If this proof for 0.999... = 1 does not convince every single one of you, please drop out and consider suicide

So you want to be a worse version of Vixra. How is that possible?

I also watch mathologer

Is there anyone who doesn't think that? That's what was taught to us in university.

You can never find a number x that 0.999...+x=1, then 0.999...=1. That's because in order to obtain 1 you would need x to be 0.0...infinitezeros...1 but that's not possible, the 1 would make it a finite number and there will always be 9s after the 1 in x

x = 0

>a + ... + b = b + ... + a
I'm not a pure mathfag but I'm pretty sure they learn the Riemann rearrangement theorem early on in Analysis.

Yes, i forgot to say x different from zero :)

1/3=0.33333333....
2/3=0.66666666...
3/3=0.999999999
But n/n is defined as 1
Therefore 0.999999....=1

Brainlet. Rearangment only apples to CONDITIONALLY CONVERGENT series. All these have positive terms so that is not the case.

If you don't believe me, find a rearrangement that results in a different number.

Yeah that's what I thought.

Just look at [math](x_n) \in1, x_i=1 \ \forall i \in \mathbb N[/math] and [math]y_n = \sum_{i=1}^{n}10^{-i} \cdot 9, (y_n)\in 0.999...[/math] obviously [math]\underset{n \to \infty}{lim}x_n-y_n=0[/math] (simple calculation.)
Then you have [math]y_n\in 1\Rightarrow 0.999...=1[/math] qed.

Simply applying the definition of equality is enough. Your proof is missing some serious steps which require some real analysis (e.g. where you are changing the order of the summation without even mentioning why that works or claiming your series converges without giving a proof).

It's easier to see that the series converges than it is to try to make people understand the messy language of limits, because people who don't agree will insist that it never actually reaches zero.

No website has made me want to puke this bad in years.

1 + 1 + ... + 2 = [math]\omega + 2 \neq \omega[/math] = 2 + 1 + ... + 1

You can't reject the existence of limit ordinals without giving up the ability to define R.
Addition of infinitely many expressions is not commutative in general. That's why the Eilenburg-Mazur swindle is nontrivial.

>Is there anyone who doesn't believe the dogma they are taught?
Whether or not I agree about the subject at hand your argument is utterly disgusting.

>people still think .9999 does not equal 1

Brainlets.

Can you even read?

>find a rearrangement that results in a different number

All you did was post a divergent series.... that has nothing to do with the proof. You cannot rearrange the ACTUAL NUMBERS used in the proof because they are convergent.

Go ahead and try to rearrange any convergent series to get a different number... you will fail miserably. And everything in the proof is convergent.

>And everything in the proof is convergent.
Then why isn't there a lemma where the convergence is established, since we both agree that convergence is a necessary condition for the proof to be sound?

But that is just stupid.
You can not explain someone why two real numbers are equal if he doesn't know what a real number is.

>because people who don't agree will insist that it never actually reaches zero.
Then these people do not believe that the real numbers exist.

That's too complicated.

The idea is much simpler.

Theorem:

Let a,b in Q. a = b if and only if for all e in Q greater than 0, |a-b| < e.

The right side of the implication is evident. Now assume that for all e in Q greater than 0 |a - b| < e and that a != b, assume b < a, then a - b = c is greater than 0 and |a - b| < c/2. Thus a contradiction.

In other words, 2 rationals are equal if their difference is arbitrarily small. Thus 0.999999... = 1.