[math]0.bar{9}neq 1[/math]

OP learn logic please your examples are retarded and false

These posts aren't related to my thread.

>I just don't care
Yes, I am aware that this is a non-issue for the real world. However, it says a lot about mathematicians.

Yes, I looked at those. And in my OP I explain why these 'convergences' are approximations.

So mockery is all you are capable of? Not an argument.

>this misunderstanding is usually a result of a failure to understand the mathematical concept of "infinity"
Finally a serious post. However, I do not see how there is a difference between English and Mathematics. "large beyond comprehension" is wrong. I'd say a good definition is that infinites are too large and infinitesimals are too small to express in numbers. Do you understand what I mean with that?

Namecalling. Feel free too explain the actual mistakes.

They're not fallacious, you're essentially just arguing against the chosen definitions/framework because you don't think 0.999...=1 is a an 'acceptable' result.

you make retarded logical leaps, 'just like 0.9 = 1' where is the implication??

'clearly f(0.9)=0' what implies this? u are assuming 0.9!=1

learn logic

>I'd say a good definition is that infinites are too large and infinitesimals are too small to express in numbers.
That's your problem right there, because even if the number of nines in 0.999... is infinitely large by your definition, it will still be less than 1 by an infinitesimal amount.
In one of the proofs for this we write
>x=0.999...
>10x=9.999...
So
>10x-x=9x=9
>x=1
When we write
>10x-x=9x=9
we use the mathematical concept of infinity, because 10x-x can only equal the number 9 if the number of nines after zero in x and 10x are the same.

'hi im faggot OP and im gonna define infinity how i want to and then get confused when mathematics isn't consistent with my arbitrary definition of infinity hudurrrrrrrr'

>you're essentially just arguing against the chosen definitions/framework
When the current framework defines that two different numbers are the same then of course I will argue against it.

That is an example why "my framework" or whatever you want to call it makes more sense in my opinion than the current one. It is not meant to be a proof. Next time think before posting.

'hrurrrdurr im stoopid'

>even if the number of nines in 0.999... is infinitely large by your definition, it will still be less than 1 by an infinitesimal amount.
Exactly.

>10x-x=9x=9
This is not mathematically correct. What you do is crop a digit. You can't make up rules that only apply to infinitely long numbers. If a mathematical function / operation can only be applied to infinites it is not correct.

You perfectly described yourself.

>When the current framework defines that two different numbers are the same then of course I will argue against it.
Why are they two different numbers a priori? Infinite decimal expansions like 0.999... has no inherit definition without refining your ideas of what infinite means. Just saying "add 9s forever" doesn't mean anything by itself.

>You can't make up rules that only apply to infinitely long numbers.
But you can you dumbass, you're confusing mathematics and physics.

Because they are different numbers, which is why you express them differently.

See:
>Yes, I am aware that this is a non-issue for the real world. However, it says a lot about mathematicians.

>you can
Okay, then prove it.

You can define anything you like in maths as long as it doesn't conflict with the basic axioms.

>Because they are different numbers
A number is not its representation. 0.999... and 1 are considered to be two notational representations of the same object, not two different numbers which are somehow equal. 1 is also represented by 1.0000, 2/2, 3/3 etc.

>Let me make a similar example to explain my reasoning. Imagine there is a distance between A and B of 1 meter. Now this procedure repeats infinitely: 1. the distance doubles 2. you move 1 meter towards B. Even after infinite repetitions you will not reach B. Thus,
>∑∞n=112n=1∑n=1∞12n=1
>is merely an approximation and not actually true.
No, you will reach B because the distance from A also grows at the same rate as the distance to B. So after the first step you will be halfway there, at the second step you will be 3/4 of the way there, etc. Your intuition failed because it is not logically rigorous. Your intuition does not trump proof.

Yes. Now prove to me how an operation which doesn't conflict with the basic axioms can only be valid for infinites.

You are just talking about semantics here and your post boils down to [math]0.\bar{9}=1[/math] because you define it to be so, which is no different than saying 1=2 because you define it to be so.

>you will reach B because the distance from A also grows at the same rate as the distance to B
Wrong. You will never reach B. The distance to B will always be 1 or 2 meters, depending on which part of the recursion you are.

>You are just talking about semantics here
That's because THAT'S EXACTLY WHAT IT IS. We define decimal expansions by infinite series which are defined by the analytic definition of limits. The consequence is that 0.999... means the same thing as 1.

find an example where this actually matters and i might start caring

You said
>Even after infinite repetitions you will not reach B.
After infinite repetitions there will be 0 meters between you and B, because 1-1/2^n approaches 0. Show your proof.

