I don't know if you're right, but I love the way you put it. I'm rooting for you!

If this is true, why does this logic fail for addition calculations, as demonstrated by explicit example?

zero does not always mean nothing and temperature is a good example.

zero is also used to show relationships between variables and between variables with constants.

I can't find an explanation exactly for multiplying more than one factor, but so far all examples I'm finding they don't round off multiple factors until the end, and I think that's probably right.

If you measured the L x W of a box very precisely 1.49m and 1.49m giving an area of one side = 2.22m^2. Then measured the height less precisely to only 1 m, so 2.22 x 1 = 2 m^3. I don't see why we would round the first two independent and more precise measurments each to 1, which would give us a final volume of 1 m^3. I think 2 m^3 is more accurate here.

You constructed the bolded, they may not be quite that way.


All in all: all rounding is a form of crime against nature. Things are how they are, how we happen to design our numbers is irrelevant. Therefore it can never be right to make a number less correct than we know them to be. If you are adding or multiplying that number to something less specified, that's a different thing, then you will have to use the least significant number for getting the number of decimals you can state are right.

Lol, rounding is a crime against nature... unless it's a round ass! Pow!

I took the maximum range of values that could possibly have resulted in those values. Can you show me how you can get a larger maximum or a smaller minimum?

Nah. It's artificial accuracy that's the crime against nature.

This much is true. There are constraints based on the number system we're using. A more formal analysis would be using epsilons and such.

This is precisely the problem I stated earlier:

Your logic may be valid, but the argument is not sound. The numbers you started with are wrong, so your calculator is wrong. So you don't really know much of anything.

Why not be explicit? I've given you an example where holding extra decimals makes things worse by creating an artificial sense of accuracy that's being used to round. It doesn't work that way.

Still, nothing says your example is correct. The range may not be symmetrically around a number. If you have 14.5, the range may be something like 13.5-15.7, or something else, 14.1-14.6, and so on. Therefore your example is not relevant for proving anything.

Just by rounding you will never get the numbers more right, in the sense of more exact. You will always get them more inexact. Of course every added decimal will have only a tenth of the impact of the previous, but the impact is still there.

The aim with a final rounding is only not to state more decimals than you know. And that depends on the factor of which you know the least significant figures.

Every rounding you do without being forced to will diminish the value of the sum/product you are getting, on average. The only rounding you are forced to do is the final rounding.

Round the sum of this addition to one signficant figure:

1.3+1.3=?

Those or you who say 3 are in my camp. Those who say 2 are not.



Lets say you have a 1, the range of which is for example about 0.5-1.5. Take this 1, and add 1.3+1.3, the latter measures being more exact, range about 1.25-1.35. Then you have 1+1.3+1.3. What's the sum? Isn't the correct answer 4? If you had rounded 1.3+1.3 to 1+1, you'd get the sum 3. You can all see that must be wrong?

You do not round intermediate terms during a series of calculations, period. You round the final result to match the precision of the source data if we know it to not be exact.

Rounding at each step introduces errors that compound.

Yes and i personally dont mind how stupid it may look to do eg 2*3.141593*6378km to get the circumference of earth at equator lol instead of 2*3.142*6378 or the period of a 300km altitude satellite 2*3.141593*((6378000+300000)^3/(6.6738*10^-11*5.97*10^24))^(1/2)=5432 sec as long as you know how to do it right if necessary (which might then also require a better theory for the gravitational effects of a not perfect sphere and other minor effects like friction from the tiny atmosphere still left etc)

The proper way is to have uncertainties for everything and do it that way and estimate how their uncertainties have propagated to the final value calculated (eg the period of a satellite at altitude h that depends on M,G,r,h) so that you know how much to keep there using the maximum accuracy for each variable/parameter/constant.

Are you quite certain you understand how significant figures work? The question of "rounding" is different from the question of "significant figures."

I see why what you're doing is wrong. I don't see how it impacts anything I've stated.

You're confusing three very distinct concepts here. There's a measurement error (+- some tolerance), there is the introduction of error based on rounding, and there is the concept of significant figures.

Pretending like your measurements are exact numbers only gives the illusion of being exact. The numbers beyond a certain point should be assumed to be random. So pretending like the digits your calculator spits out are somehow "more exact" than something else doesn't really make a lot of sense.

The aim of rounding is about estimation. The aim of significant figures is to avoid overstating the level of accuracy. Two very different ideas.

http://www.purplemath.com/modules/rounding2.htm
Looks I've got it right.

From Wiki: In this case you only know four significant figures. But if you add a multitude of these numbers, you probably will get a better result if using the raw numbers, and then rounding the sum to the nearest four digits.


Aaron. Let's forget about everything else. Do you agree with this? :
For a while it looked you don't.

Nah. You're just under the illusion of accuracy still. But what else might one expect if you're pointing to a site on remedial mathematics?

What is the number of significant digits of the approximation 3.14 to the value pi? (Hint: It's not 3.)

https://ece.uwaterloo.ca/~dwharder/N...ificantDigits/

A more interesting example involves division involving the subtraction of two numbers that are close to each other.

4.532/(2.157 - 2.124) = 137.3

But with a small variation in the denominator (in the THIRD decimal of just one of the numbers):

4.532/(2.157 - 2.129) = 161.9

It's way more complicated than what they teach you in elementary school.

FYP

No. You are in love with the idea that I am.

3.14 approximates π to almost or approximately three significant digits https://ece.uwaterloo.ca/~dwharder/N...ificantDigits/

Looks so. But how do you solve it? Lessening accuracy by rounding will not fix the problem.

The resolution is to understand that the rules of significant figures does not follow simplistic rules we teach children. There are more delicate heuristics that can be considered. For example, the last digit of the calculation should be thought of as being +- 1 for most short calculations (this would resolve the 14-15 problem).

The more accurate way to do it is to use epsilons.

I'm reminded of a numerical analysis course in which we did some simple stuff with floating point errors. We had to get over the idea that the calculator (or computer) is automatically trustworthy for as many digits as it gives you.

There's different degrees of "trust". I think all numbers they give usually have some relevance, even if really minuscule, it's just that the strings of numbers usually must be overshadowed by other considerations. One is not giving a false impression of exactness in calculations, getting significant figures and roundings right in the final answers.

But the strings of numbers are not false as such. They can "only" be interpreted wrongly.

1,000,000+0.0003 "=" 1,000,000.0003 ~= 1,000,000. But even the 0.0003 can, under some extreme circumstances, be that extra little that tips the number upwards. It will never make it smaller, almost never has an impact.

To put what plaaynde is saying to work let's take:

1,000,000 + .0003(10^10) = 4,000,000

What's the difference if we instead of multiplying the .0003 times 10^10, we just added it 10^10 times:

1,000,000 + .0003 + .0003 + .0003 .... = 4,000,000?

Those are one and the same.

7x8 = 7+7+7+7+7+7+7+7

Yeah, I didn't get the point either, as those are the exact same calculation just using different notation. We could write it many other ways too.

That's why I'm asking. It goes against the rules as I learned them and as shown in the links below. So I'm back to square one here.

I'm thinking for addition of multiple terms the rule should be to first add all terms with the most precision together, then add them to the next terms with less precision and so on, probably rounding in each spot.