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

[micromass]

You said,
Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

But you can't arbitrary use varying numbers of decimal points.
change line 2 so 10X = 9.99 that is make it correct
then 10X -X = 9.99 - .999 = 8.991
or 9X = 8.991
or X .999 back where we started.

No, you're missing that there are infinitely many numbers after the decimal point, not finitely many. The proof is completely correct.

 

[Mark44]

You said,
Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

But you can't arbitrary use varying numbers of decimal points.
change line 2 so 10X = 9.99 that is make it correct

You are mistaken. There are not "varying numbers of decimal points" anywhere in the work above. Evidently you are not familiar with the ellipsis notation, ... , which means "and continuing in the same fashion. In the 2nd line above, the notation 9.999... means that the 9's repeat endlessly.

Similarly, n = 1, 2, 3, ... is shorthand for saying that n can be any positive integer.

then 10X -X = 9.99 - .999 = 8.991
or 9X = 8.991
or X .999 back where we started.

No. .999 is very different from .999...

 

[rajeshmarndi]

Is .999... tends to 1(as it is always less than 1) or equal to 1.

 

[micromass]

Is .999... tends to 1

That is as meaningless as saying that 3 tends to 4.

 

[WWGD]

A standard nonzero Real number can not be indefinitely small, by the Archimedean property. If it can be made indefinitely small, it must equal zero ( this is not so for the non-standard Reals, where you can have infinitesimals ). The expression 1-0.99999... can be made as small as you want , so it must equal 0. Then x-y =0 implies x=y. More formally, you have an argument using the Reals as a metric space, so that d(x,y)=0 iff x=y . Then apply Archimedean principle.

 

[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).