Thanks for the clarifications, DrMatrix and curiousgeorgeordeadcat. I e-mailed the professor in question, and am awaiting his response. I’ll let you know what he has to say.
Worth the reminder that in the real world, there is no such thing as an “infinite” decimal. After a few hundred thousand decimal places or so, you have quantities way far less than the width of an electron, way too small to be measured. From a real world perspective, if we think of .3333…[100,000 decimal places]…33 , it’s not different from .3333…[100,000 decimal places]…34
since the difference cannot be measured. Thus, in the real world, there will always be a slight difference between .33333 and 1/3.
When we’re dealing with math, however, we have an abstract system, where we can indeed have infinite non-repeating decimals. In the abstract world of math, .3333… is identical to 1/3.
And, as noted, there is the simple but elegant proof already cited that .9999… = 1, namely:
X = 0.99999…
10X = 9.99999… and subtract the equations, obtaining:
9X = 9
X = 1 QED. But a six year old ain’t gonna get that, neither.
OK, so apparently I was completely wrong on this one. My professor simply intended his comment to mean that I hadn’t proved that every real number has a decimal representation.
But part of being wrong is learning things, so I think I’m going to start another thread with a question I’ve got out of all this. Looking forward to the replies…and thanks for your patience with someone who thinks he knows more than he does.
Nevermind, figured out the answer myself.
I think the answer to ?Authority’s point is that 3/5 = .6 only in that everyone agrees to drop the zeros, as they don’t affect calculations in any way.
If you start dropping 3s from the end of 1/3 = .33333… ever, you’re only approximating the actual value of 1/3. It does matter that you stopped writing the additional threes after the decimal. You’re never, ever truly done.
why did everyone ignore DynoSaur’s post… he proved the answer. Or… should i say his proof shows the answer. Or… his proof is the answer. Blah… his proof proves that 0.999… = 1
yeah, that’s what i meant.
[partial hijack]
I don’t have a PhD in math, but I’m still a little troubled…
If we assume that .9999… = 1, is 1/.99999 = 1 or > 1?
According to the prior logic, it should be = 1. But simply saying that because there is no number between them doesn’t make them equal (does it?). I can’t seem to mesh the two ideas.
[/partial hijack]
Between any two different numbers, there are always more numbers to be found. Between 1.0 and 2.0 is 1.5, between 0.0001 and 0.0002 is 0.00015, between 0.0000…<insert one million zeros>…1 and 0.0000…<insert one million zeros>…2, well, you get the point.
So if there are no numbers in between A and B, it follows that A and B are the same number. 1/0.9999… is 1 because 1 and 0.9999… are the same number.
Quoth ?authoriy:
I don’t mean that. I was referring to your earlier post, where you said
You don’t understand why they’re different, because they’re not different. That’s what I said.
So, please tell us, where is the boundary between the “real world” numbers and the “abstract” numbers. Are real numbers “real world” numbers, or only the rationals? How about negative numbers?
I always think of all numbers as being abstract, in the sense that you can show me 2 apples or 3 cars or 1/2 acre of land, but you “really” can’t show me just a 2, 3, or 1/2.
I’m sorry, but I’m still not quite there yet. For the sake of simplicity, let’s assume a world where only whole numbers exist. Using the above logic, 1=2 2=3 3=4 4=5 etc. because there are no numbers between them. We clearly know that 2 < 3, so why can it not be said that .9999… < 1? And if .9999… < 1, how can they be equal to each other?
sethdallob:
Using the standard definition of real numbers, you can prove
that any two distinct real numbers have another real number,
distinct from them, between them. That is:
If x and y are real numbers, and x<y, then there is a real
number z such that x<z<y.
And thus if x and y are real numbers, and there is no real
number z such that either x<z<y or y<z<x, then x=y.
No one has claimed a similar theorem holds for the integers.
The set of integers is not the same as the set of real
numbers, and so the fact that a similar theorem does not
hold for the integers yields no contradiction.