Significant Figures....please be gentle
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Significant Figures....please be gentle
I'm in my senior year of high school. I've taken introduction to chemistry, chemistry, conceptual physics, and currently physics and next year when I start college I will be taking a lot more math.
For the life of me I cannot wrap my feeble little mind around the concept of Significant Figures. Ever since 10th grade I've always gotten at least 510 points off every test because of sig figs and everytime I go to the whiteboard to do a problem or give an answer from my calculator I always look like a dolt because of it.
Right now I'm looking at a slew of kentic and potential energy problems(with power and work sprinkled in) that are due tomarrow and I was wondering if anybody could give and extremely dumbed down explaination on how sig figs work.
I know as somebody who likes math as a disclipine, this is unexcuseable but I implore your help.
For the life of me I cannot wrap my feeble little mind around the concept of Significant Figures. Ever since 10th grade I've always gotten at least 510 points off every test because of sig figs and everytime I go to the whiteboard to do a problem or give an answer from my calculator I always look like a dolt because of it.
Right now I'm looking at a slew of kentic and potential energy problems(with power and work sprinkled in) that are due tomarrow and I was wondering if anybody could give and extremely dumbed down explaination on how sig figs work.
I know as somebody who likes math as a disclipine, this is unexcuseable but I implore your help.
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Re: Significant Figures....please be gentle
Are you having problems understanding why sig figs work the way they do, or are you just having problems applying the rules? (You were taught the rules, right?) It would help if you gave a specific example of a problem where you got the sig figs wrong.
Re: Significant Figures....please be gentle
A good rule of thumb is to look at the data in the question and find the value with the fewest sig figs. Your answer should have exactly that many.
This is assuming you know how to count how many sig figs a given number has.
This is assuming you know how to count how many sig figs a given number has.
Re: Significant Figures....please be gentle
Blatm wrote:This is assuming you know how to count how many sig figs a given number has.
Thats the problem, right there.
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Re: Significant Figures....please be gentle
I'll try to relate how I got the notion of significant digits when it caused me a bit of confusion at first.
First, recognize that numbers, particularly measurements, are never perfectly exact, but only exact to a particular degree.
Let's say I measure the length of my monitor with precision down to the millimeter, and I come up with a length of 0.554m. Then I measure the length of my television with precision down to the centimeter, and I come up with a length of 0.55m. In that case, I can't say that the monitor is bigger than the television: the measurement of the monitor has 3 significant digits (0.554), while the measurement of the television has 2 significant digits (0.55), so if I'm going to compare them, they need to be both at 2 significant digits (yielding 0.55 = 0.55).
The general idea is that you're looking for the least precise measurement in your equationthat least precise measurement is your "weakest link", so whatever answer you get out of that equation cannot be more precise than your least precise measurement. (If you measure everything in centimeters, you can't get an answer in terms of exact millimeters. The digits are only significant out to that measurement in centimeters.)
This also applies when you're dealing with orders of magnitude 10 or above. (Just put the earlier example in terms of 554mm versus 550mm.)
Does this make sense so far? It's not the most precise explanation, but it's how I wrapped my head around the concept. KF
First, recognize that numbers, particularly measurements, are never perfectly exact, but only exact to a particular degree.
Let's say I measure the length of my monitor with precision down to the millimeter, and I come up with a length of 0.554m. Then I measure the length of my television with precision down to the centimeter, and I come up with a length of 0.55m. In that case, I can't say that the monitor is bigger than the television: the measurement of the monitor has 3 significant digits (0.554), while the measurement of the television has 2 significant digits (0.55), so if I'm going to compare them, they need to be both at 2 significant digits (yielding 0.55 = 0.55).
The general idea is that you're looking for the least precise measurement in your equationthat least precise measurement is your "weakest link", so whatever answer you get out of that equation cannot be more precise than your least precise measurement. (If you measure everything in centimeters, you can't get an answer in terms of exact millimeters. The digits are only significant out to that measurement in centimeters.)
This also applies when you're dealing with orders of magnitude 10 or above. (Just put the earlier example in terms of 554mm versus 550mm.)
Does this make sense so far? It's not the most precise explanation, but it's how I wrapped my head around the concept. KF
~Kizyr
Re: Significant Figures....please be gentle
The first significant figure is the first nonzero digit (from the left).
