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
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.
For some other reason (usually measurement accuracy) we know our data isn’t that accurate, so we remove significant figures as a convenient way to communicate the lack of accuracy
Firstly, I would like to point out that it isn’t more accurate, it’s more precise, to use more significant figures. And the reason to not use so many depends on the application. In the real world, there is rarely a need to go to lots of significant figures, and also a lot of things can’t measure to such precision anyway.
5.95 is a shorthand for 5.9532 +- 0.00XX
It’s just to save space, and to prevent someone from thinking the 32 has any real meaning.
Yes it’s a better guess than 5.9500 but we already assume 5.95 means 5.95XX and not followed by zeros.
Well sig figs has everything to do with accuracy. Scientist want to convey the accuracy of calculated numbers. Thus, your answer can only be as accurate as your least accurate measurement. We cut those extra numbers off not to be more accurate but to remain equally accurate as before.
The Sun is a long way away, the average distance from Earth is more or less defined by one astronomical unit, which is ninety-two million nine hundred and fifty-five thousand eight hundred and seven miles.
That’s a long way, imagine if we could drive a car at 100mph. How long would it take us to travel ninety-two million nine hundred and fifty-five thousand eight hundred and seven miles? The distance between the Earth and the Sun varies and sometimes it’s more than ninety-two million nine hundred and fifty-five thousand eight hundred and seven miles.
Notice how the above obscures the key information. To focus on what is *significant*, we use significant figures. If I talk about the distance to the Sun being 93 million miles, sure I’ve thrown away some accuracy, but it allows us to focus on the concept, making it easier for us to absorb and make comparisons.
because the values you input in reallife-experiments/observations arent the fixed values you’re thinking off.
when an experiment got 1.23 as its result it isn’t EXACTLY 1.23 (as in 1.23000…..000), but it is possibly 1.2298 +/-0.0015, so so rounding it up to 1.23 and saying it is that value is the right thing to do because even if you go 2-3 standard deviations away, you still land at something that rounds back to 1.23
but when you put in those 1.2298 +/-0.0015 into your equation properly you will get to something like 5.9532 +/- 0.0060 (error margins just randomly chosen/estimated, didnt bother calculation, those are just for visualization purposes here)
so when you then present your result as 5.9532 when it could easily also be 5.9549 or 5.9511 would be miss-representing the accuracy of your experiment.
and even outside of experiments, when we deal with abstracted math models, it generally makes little sense to bother with additional digits if you only care about the one before the decimal point for example.
If it’s pure math, you can keep those digits.
If it’s anything applied, like a measurement, then you can’t keep those digits because you don’t actually know that precision. Your measurement tool is accurate to X.xx; how can it possibly verify that the math is correct to X.xxxx? This is to prevent mathematics from creating an impossible situation, where a mathematically derived precision is unreachable by physical means.
In some fields of engineering, you can’t go ahead if you measure something is 3.79 but the specification requires 3.790. You need to meet/match specification, and specifications can only be as accurate as the measurement tools used to derive them.
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.
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