I What is the proof of the rules of significant figures?

AI Thread Summary
The discussion emphasizes that the rules of significant figures serve as a practical guide for understanding measurement precision rather than strict mathematical laws. When multiplying or dividing, the result should retain the same number of significant figures as the least precise measurement, reflecting the uncertainty inherent in the measurements. The example illustrates that a measured value, like 2.3, indicates confidence only in the digits presented, with potential variations that cannot be precisely determined. The overall error in a calculated value is dominated by the least precise term, underscoring the importance of significant figures in conveying meaningful precision. Ultimately, significant figures help communicate the reliability of measurements in practical applications.
fxdung
Messages
387
Reaction score
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
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
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:
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
25
Views
283
Replies
11
Views
1K
Replies
4
Views
8K
Replies
1
Views
1K
Replies
12
Views
2K
Replies
4
Views
3K
Replies
41
Views
13K
Back
Top