A few examples: (first significant digit in bold)
300
20.76
0.10
0.00237
taking the above examples to 2 significant figures:
300
21
0.10
0.0024
ie rounding to the second significant figure (the one after the first ). Last point is that leading and trailing zeroes that are important are still included, even if not part of the desired (in this case 2) significant figures (same examples as above, just bolding the zeroes I mean):
300
21
0.10
0.0024
A few examples: (first significant digit in bold)
300
20.76
0.10
0.00237
taking the above examples to 2 significant figures:
300
21
0.10
0.0024
ie rounding to the second significant figure (the one after the first ). Last point is that leading and trailing zeroes that are important are still included, even if not part of the desired (in this case 2) significant figures (same examples as above, just bolding the zeroes I mean):
300
21
0.10
0.0024
Re: Significant Figures....please be gentle
I think that a lot of people don't understand the rules for zeros so I'll talk about them.
The significant figures of a number are the figures where the measurement is contained so to speak. For example, if something is 0.0023 kg, then it is obviously a small object that we used a small instrument to measure. The instrument measures in grams and it has 2 sig figs. It couldn't say 0.3019 kg because we are measuring something in the range of 0.0010 to 0.0100 kg. The zeros don't count because then we could change the sig figs based on which units we use, for example 2.3 g needs to have the same number of sig figs. At the same time, trailing zeros do count if they were a measured zero. For example, if we have something that weighs 20.00 kg, then we count the two extra zeros because the number of sig figs shouldn't change if there is only a very small change in the measurement (unless were going from 99.9 to 100.0 or something). If we had 19.93 kg then it's essentially the same thing so we should still have 4 sig figs, and indeed we do have 4 sig figs in both 19.93 kg and 20.00 kg. The convention is that the zeros are not counted as significant if they start in the middle of an integer, for example 4000 g has 1 sig fig. If we wanted to indicate three sig figs then we'd say 4.00 x 10^3 g or 4.00 kg.Other than that, just could the number of digits from the first digit to the last digit.
The significant figures of a number are the figures where the measurement is contained so to speak. For example, if something is 0.0023 kg, then it is obviously a small object that we used a small instrument to measure. The instrument measures in grams and it has 2 sig figs. It couldn't say 0.3019 kg because we are measuring something in the range of 0.0010 to 0.0100 kg. The zeros don't count because then we could change the sig figs based on which units we use, for example 2.3 g needs to have the same number of sig figs. At the same time, trailing zeros do count if they were a measured zero. For example, if we have something that weighs 20.00 kg, then we count the two extra zeros because the number of sig figs shouldn't change if there is only a very small change in the measurement (unless were going from 99.9 to 100.0 or something). If we had 19.93 kg then it's essentially the same thing so we should still have 4 sig figs, and indeed we do have 4 sig figs in both 19.93 kg and 20.00 kg. The convention is that the zeros are not counted as significant if they start in the middle of an integer, for example 4000 g has 1 sig fig. If we wanted to indicate three sig figs then we'd say 4.00 x 10^3 g or 4.00 kg.Other than that, just could the number of digits from the first digit to the last digit.
Re: Significant Figures....please be gentle
One thing that always helped me was converting the number into scientific notation and counting the digits. (And only clip trailing 0s if there was no decimal point in the original number)
 phlip
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Re: Significant Figures....please be gentle
As has been mentioned, the important thing to consider is the precision of a measurement. If your measurements are extremely precise, but you don't give as much precision in the answer, then you've lost information (picture someone looking at an atomic clock and announcing the time to be "about noonish")... but if your answers are more precise than your measurements, then you've invented precision without increasing accuracy (think of the joke about the museum guide saying that the dinosaur bones are 70,000,003 years old, since he was told they were seventy million years old when he started working there three years ago).
If you have an object, and someone tells you it's 147g, that's probably rounded to the nearest gram (if they'd rounded it to the nearest tenth of a gram, and it happened to be exactly 147, they'd still write 147.0g). So it's actual mass could be 147g plus or minus half a gram.
Now, if you have another object, that weighs 24.2g... that's going to be precise give or take 1/20g (if it was more than 1/20g out, it'd round to 24.1g or 24.3g instead)... we have one decimal place of precision.
So suppose we want to find the total mass of the two objects... well, we can just add them together and get 171.2g... but we also have to add the error margins together, and we get plus or minus just over half a gram. Now, we're only really concerned with the order of magnitude of the error, so it's close enough to just "half a gram"... so since that's our error range, we should round our result to the nearest gram, which is 171g, since that's the only part of the result we have confidence in.
So, in general, when adding or subtracting, your result should have the same number of decimal places as your leastprecise input.
However, multiplication is a bit different, and this is where the significant figures come in. Say you have a rectangle, and you measure the two sides... one is 12m (ie precise to the nearest metre  plus or minus half a metre), the other is 2.58m (ie plus or minus 1/200m), and we want to find the area.