Then this definition is a bad one. I guess it takes a mathematician to define that an approximation is an equation. And even if you follow this definition, you should always be aware of the approximation.

In physics and engineering competent people are always aware that they deal with approximations.

I think he's "trolling" us Veeky Forums, no one can be this dumb.

If [math]1/3 = 0.\overline{3}[/math] and [math]1/3 * 3 = 1[/math] then if [math]1/3 = 1/3[/math] and [math]0.\overline{3} = 0.\overline{3}[/math] then it's reasonable that [math]0.\overline{3} * 3 = 1[/math]

How is it a bad one? It is a non-issue physically as you said, and you don't need them to be different to do analysis (in fact you have to go out of your way to accomodate it, all because it doesn't satisy a hand-wavy idea of what an infinite decimal expansion should mean). You think "0.999..." should definitively denote a different object from 1, but you don't really have a good reason why.

youtu.be/akNNPBrfZQ4

[math]\frac{1}{3}=0.\bar{3}[/math] is an approximation and not actually true.

Try to repeat your "proof" with base 3. You can't. Because in base 3 [math]\frac{1}{3}=0.1[/math] is NOT an approximation and thus the approximation error vanishes.

Let me repeat myself:
>Yes, I am aware that this is a non-issue for the real world. However, it says a lot about mathematicians.
So basically mathematicians are too stubborn to admit or too unintelligent to realize that this is an approximation. This is what bothers me.

It's not an approximation.

>base 3
1/10 * 10 is 1, in all bases I can think of

An approximation of what, exactly?

Exactly this is the issue. I say it is an approximation because those are two different numbers. You claim that they represent the same number and the only "proof" you have is "because we define so".

how are they different numbers?

What a thematic post number

The same reason why 1 and 2 are different numbers.

because we define 1 to be {0} whereas 2 is defined as {0,1} haha

1 and 2 are defined to be different numbers since 2 is the successor of 1. I could make a thread sperging about how 1=2 as that's what you're doing.

troll thread, no one can legitimately be this fucking stupid

Actually it's a TeX practice thread disguised as a troll thread.

Your example of going from A to B is not an accurate representation of ∑1/2^n. It's
f(n) = 2*(f(n-1)) - 1, f(0) = 1. It's pretty obvious if you look at any case between n = 1 and infinity.

You claim they represent different numbers and the only "proof" you have is "because they look different".

[math]1/3=0.\bar{3}
2/3=0.\bar{6}
3/3=0.\bar{9}=1[/math]

That not "intuitive" enough for you?
Okay, how about
[math]1-0.\bar{9} = 0.\bar{0}[/math]

What's [math]0.\bar{0}[/math]? An infinite string of zeros. Or also known as, fucking zero.
[math]a-b = 0 \iff b=a
\therefore 0.\bar{9} = 1[/math]

Excuse my shitty TeX if I fuck it up pls

TL;DR LEARN TO INTO INFINITY

0.999 is converging on one so for all intensive purposes can be seen as one however true mathematicians know this is simply not the case.

Whoops I'm retarded. First one should be
[math]1/3=0.\bar{3}[/math]
[math]2/3=0.\bar{6}[/math]
[math]3/3=0.\bar{9}=1[/math]

>What is convergence
>What is rounding
Go back to school kid

first we need a construction of the floating point numbers. plz provide one. then we can continue.

im not asking this to annoy you but it appears to be the fundamental problem to understand this issue. what does it mean to write numbers as decimals instead of fractions?
e.g. what does 0.123421 mean?

i can write it as 123421/1000000 if i follow common rules.

now 1/1 is 10/10 = 9/10 + 1/10
Repeat the same with 1/10:
1/10 = 9/100 + 1/100 (keep in mind that 9/100 == 0.09)

And so on.
what you get it 1 = 0.999...

Don't confuse the poor OP, he's fresh out of undergrad mechanical engineering and legitimately thinks every number has a fuzzy decimal tail that gets significant-digit'd out of existence as soon as you write it on paper.

It's clear from OPs posts that he can't even comprehend the concept of infinity, since he conflates a finite yet indeterminate point on an infinite axis with infinity. OP gb2 physics 112 and report back once you've taken freshman or even high school calculus.

learn2calculus

They are trivially the same if you define the reals as the set of all Dedekind cuts. There is no rational less than 1 and greater than 0.9 repeating.

If you wish to define a notion of real number such that they are not equal, you are free to do so. But it will not have the properties of an ordered field where every bounded subset has a supremum.

will 0.999... ever stop being a free hundred replies

It's a well known result that a system cannot prove itself consistent