DanishDynamite
Penultimate Amazing
- Joined
- Aug 10, 2001
- Messages
- 10,752
xouper:

If only you could be as reasonable in gun control debates.Agreed.
If only you could be as reasonable in gun control debates.Agreed.
0.0000...1 is meaningless, because the ... means an infinite number of zeroes, and you can't put a 1 on the end of an infinite number of zeroes. So what you actually have is 0.0000... . Which does equal 0.geni said:Are you saying that 0.0000...1=0
( it all used to be so simple sop)
I did wonder if the answer was along thoes lines.xouper said:Neither of those (0.000...1 and 0.999...8) are real numbers according to any of the accepted definitions of real numbers, so it is not meaningful to ask if a real number is equal to one of those things you asked about.
I'm doing the best I can considering that we do not agree on the initial axioms in the gun control issue.DanishDynamite: xouper: If only you could be as reasonable in gun control debates.![]()
LuxFerum said:when exactly the result will be zero??Code:1-0,9 =0,1 1-0,99 =0,01 1-0,999 =0,001 1-0,9999 =0,0001 1-0,99999 =0,00001 1-0,999999 =0,000001
never!!!!
That's not a bad question. Here's a hint.xouper: Neither of those (0.000...1 and 0.999...8) are real numbers according to any of the accepted definitions of real numbers,
geni: Purely out of interest whay are they not real?
Both.xouper said:I have to ask the obvious question here to LuxFerum, Zep and Suggestologist. What is your objective for participating in this thread? Are you just here to post your mistaken notions, or are you interested in learning why you are wrong?
My highlighting.Theorem:
If |r| < 1, then the geometric series
a + ar + ar^2 + ... ar^(n-1) + ...
converges to the sum a/(1-r). If |r| >= 1, the series diverges unless a = 0. If a = 0, the series converges to the sum 0.
Zep said:Both.
Herewith a formal theorem for the sum of a geometric series, from Calculus and Analytical Geometry, George B Thomas Jr, MIT, Addison-Wesley Publishing, 1975 - p624:
My highlighting.
The wording, please note, is "converges" and "diverges," and is specifically NOT "is equal to".
So may we now cease with the personal attacks on our respective mathematical knowledges? I may be old and rusty, but I did graduate too.
Sorry, but you have misinterpreted what the book says. The terms "converging" and "diverging" describe the behavior of the series, not the sum. The theorem says the sum is exactly equal to the series if it is a converging series. I just looked in one of my old calc texts (1995) and it specifically says a coverging geometric series equals the sum a/(1-r).Zep:My highlighting.Theorem:
If |r| < 1, then the geometric series
a + ar + ar^2 + ... ar^(n-1) + ...
converges to the sum a/(1-r). If |r| >= 1, the series diverges unless a = 0. If a = 0, the series converges to the sum 0.
The wording, please note, is "converges" and "diverges," and is specifically NOT "is equal to".
I would respectfully disagree with Cabbage, and submit the following quote from the same reference, ibid, p621 (Chapter 18 - Infinite Series):slimshady2357 said:
Have you read Cabbage's post near the end of the 4th page? The long one that talks about how decimal notation is defined?
0.999... is defined to be EQUAL TO the limit of it's expansion.
0.999... is actually equal to the limit of the series. It's EQUAL to the value the series converges to, that is, 1.
What do you think of Cabbage's post? Where do you disagree with it?
Just asking, no ill will or insulting tone intending
Adam
So this would suggest to me that the limit of the sequence 0.9 + 0.09 +... would indeed converge on 1, and the formula derived previously is used to calculate what this limit value is. However, it also indicates that this is a convergence towards a limit value, not an equality.Limit of a sequence
A sequence {(n, an)} may have a different value an for each different value n. But it may also happen that as n increases, the different an's tend to cluster around some fixed number L. If there is such a number L, with the property that |L - an| is arbitrarily small for all sufficiently large values of the index n, we say that an converges to L as limit, and write
lim an = L
n-> infinity
By the phrase "|L-an| is arbitrarily small for all sufficiently large values of n," we mean that to any positive number E there corresponds an index N such that
|L - an| < E for all n > N
That is, all terms after the Nth lie within the distance E from L. If no such limit exists, then we say that the given sequence diverges.
Please do not confuse the sequence with the series.Zep: So this would suggest to me that the limit of the sequence 0.9 + 0.09 +... would indeed converge on 1, and the formula derived previously is used to calculate what this limit value is. However, it also indicates that this is a convergence towards a limit value, not an equality.
LW said:
For starters, "infinity" is not a number. There are actually an infinite number of different orders of infinity.
Aleph_naught, since the cardinality of the infinite sequence { .9, .99, .999, ... } is the same as the cardinality of the natural numbers.Suggestologist: Which one are you using to take a limit to "infinity"? Is it omega, or aleph_naught?
I haven't confused them, but I will simply quote more of my textbook that says,xouper said:Please do not confuse the sequence with the series.
The infinite sequence R = {.9, .99, .999, .999, ...} converges on a limit.
The infinite series S = .9 + .09 + .009 + .0009 + ... is exactly equal to one.
In other words, the series S is the limit of the sequence R.
lim R = S.
1 = 0.999...