Neil degrass tyson

...

What is the difference between 1and 0.9999...?

Are you so inanely retarded that you overlooked how you're adding two distinct values to sum 1, even at infinity?

There are an infinite amount of numbers larger than [math]0.\bar{9}[/math] but less than 1.
Math needs reformation to show them, but they are proveable to exist through infinite sum comparisons via calculus.

the difference is something so fast or so small that we cannot begin to comprend.

>When you make an ez bait thread and it works far better than expected

.99999... = 1
The statement of the title, is, in fact, meaningless, because it tacitly assumes that we can add-up "infinitely" many numbers, and good old Zenon already told us that this is absurd.

The true statement is that the sequence, a(n), defined by the recurrence

a(n)=a(n-1)+9/10^n a(0)=0 ,

has the finitistic property that there exists an algorithm that inputs a (symbolic!) positive rational number ε and outputs a (symbolic!) positive integer N=N(ε) such that

|a(n)-1|N .

Note that nowhere did I use the quantifier "for every", that is completely meaningless if it is applied to an "infinite" set. There are no infinite sets! Everything can be reduced to manipulations with a (finite!) set of symbols.

The difference is finite.
1/2^n will always be less than 9/10^n for any partial sum of cumulation addition that is a test of any and every value substituting n. So for [math]\stackrel{x}{y} \rightarrow \infty , \frac{x}{(x+1)^n} < \frac{(x+y)}{(x+y+1)^n}[/math] shows that for any sum to n, these values are not equal - but unreformed math would cheat via laziness to claim all these values simultaneously simply equate to [math]0.\bar{9}[/math]. Infinity imposes brainlet problems.

there are candidate ways to write what the numbers truly may be, such as [math]0.\bar{9}_{\frac{1}{2}} < 0.\bar{9}_{\frac{9}{10}}[/math], where a truly maximal representation of the number might be [math]0.\bar{9}_{\frac{\infty}{\mathclap{\infty}o}}[/math]

so your worry is that b>1 ?
that's a first

feel free to show where it drifts away from 1

>[math]\frac{1}{10}\big( 9+b \big) = b[/math]
>[math]\frac{9}{10} = \frac{9}{10} b [/math]
>[math]1 = b[/math]

learn to do math lmfao

There is nothing wrong with that part of his proof though. The error is assuming [math]b[/math] is a finite number. This has to be proved.

and the claimed error is...?

you are the first person in history claiming that 0.999.... > 1

Feel free to stop posting you utter retard. I've addressed your retard math in virtually every thread you've posted it over the past couple of months, including your "feel free" replies and "which line doesnt equal 1".

You are an utter fucking tard cause your math says [math]\frac{\bar{9}}{1\bar{0}} + \frac{1}{1\bar{0}} = 1[/math] which directly admits [math]0.\bar{9}[/math] + "some value" = 1; 1 - "some value" = [math]0.\bar{9}[/math], and that [math]0.\bar{9} \neq 1 [/math], all the while you claim the opposite of what the math shows, that 0.999... = 1 even though it doesnt.

Subhuman brainlet.

+2+3+4... = -1/12
.99999... = 1
both right tho

It makes sense intuitively. I'm just speaking about the rigor of the proof, not whether the general idea is logically consistent.

so you can't show any point where it happens,
good

[math]0.\bar{9} + x = 1[/math] is your claim.

What is the value of x?

[math] x = \frac{1}{\infty}[/math]

zero

No, its [math] \frac{1}{number}[/math].
You've already shown the pattern of work continues as [math]+\frac{1}{10}, +\frac{1}{100}, +\frac{1}{1000}, +... [/math] all the way to [math]..., +\frac{1}{1\bar{0}} || +\frac{1}{\infty}}[/math]. I know you don't realize this, but these are non-zero values.

[math]\frac{1}{1\bar{0}} or \frac{1}{\infty}[/math]

0.99999... IS NOT A NUMBER. A NUMBER MUST HAVE A BEGINNING AND AN END.

omg it's shitlatexman again

1/9 is not a number
ok

Could you please share the form of these numbers? Is it possible to do so? It's not that I think you're wrong, but I am not convinced.

