Originally Posted by

**C_ntua**
Yes, I would disagree on that part even though I agreed at first

The "..." is a bit tricky symbol and confuses things. Since you put "x" you assume something like this:

Code:

0.999..... = Σ(x=1...inf)(9*10^-x)

which is fine, but what you do with the above representation is that somehow you use only the last "x". That is how I see it. Where there is a huge difference

x->inf means that x goes to infinity

x=1...inf means that you have a put one by one numbers from 1 to infinity

If you want to use limits you would do:

Code:

lim(Z->inf)(Σ(x=1...Z)(9*10^-x))

You see that there is a separation on the symbols, mixing x with Z here is what makes this assumption wrong.

The symbols "lim" and "->" follow certain rules and those should be used exactly as they are and not just because they seem right. So the starting point is

Code:

0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ...

and you should add limits if you want by starting there.

Actually, the two are identical:

Code:

lim(Z->inf)(Σ(x=1...Z)(9*10^-x))
0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

Except that I was too lazy to get the sigma symbol here, so I simply wrote it out with "+ ... +". Let me proof it again for you using the sigma notation.

But first let me discuss this part of what you said:

so you are not getting closer and closer since you will be adding zero. True?

In other words the difference will always be "0.00..001". The more 9s you add the more 0s you can add and you are always on the same point.

But:

Code:

lim(x->inf)(9*10^-x)

Is not a very small number, you have to understand. It's actually 0, per definition of the limit. If x grows towards infinity "9*10^-x" grows smaller. The "9*10^-x" term grows to 0 but never actually becomes 0, meaning the LIMIT is 0.

Now let me proof it using your notation (which is the same except for notational differences). If you still disagree with something please show me exactly what, because I am curious about all this.

Code:

0.9999... = lim[Z->inf]( Σ[x=1...Z](9*10^-x) )
1 - 0.9999... = 1 - lim[Z->inf]( Σ[x=1...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1 - Σ[x=1...Z](9*10^-x) )

There, same three steps as before. Now we start taking terms from the sigma and combining it with the "1" (notice the change in the "1" in the sum).

Code:

1 - 0.9999... = lim[Z->inf]( (1 - 9*10^-1) - Σ[x=2...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1*10^-1 - Σ[x=2...Z](9*10^-x) )

And then we take another term:

Code:

1 - 0.9999... = lim[Z->inf]( (1*10^-1 - 9*10^-2) - Σ[x=3...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1*10^-2 - Σ[x=3...Z](9*10^-x) )

We repeat this until we get to:

Code:

1 - 0.9999... = lim[Z->inf]( 1*10^-(Z-1) - Σ[x=Z...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1*10^-(Z-1) - 9*10^-Z )
1 - 0.9999... = lim[Z->inf]( 1*10^-Z )

As said before, that limit is 0! So:

Code:

1 - 0.9999... = 0
1 = 0.9999...

If you still disagree with something, let me know.

Edit: Okay so you already agreed on it by the time I finished writing it... See that it's the same thing I did before except being too lazy to find out how to use the sigma symbol in here.