.9999999 = 1

[Burke]

That is actually a unanimously accepted proof, it is just a fact. I think you might not believe at first sight because you are under the notion that any given number has only one decimal expansion, which is not the case. It is as easy to prove in base-3 as it is in base-10. Defining pi would have nothing to do with this, as it only shows that pi is defined and you can represent it with decimal notation. I could define pi as a continued fraction if I like or from bessel functions like Ramajan did: these represent the same number, but will likely have differing based off of convergence.

As a challenge, if .999... is not equal to 1, what is the difference of the two of them (i.e., what is 1-.999...). Recognizing what

lim 1/(10^N) is as N approaches infinity should help you in this endeavor.

[seiki]

If I went with nanicoar then I presume you would tell me 1 - .999... = what? .oooooooooo...1? It would not be a valid answer ever to throw a 1 at the end of the repeating 0's because this would end the whole sequence of endlessly repeating 0's.

[clintjm]

Originally Posted by nanicoar

You can easily define a math where the infinitesimal exists, and consequently show that .999... doesn't equal 1. Proof of nonexistence is only valid within the relevant universe of discourse, and it is unfortunate that you went through a proof without defining its limitations.
My own opinion on the matter is that if your math contains pi, 0.999... doesn't equal 1.

At the same time it can be so close to 1 by infinity that the human mind can't distinguish it from 1.

[seiki]

Originally Posted by nanicoar

You can easily define a math where the infinitesimal exists, and consequently show that .999... doesn't equal 1. Proof of nonexistence is only valid within the relevant universe of discourse, and it is unfortunate that you went through a proof without defining its limitations.
My own opinion on the matter is that if your math contains pi, 0.999... doesn't equal 1.

I never used pi.

[Aquilus]

A pretty old one but quite a good if you consider this uses only basic algebra.
a=b
a² = ab
a²+ a² = a² + ab
2a² =a² + ab
2a² - 2ab = a² + ab - 2ab
can also be written as:
2(a² - ab) = 1(a² - ab)
cancelling (a² - ab):
2 = 1

[seiki]

Now that is some cosmic shit right there.

[seiki]

Originally Posted by Aquilus

A pretty old one but quite a good if you consider this uses only basic algebra.
a=b
a² = ab
a²+ a² = a² + ab
2a² =a² + ab
2a² - 2ab = a² + ab - 2ab
can also be written as:
2(a² - ab) = 1(a² - ab)
cancelling (a² - ab):
2 = 1

Wait. You divided by 0. Invalid answer.

[Aquilus]

Originally Posted by seiki

Wait. You divided by 0. Invalid answer.

that's right^^ but for such a simple task not too bad huh?
there are some other, way longer versions of it so in basic check every step if it's right but don't keep in mind that a=b in the beginning.

[seiki]

I often forget that removing the same thing from both sides or cancelling out is division. I just see it as cancelling out like it is its own thing. But it would be perfect otherwise.
I do not think it can be allowed because people's minds would be literally blowing.