The Packbats' Weblog

(Apologies for the repeated comments. Free LJ accounts apparently aren't cool enough to edit their comments, and LJ's comment preview feature bites hard, so I have to try a couple times to find the appropriate tags. Sorry.)



Exactly. If, at any point in working with a non-terminating decimal, you choose to operate on the number as though it did terminate, then you're working on an approximation.



Sorry, let me back up for a second.



I think we can agree that 1/3 != .333 (i.e., without the ... notation), right? We can represent 1/3 as .333... repeating infinitely. But, then what happens if you perform an operation on such a number?



You might say that 3 * .333... must be equal to .999... infinitely, but you've just induced a rounding error; to get that result, you had to behave as though .333... terminated at some point.



Or, let me try yet another tack:

[1] 1/3     = 10/30
[2]         = 9/30 + 1/30
[3]         = 9/3 * 1/10 + 1/3 * 1/10
[4]         = 3 * 1/10 + 1/3 * 1/10
[5]         = 3 * 1/10 + 3 * 1/10 * 1/10 + 1/3 * 1/10 * 1/10
[6]         = 3 * 1/10 + 3 * 1/102 + 3 * 1/103 + ...
[7]         = .3 + .03 + .003 + 1/(3 * 104)
[8]         = .333 + 1/(3 * 104)
[9] 3 * 1/3 = (3 * .333) + (3 * 1/(3 * 104))
[10]        = 1

The math I'm presenting here is as algebraically correct as what anybody else has presented, yet it arrives at a different conclusion.



I think you have to be really careful if you're dealing with infinitely repeating decimals and you're concerned about accuracy.