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October 3rd, 2016, 01:45 AM   #21
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Quote:
 Originally Posted by zylo The difference between .5000... and .4999... approaches 0 as n approaches infinity, but it never equals 0. So .49999999....... never equals .50000000..... But, you say, if .499999... doesn't equal .500000... in the limit as n approaches infinity, there must be a number between them. NO. because in the limit as n approaches infinity they are real numbers and the real numbers are complete: there is no real number between them.

$\displaystyle 0.5000... = 0.5$
$\displaystyle 0.4999... = 0.4\dot{9} = 0.5$

So the two values you quoted are equal to each other.

I think you're trying to say something like this:

Define two functions:

$\displaystyle f(n) = \frac{2}{5} + \sum_{i=1}^{n} \frac{9}{10^i}$
$\displaystyle g(n) = 0.5$

As $\displaystyle n \rightarrow \infty$, $\displaystyle g(n) - f(n) \rightarrow 0$.

That's fine. However, if you set n to be infinity, what happens? $\displaystyle g(\infty) - f(\infty) = 0$

But then you tell me that n cannot be equal to infinity. Why not?

October 3rd, 2016, 08:46 AM   #22
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Quote:
 Originally Posted by Benit13 $\displaystyle 0.5000... = 0.5$ $\displaystyle 0.4999... = 0.4\dot{9} = 0.5$ So the two values you quoted are equal to each other. But then you tell me that n cannot be equal to infinity. Why not?
1) You are assuming what you are trying to prove.

2) infinity is not a number.

You conveniently ignored my following post:
$\displaystyle \lim_{n\rightarrow \infty}$1/n =0 but 1/n never equals zero.

Suppose you listed the n-place decimals in [0,1) in order:

.0000000
.0000001
.0000002
..............

As n approaches $\displaystyle \infty$, they approach a countably infinite list of the real numbers.
If you let n= $\displaystyle \infty$, they would all be zero by your thinking, because
.00000.....1 = 0 when n = $\displaystyle \infty$
.00000.....2 = 0 when n = $\displaystyle \infty$
.............

$\displaystyle \lim_{n\rightarrow \infty}$.000....1 = 0, but .000....1 never equals 0.

Last edited by skipjack; October 4th, 2016 at 03:04 PM.

October 3rd, 2016, 09:12 AM   #23
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Quote:
 Originally Posted by zylo As n approaches $\displaystyle \infty$ they approach a countably infinite list of the real numbers.
Continually adding further rationals to a list of rationals doesn't imply that the set of all the irrationals is countable.

October 3rd, 2016, 09:38 AM   #24
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Quote:
 Originally Posted by zylo 2) infinity is not a number.
This is correct. $\frac25+\sum \limits_{n=2}^\infty \frac9{10^n}=0.5$ does not depend on evaluating the sum at $n=\infty$, but it does represent the value of the infinite decimal.

Quote:
 Originally Posted by zylo As n approaches $\displaystyle \infty$, they approach a countably infinite list of the real numbers.
No, for all finite $n$, they are a finite list of rationals.
The number of digits cannot "approach infinity". It is either finite or it is not. The length of the decimal is not convergent so it doesn't converge to $\infty$.
Indeed the phrase "approaches infinity" in the context of limits has a precisely defined meaning that does not use the concept of "infinity". Your attempt to use the phrase as in some way related to decimals of infinite length is an inaccurate blurring of precise definitions.

Last edited by skipjack; October 4th, 2016 at 03:05 PM.

October 4th, 2016, 03:29 AM   #25
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Quote:
 Originally Posted by zylo 1) You are assuming what you are trying to prove.
I'm not trying to prove anything, I'm just quoting the result. What assumptions am I making that prevent me from putting infinity as a limit of a summand? If you mean this...

Quote:
 2) infinity is not a number.
then that's fine; by specifying the upper limit on the summand as infinity, I am specifying that the sum includes all terms in the series. Although one cannot actually go and compute the summand by manually adding all of those terms, it doesn't mean we cannot know the result of such an exercise. Maybe this is what you don't like? It's like Zeno's paradox again.

Quote:
 You conveniently ignored my following post: $\displaystyle \lim_{n\rightarrow \infty}$1/n =0 but 1/n never equals zero.
I didn't ignore your post... it's just that you are asserting that 1/n never equals zero, not demonstrating it. You need to demonstrate to me why n cannot be infinity and saying that infinity is not a number is not good enough evidence.

Quote:
 Suppose you listed the n-place decimals in [0,1) in order: .0000000 .0000001 .0000002 .............. As n approaches $\displaystyle \infty$, they approach a countably infinite list of the real numbers.
No they don't... they approach a countably infinite list of rational numbers.

Quote:
 If you let n= $\displaystyle \infty$, they would all be zero by your thinking, because .00000.....1 = 0 when n = $\displaystyle \infty$ .00000.....2 = 0 when n = $\displaystyle \infty$ .............
What on Earth is .00000.....1? That doesn't make sense. Whatever it is, it's certainly not not part of the standard decimal number system. The ellipsis (...) denotes an infinite string of digits and an infinite string of digits does not terminate; the whole point of the concept of infinity is that infinite things don't terminate, so by putting a 1 at the end of the number, you are contradicting yourself. It makes no sense to query what their values might be because they are not even valid representations. It's like asking what the wingspan is of an elephant.

As CRGreathouse argued in previous threads, if you want to mess around with your own number system, that's fine, but if you do that you are no longer talking about the reals or the decimal number system. You can even create your own system that includes weird things like 0.000...1 or $\displaystyle 0.\dot{0}5$ or whatever, but those numbers are certainly not decimal numbers.

Last edited by skipjack; October 4th, 2016 at 03:08 PM.

October 4th, 2016, 04:56 AM   #26
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Quote:
 Originally Posted by Benit13 Whatever it is, it's certainly not not part of the standard decimal number system.
Oops! This should read "Whatever it is, it's certainly not part of the standard decimal number system."

October 4th, 2016, 07:40 AM   #27
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Quote:
 Originally Posted by zylo As n approaches $\displaystyle \infty$ they approach a countably infinite list of the real numbers.
As n increases, more rationals are listed, and there is no upper bound to how many are listed, but each of them is an n-decimal approximation of infinitely many reals, so it doesn't follow that the reals are countable.

Quote:
 Originally Posted by zylo If you let n= $\displaystyle \infty$, they would all be zero by your thinking, because .00000.....1 = 0 when n=$\displaystyle \infty$ .00000.....2 = 0 when n=$\displaystyle \infty$
They wouldn't all be zero. The last in the list for any value of n is the decimal containing n nines, which approaches 1 as n tends to $\infty$.

 October 4th, 2016, 08:06 AM #28 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 861 Thanks: 68 The n-place decimals in [0,1) are unique, rational and countably finite. As n $\displaystyle \rightarrow \infty$ they remain unique, and become real and countably infinite.
 October 4th, 2016, 08:22 AM #29 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 861 Thanks: 68 .00....01 and .00....02 are not the same no matter how many decimal places you take. .49999.... and .50000.. are not the same no matter how many decimal places you take.
October 4th, 2016, 08:28 AM   #30
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Quote:
 Originally Posted by zylo .00....01 and .00....02 are not the same no matter how many decimal places you take. .49999.... and .50000.. are not the same no matter how many decimal places you take.
Quick question... did you read my post?

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