Fractional Parts

[Cptn]

If I take any whole number, divide it into an arbitrary amount of fractional parts, round traditionally those fractional parts, add them back together, I will end up with my original number.

Is this correct?
If so, is there a name for this rule?

*Edit

Given that each fractional part is unique.

For example:

51267 x 0.5 = 25633.5 = 25634
51267 x 0.2 = 10253.4 = 10253
51267 x 0.3 = 15380.1 = 15380
.................................-------
..................................51267

 

[JeffM]

If I take any whole number, divide it into an arbitrary amount of fractional parts, round traditionally those fractional parts, add them back together, I will end up with my original number.

Is this correct? NO
If so, is there a name for this rule?

\(\displaystyle 1 = 0.2 + 0.3 + 0.4 + 0.1,\ but\ 0.2 \approx 0,\ 0.4 \approx 0,\ 0.3 \approx 0,\ and\ 0.1 \approx 0\ and\ 0 + 0 + 0 + 0 \ne 1.\)

This is an example of round-off error. Rounding always has the potential to create an error. In the study of numerical methods, methods for minimizing potential round-off (or truncation) error are given, but you cannot eliminate the potential for some amount of error when you do rounding. If I remember correctly, the primary method for minimizing potential round-off error in a sum was to delay rounding until the sum was complete. Another method important when using floating point arithmetic was to sort the summands in ascending order before adding them.

In less formal settings, if the rounded numbers do not add up to the rounded total, adjust the rounding of one or two of the addends so the rounded sum comes out right.

 

[Cptn]

Thanks for the help Jeff.

Sorry if I didn't word my problem clearly. Math hurts my head.

What if I'm multiplying like in the example above? Is there still possibility of error?

Thanks again.

 

[HallsofIvy]

Still, the answer is no, you will not necessarily get back the original value. A variation of your example:

51267 x 0.1= 5126.7=> 5127
51267 x 0.1= 5126.7=> 5127
51267 x 0.8= 41013.6=> 41014 and those add to 51268.

The difference is that we had to round up on each of those. We could also find examples where we round down and get a sum to 51266. If you round correctly, you should get the original number or the original number plus or minus 1.

 

[Cptn]

I understand. Thank you HallsofIvy.

One final take on the same problem, if I may. What if the multipliers were each unique? In other words, there would never be two of [51267 x 0.1].

 

[JeffM]

I understand. Thank you HallsofIvy.

One final take on the same problem, if I may. What if the multipliers were each unique? In other words, there would never be two of [51267 x 0.1].

The point is that when you round, you introduce an error into the computation, either to simplify the computation or to make the numbers more comprehensible. Now presumably that error is small enough that it is not important, but the error is there and so has a POTENTIAL effect on the computation. How big that potential is depends on the type and number of computations being done, but the potential is never zero. There is never a guarantee that your known error will not have an effect on the answer. Make sense?

\(\displaystyle 41,202 * 1.2 = 49,442.4 \approx 49,442.\)

\(\displaystyle 37,741 * 1.4 = 52,837.4 \approx 52,837.\)

\(\displaystyle 28,554 * 1.8 = 51,397.2 \approx 51,397.\)

\(\displaystyle 49442.4 + 52837.4 + 51,397.2 = 153,677 \ne 153,676 = 49,442 + 52,837 + 51,397.\)

As a practical matter, I'd suggest not rounding until you have finished a computation. With modern calculators and computers, the need to simplify computations is far less pressing than it was forty or fifty years ago so do your computations with exact numbers. If you need to round for purposes of comprehensibility, wait until you have the final result.