I really must know the answer to this important mathematical question

No, I don't think it does, because you have to put x in a finite state to use it in such a way. 10^∞ is the same as ∞ in the same way that 10^0 is 0, but a number that approaches either infinity or zero is impossible to use.

Your logic relies on semantics. Taking 10^1 and making it 10^2 is an approach to infinity. You have moved closer to infinity. However, those are two different values. If x has values 1 and 2 then x approaches infinity. It doesn't matter how many values x has, or how great those values are. If it has more than one finite value, it "approaches infinity".​

 

I think it was percent of health remaining X bonus from type of ball (Values 1 through to 4, not counting special balls), over total health, then X any status such as sleep/paralysis = Catch rate/chance

 

Squeezing in one last lost before I'm off for good tonight.

Not all limits are infinity. You just don't seem to like the concept of limits and most people seem to grasp that something drifts off endlessly before coming to grips with the fact that it could be going somewhere. The real limit I'm trying to make from this is that 10^-x approaches 0. We know that converges because the numbers keep decreasing but I know it will never reach -1, so it has to converge to a value greater than that. So where does it end?

 

The mathematician in me cried a little bit. Anything raised by zero is ONE!

Now I'll resolve this little debate you and Mixt have been having.

1/3*3 = 3/3 = 1
1/3 = 0.333...
0.333 * 3 = 0.999...
Therefore, 0.999... = 1

Does this proof satisfy you? The logic is sound. How about this one?

x = 0.9999999999...
10 x = 9.999999999...
So, then 10x - x = 9.999999999... - 0.999999999... = 9
Which implies 9x = 9
Which implies x = 1
Therefore anything that approaches 0.9999999 infinitely is equivalent to 1. So then I can say that 1.999999999 is equivalent to 2 and so on forth.

Like Mixt said, it's simple limits. It's like the first thing you learn in Calculus.

 

Clearly you don't know your logic sir, or you would've realized that z = (2) + (2) also is equivalent to z = (Fish)

As 2+2 clearly equals fish, ergo 1 + 1 = 4 or Fish. It wouldn't have been counted correctly without both answers.

 

It doesn't end.

.111111... > 0
.111111... < .211111...

And

.999999... < 1
.999999... > .899999...

That's all. There are an infinite amount of added 1's between .1 and 0 and an infinite number of added 9's between .9 and 1, so they will never, ever, reach their destinations.

Meaning it is being raised by nothing? Sorry, it was late.

The logic is not sound because if .999999... ever reaches 1, then it literally is 1, and the notation .999999... is incorrect. The ... directly implies that there is no limit. If you apply a limit, then it is no longer .999999....

Furthermore, in order to use .999999... in an equation like you did, you have to set it into a finite state. Otherwise each usage of it could be any finite state. You're trying to use a function or an array in place of a concrete variable.

var num = 0.999999999999999999

That you can use. You can go:

x = 0.999999999999999999

And that will be usable in an equation because it is a single, finite number.

What you are using is something like this:

function infNum() {
for (x = 0; x < 1; x + 0.000000000000000000000000000000000000000000000000000000000000000001) {
document.write(x);
};
};

So you are doing:

x = infNum();

Or:

x = [ 0.000000000000000000000000000000000000000000000000000000000000000001, 0.000000000000000000000000000000000000000000000000000000000000000002, 0.000000000000000000000000000000000000000000000000000000000000000003, 0.000000000000000000000000000000000000000000000000000000000000000004,
0.000000000000000000000000000000000000000000000000000000000000000005, ... ];

That doesn't work in equivalent exchanges. Let me show you why:

x = [ 0.000000000000000000000000000000000000000000000000000000000000000001, 0.000000000000000000000000000000000000000000000000000000000000000002, 0.000000000000000000000000000000000000000000000000000000000000000003, 0.000000000000000000000000000000000000000000000000000000000000000004,
0.000000000000000000000000000000000000000000000000000000000000000005, ... ]

10x = Which of [ 0.000000000000000000000000000000000000000000000000000000000000000001, 0.000000000000000000000000000000000000000000000000000000000000000002, 0.000000000000000000000000000000000000000000000000000000000000000003, 0.000000000000000000000000000000000000000000000000000000000000000004,
0.000000000000000000000000000000000000000000000000000000000000000005, ... ] ? * 10

Nope, I'm afraid I don't remember learning this in Calculus.​

 

The above example in programming is wrong because you have finite digits however you need to create a series to be accurate in this case.
Anyways... this 0.999 youre talking about is the series of Σ(9/10^n), n=1->oo (which is, 9/10 + 9/100 + 9/1000 + 9/10000 +...) therefore

http://upload.wikimedia.org/wikipedia/en/math/5/6/9/56949181a290ce561f27bd550a720392.png

 

Going from the simplest mathematical question to an interesting debate in regards to the equivalence of a value using a large number of mathematical proofs.

This thread is why I love the Spam Zone.

 

Did this seriously become a heated debate about asymptotes and infinity ...

wow KHV ...

wow ...

 

I miss chunking <3

 

I am aware of the first bit. However you simply cannot express infinite digits in programming (as far as I know) because they are impossible numbers. That is what I was trying to convey. If I were able to use an infinite number to express an infinite number, then the expression would not have been necessary.

The problem with the logic is right there in the equation. Tell me if I'm wrong about this, but you are using a value that references and edits itself.

.9999... = .9999... with an added 9

Isn't that correct?​

 

No, you have to distinguish infinity from numbers. Infinity is not a number, you can't add one more 9 in a number with infinite digits and make it bigger than before. 0,999... is a series as I said earlier and this is the only way it can be accurately expressed. The other proofs you mentioned earlier that it equals 1 are not exactly valid.

 

Okay, then I don't understand how you can use a series in an equation. If no number in the series .99999... is equal to 1, then the series itself cannot be.

Am I missing something?​

 

One word: convergence.

 

Explain in great detail. I do not understand why I should accept that .99999... has a limit at all when it will logically never reach said limit.​

 

I took 3 times differential calculus 1 before I passed it, if I am to explain it "in great detail" I might as well write a book about it and publish it now.
You lack basic knowledge in maths (limits and their properties- infinity, convergence/divergence) and I am not the one to teach you them.

 

Psst chill dude, don't be so sure about that.

 

Then I am afraid that I cannot take your word for it. I will not listen to an appeal to authority because that is fallacious. I would rather have it reasoned out and explained because that is what you do with logical concepts. You do not simply take someone's word for it what the answer is.

If you are not willing to explain further when asked, then please do not attempt to in the first place.

An argument left uncountered, or countered badly, wins by default.​

 

Interestingly enough though, if I recall correctly, 2^infinity > infinity in one notation of infinity but 2^infinity = infinity other times ... which is werid but also interesting I guess? *shrug*.