The more significant figures there are in a measurement, the more precise the measurement is considered to be. However, the uncertainty of a measurement increases as the number of significant figures decreases.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
800
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
614
  • Introductory Physics Homework Help
Replies
6
Views
635
  • General Math
Replies
5
Views
1K
  • General Math
2
Replies
47
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
797
  • General Math
Replies
4
Views
7K
Replies
1
Views
5K
Back
Top