Perhaps. Maybe something like how we can 'rank' the cardinality of infinite sets.

Maybe either of you two can see the error in my reasoning.

Claim: [math]1-0.99\overline{9}=0[/math]

Proof: Let us suppose that [math]0

>-1/12
are you trolling or serious right now?

Different poster. I think I ran into this guy in a different one of these sub threads and I hashed out his idea on my own time.

The idea falls from this:
[math]\lim_{n\to\infty} \dfrac{1}{10^n} = 0.0...01\\

\lim_{n\to\infty} \dfrac{k}{10^n} = 0.0...0k[/math]

It is problematic for many reasons. One is that it invalidates the identity principle (a+0=a) which makes it a different set from the reals(at least).

>non-zero
sure buddy

wolframalpha.com/input/?i=1/infinity

fight it out with w-a

Infinity is a broken idea in math the way it is used. It is implied to be a number when it actually functions as a set of numbers, so that infinity + 1 or infinity + infinity still evaluates to infinity. In this sense, infinity is a set or a range of values rather than a specific number, which is why it would make sense that [math]0.\bar{9}[/math] does not invoke any specific number and that there could be numbers greater than it, as numbers that would exist in the infinite range would have comparable unequal values just the same as the countable unique numbers before infinite uncountability.

So if we take for example [math]0.\bar{9}[/math], this isn't a single number, and as a classical math number may as well be as alien to the concept of numbers as infinity itself is, and this idea holds partially true for any other repeating decimal. There also exists no natural way to result 0.999... short of directly crafting it with an infinite sum, and heres where it can be shown with a different way of writing these repeating decimals.
You can say [math]\frac{1}{3} = 0.\bar{3}[/math] and that [math]0.\bar{3} × 3 = 0.\bar{9}[/math] so [math]0.\bar{9}[/math] must equal 1, but that is being a little disingenuous. That we can evaluate [math]\frac{1}{3}[/math] as 0.333... is cheating, in a way. For any non-repeating decimal, the fraction and the decimal both have finite values attributed to them so for 1/3 with its finite values to evaluate as 0.333..., a non-finite string of 3's, is actually an accident. 1/3 as a decimal should really read as [math]0.\bar{3}_{\frac{1}{3}}[/math], so instead of the unintuitive classic equation [math]0.\bar{3} × 3 = 0.\bar{9} = 1[/math], it can be instead intuitively solved as [math]0.\bar{3}_{\frac{1}{3}} × 3 = 0.\bar{9}_{\frac{3}{3}} = 0.\bar{9}_{1} = 0.\bar{9}_{\stackrel{\leftarrow}{1}} = 1[/math]

sorry im bad at LaTEX

the 0.999...=1 thing seems obvious to me but I've never seen a convincing proof of the -1/12 meme. Does anyone have something remotely convincing?

Its that infinity, or infinity sums, or an infinite'th step, these ideas have no real numerical value.

We can take the convergent sums of
[math]A= \sum_{n=1}^{\infty} \frac{1}{2^n}, B= \sum_{n=1}^{\infty} \frac{9}{10^n}[/math] to possibly evaluate that both A and B converge to 1, but for every partial sum A is less than B. If they were meant to reach the same singular, finitely attributal value of [math]0.\bar{9}[/math] or 1, they would actually accomplish this at different rates and different times, where this idea of rate or time is completely absent from the answer and the variables of the classical equations, but if it were present would be able to show what we can already evaluate from any amount of partial sum comparisons: that A is less than B.

This holds because infinity is not a singular number but a range of numbers, and A evaluates in that range ahead of B, so A must be a value less than B.

A problem like this probably wouldn't even exist if infinity hadn't been misused the way it has been in classical math. If its a number bigger than all other numbers, infinity+1 should classically evaluate to being larger than infinity, but instead it just dynamically redefines the definition of infinity to say infinity + 1 still equals infinity. It is quite paradoxical that infinity would be called a number, yet that number would be any uncountable number. We don't treat numbers this way and infinity fucks everything up by getting different rules. [math] 1 + 1 \neq 1, \infty + 1 \neq \infty[/math]