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

[HallsofIvy]

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.

 

[WWGD]

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.

 

[Intrastellar]

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 ?

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).