Hey, when we solve for significant figures, why do we completely get rid of the remaining decimals even though hey have more accurate information?
Ex. 1.23*4.84=5.9532 but we would make it 5.95 based on Sig figs, even though those last two decimals are closer to the answer. Why is this? I know it’s less accurate, though it seems like we’re losing valuable accuracy (even if it’s not perfect, it should be closer)
In: Mathematics
I would say it depends on what you are doing with the number. If you are in the middle of calculations, use as many significant figures as you can. Ideally use algebra or a computer so you don’t have to round at all. Rounding in the middle of calculations can introduce extra error.
If you are publishing a result, use as many significant figures as your least accurate measurement. If you were measuring something with a ruler, it would be misleading to claim that it was 4.837905 mm long. You have no idea what those numbers are after the first 1 or 2 places.
Because more accurate information is precisely (no pun intended) what they DON’T have.
How accurate, really, were your initial numbers? And can you actually trust those last digits of your result – or are they basically just random digits, in roughly the right numerical area, giving you a false feeling of accuracy?
Errors compound. Put simply – **small errors get bigger when you mix them together**. If your numbers were, say, actually 1.23 ± 0.004 and 4.84 ± 0.004 (in other words, merely correct to roughly 2 decimal places) – the actual product could be anywhere between 5.928936 and 5.977496. It makes no sense to think that you “know” the result to 4 decimal places; actually, you don’t even know it to 1 place. About the best you can say is that it’s a few hundredths under 6.
(Yes, if we’re just talking theoretical numbers in a theoretical context, the result is precise. But for real world stuff, the end result is no better than the accuracy of your measurements – and you’d better allow for the margins of error, or you could end up with a result that is, basically, little more than garbage.)
Significant figure is a way to estimate measurement accuracy. The question is not what the math says, but how sure you are of your measurement. 1.1 implies you have the measurement correct within 0.05 units. A measurement of 1 millimeter implies correctness within 0.5 millimeters. Doing math on the numbers doesn’t change the error. So we keep the number of significant digits the same to estimate error.
Lets say I have a scale. The smallest thing it can measure is 0.01 grams. So, if I put one gizmo on the scale, and the scale says 1.23 grams, I don’t really know anything about the digits beyond the 3. Maybe the true weight of the gizmo is 1.234 grams, but the scale isn’t accurate enough so it just says 1.23 grams.
So, when I multiply the 1.23 by 4.84, I get 5.9532 grams. If I use that number, it looks like I know exactly how much it weighs down to the 0.0001 gram. But I don’t! Maybe the true weight was 1.234, so when I multiplied it, the true value should have been 5.97256, not 5.9532. The first digit 5 is correct, the one-tenths place 9 is correct, the one-hundredths place 5 is not quite correct, and the remaining digits, 32, are total nonsense. If I leave them in, it looks like I know what the weight is down to the ten-thousandth of a gram, when I don’t even know the hundredth of a gram exactly. So, we drop them, and keep only the numbers we are at least somewhat confident about.
> Hey, when we solve for significant figures, why do we completely get rid of the remaining decimals even though hey have more accurate information?
They don’t have more accurate information. They have more precise information but precision isn’t useful if you’re not actually on target.
> Ex. 1.23*4.84=5.9532 but we would make it 5.95 based on Sig figs, even though those last two decimals are closer to the answer.
But they are not closer to the answer.
1.23 means something between 1.225 and 1.235
4.84 means something between 4.835 and 4.845
So that means that 1.23 * 4.84 could be anything between ~5.92 (1.225 * 4.835) and ~5.98 (1.235 and 4.845). The answer of 5.95 is a lot closer to reflecting this reality than 5.9532 is.
Assuming you actually maintained your significant digits, it comes down to the required level of accuracy. If I am building a jet engine where precision tolerances are needed I should probably be careful to keep good control of my significant digits. If I am framing a deck on my house I probably don’t need to keep track of measurements in the ten-thousandths.
What I think a lot of the answers you’re getting have touched on, but maybe not stated explicitly, is the idea that the extra digits would convey a *false* sense of precision. Every measurement we make also has errors involved.
Let’s use your example, in a classic setting I would give to students. You have a rectangle and want to get its area by measuring the width of two sides using a ruler. In truth, your ruler has an accuracy of +/- 0.005, you can roughly tell whether the edge is closer to 1.23 or 1.24 but no more than that.
Based on this, you know that the area must be between (1.225 * 4.835 = 5.922875) and (1.235 * 4.845 = 5.983575) or (5.95 +/- 0.03) which shows how errors compound. Attaching more digits to the end accomplishes nothing, you can’t make the result more accurate.
However, attaching those extra digits can create the impression that you have a more precise result than you really do. Again, using your numbers, if we multiplied (1.2300 * 4.8400 = 5.9532) then it looks like those zeroes aren’t doing anything, but they are. Those zeroes tell us that our ruler is accurate to within +/- 0.00005 and that the result is actually between (5.9528) and (5.9535) which is far more precise than when we only knew two digits to the right of the decimal.
If I stepped on a scale that only weighed me to the nearest hundred pounds, it might say I’m 100 pounds. If I wanted to know how much two of me would weigh, I could multiply that by 2 and get 200.00000… lbs. I might actually weigh 145 lbs so the real answer would be 290 lbs. Because it’s a real world measurement, there’s inherent uncertainty in the starting measurement that carries over to the answer.
Two things:
1. 1.23 x 4.84=5.9532 should only be taken to five significant figures if the first two numbers are also accurate to more than five significant figures. So it’s 1.23000 x 4.84000. If the first two are only stated to three significant figures you can’t, or shouldn’t, assume the values are accurate to an additional significant figures. If the first values were actually 1.2349 and 4.8449 but rounded to the values you state then the answer would be 5.9830. Even taking the answer to three significant figures is a bit generous…
2. Even when you do know the answer to many significant figures past a point the increased accuracy starts to lose value. For example NASA uses a value of Pi to 15 decimal places, even though it is known to trillions of decimal places. This is because 15 decimal places is enough to plot a billions of miles long space flight to within a couple of inches, and by then the error bars for other factors are much larger than the course projection errors.
The question is what do 1.23 and 4.84 mean. Is an exact value so 1.23 = 1.230000000000000…. or is a measurement where the instrument only show 2 decimals?
A 2 decimal measurement would have an accuracy of +-0.005 because the real value might be any number between 1.235 and 1.225
4.84 could between 4.845 and 4.835
If you measure an area the maximum area is 1.235*4.845=5.983575 and the minimum is 1.225*4.835=5.922875
So you can see that the area is between 5.983575 and 5.922875 and 5.9532 is the average of the two numbers. Because you only measure with 2 decimals accuracy you do not know the exact area just the range.
So the simple way to handed this is to count significant digits and give the answer as 5.95 because if you add more decimals the implication is that your measurement is accurate to the last decimal.
So 5.95 is an approximation
You can do better as 5.983575/5.95 =1.0056 and 5.983575/5.95=1.0056
So you can say that the value is 5.95+-0.0056
If 1.23= 1.2300000000000… and 4.84=4.840000000… then the exact result is 5.9532
For any measurement of the real world, there is always a level of accuracy. So 1.23 and 1.2300 is a way to describe the accuracy in the measurement.
If you do a long calculation with multiple steps you should keep all the decimals but round the output to the appropriate number of decimals. It can be hard to know what the appropriate error. You might later learn how the propagation of uncertainty works mathematically.
The important part is that the final result shows not imply that the input measurement had higher accuracy than they had in reality. The number of decimals is an implicit way to describe accuracy.
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