Magnus Anderson wrote:gib wrote:Except that \(0.9 + 0.09 + 0.009 + \cdots\) does not equal \(\infty\).

Noone said that.

Magnus Anderson wrote:\(0.9 + 0.09 + 0.009 + \cdots\), though it is an [in]finite sum, does not.

That's something that has to be proven. One way to prove it is by doing arithmetic with infinite sums. If you can't do arithmetic with \(1 + 1 + 1 + \cdots\), because it equals to \(\infty\) and because \(\infty\) is not a quantity but a quality, what makes you think you can do arithmetic with \(0.9 + 0.09 + 0.009 + \cdots\)?

I need to be absolutely clear about what you're saying. Are you

actually saying that:

\(0.9 + 0.09 + 0.009 + \cdots\) = \(0.\dot9\) = \(\infty\)

Magnus Anderson wrote:What I'm saying is that you can't do arithmetic with \(\infty\), but you can do as much arithmetic as you want with an infinite number of terms.

But \(\infty\) can be represented as an infinite number of terms.

You can do arithmetic with finite terms in order to

create an infinite sum. But once it's created, you can't do arithmetic with that.

I'll explain (like that will have any effect):

You can do arithmetic with a series of 1s:

1 + 1 + 1 = 3

You can do arithmetic with a larger series of 1s:

1 + 1 + 1 + 1 + 1 + 1 = 6

There is no limit to the number of 1s you can do arithmetic with. You can do arithmetic with an infinite number of 1s:

1 + 1 + 1 + ... = \(\infty\)

But now you can't take

that and do

further arthemtic with it:

(1 + 1 + 1 + ...) + n <-- Can't do.

\(\frac{(1 + 1 + 1 + ...)}{n}\) <-- Can't do.

\((1 + 1 + 1 + ...)^{n}\) <-- Can't do.

^ Note that not all infinite sums give you infinity. So we can do the same thing with the 9s:

0.9 + 0.09 + 0.009 = 0.999

0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + 0.000009 = 0.999999

0.9 + 0.09 + 0.009 + ... = \(\dot9\)

^ Unless you think \(\dot9\) = \(\infty\), you have not derived \(\infty\) here. Therefore, you

can do further arithmetic with it.

Magnus Anderson wrote:The problem, Magnus, is that (1 + 1 + 1 + ... ) + 1 doesn't mean anything different than 1 + 1 + 1 + ... So yeah, they're equal.

Fine. In that case, you have to accept the conclusion that 1 = 0.

Ooooor... I can just avoiding doing arithmetic with it.

Magnus Anderson wrote:Well, in that case, you disagree with this Wikipedia proof because it does arithemtic with infinite sums.

I didn't say you can't do arithmetic with infinite sums, I said if the infinite sum

gives you \(\infty\), THEN you can't do arithmetic with it. \(0.\dot9 \neq \infty\). (<-- Is this seriously lost on you?)

Magnus Anderson wrote:\(\infty + 1 = \infty\) // subtract \(\infty\) from both sides

\(\infty + 1 - \infty\ = \infty - \infty\) // substitute \(\infty - \infty\) with \(0\)

\(1 = 0\)

...

But where's the mistake?

I agree with step 1. The mistake is not there. It's not in

any of the steps. It's the fact that you

took steps to begin with. You should have stopped at \(\infty + 1 = \infty\). Why? Take a guess, Magnus. Because you can't do

what with infinity?

I will say this: Another rendition of what I'm trying to say is that you

can do arithmetic with infinity, but then different rules apply. So the rule that adding 1 to something gives you a greater number no longer applies. Instead, adding 1 to \(\infty\) still gives you \(\infty\).

So don't do arithmetic with \(\infty\) or accept that different rules apply. <-- Take your pick.