Now, again, we can simply multiply them together to get 30.96m², but we also need to know the error. This error is a little more complicated to calculate, but we only need to approximate the error, and it's close enough to just add the percentage errors together. That is, instead of adding together the errors as fractions of the unit (like we did when adding), we add the errors as fractions of our values. Now, 12 plus or minus a half is an error of about 4%, 2.58 plus or minus 1/200 is an error of about 0.2%. So we add those together, get a bit over 4%. So our precision is give or take about 1m², but (as before), we're only interested in the order of magnitude... so we round it off to the nearest square metre  31m²... there's no point to noting the later digits, since (given the precision of our measurements) it could be anywhere between about 30 and 32m².
As with before, we can guess the error well enough based on the precision of the numbers... like how no decimal places means give or take a half, one decimal place is give or take 1/20, etc... we can see that two significant figures is give or take about 1%5%, three significant figures is give or take about 0.10.5%, and so on. It's only rough, because, say, 96.2 is more precise than 12.4, even though they're both 3 significant figures (since the error is smaller as a percentage of 96.2 than it is of 12.4), but, again, we only need an approximation of the error, so this is good enough.
So, where with addition and subtraction we took the number with the least decimal places, with multiplication and division, take the number with the least number of significant figures, and use that as the number of significant figures in your answer.
Decimal places are easier to understand than significant figures are... but Physics has a lot more multiplications and divisions than it does additions and subtractions, so significant figures will get a lot more use than decimal places will.
tl;dr: The important concept isn't significant figures but precision and percentage error... counting significant figures is just a really quick way of approximating that percentage error.
If you have an object, and someone tells you it's 147g, that's probably rounded to the nearest gram (if they'd rounded it to the nearest tenth of a gram, and it happened to be exactly 147, they'd still write 147.0g). So it's actual mass could be 147g plus or minus half a gram.
Now, if you have another object, that weighs 24.2g... that's going to be precise give or take 1/20g (if it was more than 1/20g out, it'd round to 24.1g or 24.3g instead)... we have one decimal place of precision.
So suppose we want to find the total mass of the two objects... well, we can just add them together and get 171.2g... but we also have to add the error margins together, and we get plus or minus just over half a gram. Now, we're only really concerned with the order of magnitude of the error, so it's close enough to just "half a gram"... so since that's our error range, we should round our result to the nearest gram, which is 171g, since that's the only part of the result we have confidence in.
So, in general, when adding or subtracting, your result should have the same number of decimal places as your leastprecise input.
However, multiplication is a bit different, and this is where the significant figures come in. Say you have a rectangle, and you measure the two sides... one is 12m (ie precise to the nearest metre  plus or minus half a metre), the other is 2.58m (ie plus or minus 1/200m), and we want to find the area.
Now, again, we can simply multiply them together to get 30.96m², but we also need to know the error. This error is a little more complicated to calculate, but we only need to approximate the error, and it's close enough to just add the percentage errors together. That is, instead of adding together the errors as fractions of the unit (like we did when adding), we add the errors as fractions of our values. Now, 12 plus or minus a half is an error of about 4%, 2.58 plus or minus 1/200 is an error of about 0.2%. So we add those together, get a bit over 4%. So our precision is give or take about 1m², but (as before), we're only interested in the order of magnitude... so we round it off to the nearest square metre  31m²... there's no point to noting the later digits, since (given the precision of our measurements) it could be anywhere between about 30 and 32m².
As with before, we can guess the error well enough based on the precision of the numbers... like how no decimal places means give or take a half, one decimal place is give or take 1/20, etc... we can see that two significant figures is give or take about 1%5%, three significant figures is give or take about 0.10.5%, and so on. It's only rough, because, say, 96.2 is more precise than 12.4, even though they're both 3 significant figures (since the error is smaller as a percentage of 96.2 than it is of 12.4), but, again, we only need an approximation of the error, so this is good enough.
So, where with addition and subtraction we took the number with the least decimal places, with multiplication and division, take the number with the least number of significant figures, and use that as the number of significant figures in your answer.
Decimal places are easier to understand than significant figures are... but Physics has a lot more multiplications and divisions than it does additions and subtractions, so significant figures will get a lot more use than decimal places will.
tl;dr: The important concept isn't significant figures but precision and percentage error... counting significant figures is just a really quick way of approximating that percentage error.
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 Yakk
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Re: Significant Figures....please be gentle
Note that Sig. Figures are a very rough way of doing this. Imagine you have 11. times 11. times 11. Using significant figures, we end up with 1.3 E3.
