No. There are no two numbers in the real number system that are infinitely close. And even if there were, .9… would still be equal to 1.
Completeness gives that any decimal expansion converges, but if you already have a number then you know that it’s decimal expansion converges to it… 0.99… = 1 in the rationals as well, it’s just not as useful to have a decimal expansion on the rationals.
In fact, it should be true in any archimedean field (come to think of it, it’s actually equivalent to the field being archimedean. That’s mildly neat.)
But it doesn’t matter since you can’t have .333… of a penny. The best you can do is either pay the guy $99.99 (33.33 each) or else one of you has to cough up an extra .01.
Or you could just not watch him eat his poop, which would be less fun.
The result follows from the fact that any monotonically increasing sequence converges to its least upper bound, which is a form of completeness–in an incomplete ordered metric space (such as Q), it can fail.
You’re right, it doesn’t strictly depend on completeness, but the only people who’d be in a position to see that aren’t going to argue the result anyway.
It’s all been dooooonnnneeee before!
Hmm. I wouldn’t say that a monotonically increasing sequence converges to it’s least upper bound depends on completeness at all. It’s a true statement in the rationals as well, as long as the sequence does indeed have a rational least upper bound. (And conversely, convergence of the sequence in the rationals implies that the sequence has a least upper bound in the rationals). The sequence converges to a rational, so no form of completeness is needed.
All that being said, it really doesn’t matter. The point remains that the original question is a basic misunderstanding of limits.
Right…what I should’ve said is that completeness implies that every bounded, monotonically increasing sequence converges. That’s what I meant.
Nobody in this thread or the other ever addressed it quite this way, so I’ll throw this out to chew on, too, for one more piece of supporting evidence.
You can’t calculate 0.33333… x 3 without calculus. Why not?
When you use regular multiplication, you have to start by multiplying the factor of 3 by the rightmost digit of the multiplicand. However, 0.33333… doesn’t have a rightmost digit because the series is infinite. So how are these guys performing that operation, anyway?
It seems to me that they’re just multiplying 3 by the digits of 0.33333… that they can see and leaving the … intact. Bzzzt, even grade-school multiplication doesn’t work like that.
Anyway, these math threads are always interesting to observe. Carry on.
I thought of a way to debate the “.999… is the closest number to 1 without being 1” argument on my way home last night.
I know, the statement itself violates the rules of real numbers, so the entire debate is pointless, but hear me out.
If .9… is infinitely close to 1 without being 1, then
.3… is infinitely close to 1/3 without being 1/3
The same rules that apply to .9… must apply to .3…
So, .3… is the closest number to 1/3 without being 1/3. Now multiply it by 3, and you get .9…, which is just as close to 1 as .3… is to 1/3. But the “space between” .3… and 1/3 is thrice the size of the “space between” .9…, due to the fact that we multiplied by 3, and therefore shrunk the difference accordingly.
Now back to one third: .3… cannot be the “closest number to 1/3 without being 1/3”, because you could subtract the difference between 1 and .9… from 1/3, which would be a number between .3… and 1/3.
All of this demonstrates faulty logical conclusions based on the premise that .9… approaches, but does not reach, 1. Therefore, the premise must be flawed, and so the only logical conclusion is that .9… is exactly equal to 1.
I don’t think I explained it very well. Obviously, IANAM, but I just wanted to participate.
There should be a sniglet for the one guy in every .9…=1 thread who never grasps the concept.
In maths indirect proof/reductio ad absurdum/proof by contradiction is quite often used, for example such a proof was used to prove that there is no largest prime.
Indeed on one of the many threads on this subject I gave my own such offering: if 0.9… is infintely close to 1 then the postive square root of this number must be even closer.