The Mystery of .999... and .333...: A Math Puzzle

In summary, the conversation discusses the concept of .999... being equal to 1. This is proven through various methods, including the fallacy of truncation and the use of geometric series. It is also noted that the size of the set of real numbers is undecidable, with the possibility of being greater than aleph-1.
  • #36
rajeshmarndi said:
Is .999... tends to 1(as it is always less than 1) or equal to 1.
You mean the sequence 0.9, 0.99, 0.999, ... tends to 1.

0.999... is, by definition, the limit of that sequence so is equal to 1.
 
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  • #37
HallsofIvy said:
You mean the sequence 0.9, 0.99, 0.999, ... tends to 1.

0.999... is, by definition, the limit of that sequence so is equal to 1.

Or, it is a monotonic Cauchy sequence, so it converges to its LUB. Not too hard to show 1 is the LUB.
 
  • #38
Tom_K said:
Yes, I accept that it is true, (by definition) I just don’t accept that it is proven by the method of truncating the repetend.

This is a little better, but it still won’t stand up as a rigorous proof:
The sum of a geometric series is equal to: a(1-r^n)/(1-r)
Where a is the first term (0.9 in this case), r is the common ratio (0.1)
if r is less than one, as n goes to infinity the sum becomes just a/(1-r)
In this case, Sum = 0.9/(1-0.1) =1, therefore 0.999… = 1
Have you ever seen the proof of the geometric series formula ?
rajeshmarndi said:
I just show how .999... = 1

Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1
The above proof IS in fact a proof using the geometric series (with a minor modification).
 
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