However, this could be anywhere from 10.51 x 10.51 x 10.51 = 1160.935651 to 11.49 x 11.49 x 11.49 = 1516.910949 (and slightly beyond).
So your answer  1300 to two significant figures  isn't actually a sufficient bound on your error. The answer is actually 1100 to 1500 to two significant figures.
However, that approach is more complex and harder to deal with. And it is "rare" in a sense that you multiply together multiple values that are all identically biased away from the centre of the possibilities represented by their significant figures.
So you punt.
You keep track of significant figures as you do math. And you follow relatively arbitrary rules to determine how many significant figures there are in the output.
Scientific notation happens to give you a better idea of how many significant figures there are than conventional notation. 1500 has, by convention, two significant figures. 1500. has, by convention, 4 significant figures. You have to use scientific notation to say it has 3 significant figures  1.5 E3 has 2 significant figures, 1.50 E3 has 3 significant figures, 1.500 E3 has 4 significant figures, and 1.5000 E3 has 5 significant figures.
In scientific notation, how many significant figures is clear. The convention in 'nonscientific notation' ends up being:
If there is no decimal place, count the number of digits from the leftmost digit to the rightmost nonzero digit.
If there is a decimal place, count the number of digits from the leftmost nonzero digit to the rightmost digit.
So, with the significant figures highlighted, we have:
15000
15000.
15010
0.000015010
1.000015010
0.01
1.57200E7
1.50E7
1.05E7
1.5E7
However, this could be anywhere from 10.51 x 10.51 x 10.51 = 1160.935651 to 11.49 x 11.49 x 11.49 = 1516.910949 (and slightly beyond).
So your answer  1300 to two significant figures  isn't actually a sufficient bound on your error. The answer is actually 1100 to 1500 to two significant figures.
However, that approach is more complex and harder to deal with. And it is "rare" in a sense that you multiply together multiple values that are all identically biased away from the centre of the possibilities represented by their significant figures.
So you punt.
You keep track of significant figures as you do math. And you follow relatively arbitrary rules to determine how many significant figures there are in the output.
Scientific notation happens to give you a better idea of how many significant figures there are than conventional notation. 1500 has, by convention, two significant figures. 1500. has, by convention, 4 significant figures. You have to use scientific notation to say it has 3 significant figures  1.5 E3 has 2 significant figures, 1.50 E3 has 3 significant figures, 1.500 E3 has 4 significant figures, and 1.5000 E3 has 5 significant figures.
In scientific notation, how many significant figures is clear. The convention in 'nonscientific notation' ends up being:
If there is no decimal place, count the number of digits from the leftmost digit to the rightmost nonzero digit.
If there is a decimal place, count the number of digits from the leftmost nonzero digit to the rightmost digit.
So, with the significant figures highlighted, we have:
15000
15000.
15010
0.000015010
1.000015010
0.01
1.57200E7
1.50E7
1.05E7
1.5E7
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Significant Figures....please be gentle
Yakk wrote:Note that Sig. Figures are a very rough way of doing this. Imagine you have 11. times 11. times 11. Using significant figures, we end up with 1.3 E3.
However, this could be anywhere from 10.51 x 10.51 x 10.51 = 1160.935651 to 11.49 x 11.49 x 11.49 = 1516.910949 (and slightly beyond).
Wait, this part here just confused me. Wouldn't it be 10.50 x 10.50?

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Re: Significant Figures....please be gentle
Kurushimi wrote:Wait, this part here just confused me. Wouldn't it be 10.50 x 10.50?
Could be. Depends on what rounding convention you're using for halfway cases. (Tiestoeven, for example, would have the behavior that Yakk described. Tiesroundup would be like you say). It doesn't really matter, to be honest.
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Re: Significant Figures....please be gentle
stephentyrone wrote:Kurushimi wrote:Wait, this part here just confused me. Wouldn't it be 10.50 x 10.50?
Could be. Depends on what rounding convention you're using for halfway cases. (Tiestoeven, for example, would have the behavior that Yakk described. Tiesroundup would be like you say). It doesn't really matter, to be honest.
Do you need a rounding convention if there are numbers larger than 10.5 by an arbitrarily small amount?
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 Yakk
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Re: Significant Figures....please be gentle
I picked numbers that where less than .5 away in order to avoid talking about the problem of rounding conventions, which is orthogonal (ie, mostly irrelevant to this discussion).
The words (and slightly beyond) where intended to point out that this wasn't the most extreme values in question that could be used.
It appears I have failed. Is there an alternative to seppuku?
The words (and slightly beyond) where intended to point out that this wasn't the most extreme values in question that could be used.
It appears I have failed. Is there an alternative to seppuku?
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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