0.999... = 1 - General - Facepunch Forum

[QUOTE=MountainWatcher;35319533]Okay, then let me ask you a question. if I use induction I can say that, for 1/n, for n = 1, the distance from the function to the asymotote is non-zero. I can also say that for 1/(n+1), the distance is also non-zero. I multiply each side by n, and divide it by 1/(n+1), which leaves a x n/(n+1), a =|= 0. So, shouldn't I be able to say that it never reaches 0? But then again I) can say the limit of n isn't infinity, so clearly I'm wrong. but why?[/QUOTE] Induction is a proof for all [i]natural numbers[/i]. Infinity is not in the natural numbers. For example you might prove that 1 + .... + 1 is finite by induction for any [I]finite number[/I] of 1's, but this does not mean 1 + 1 + .... is finite!

[QUOTE=MountainWatcher;35319856]I'm asking why I can't extend that to infinity as well, not just the naturals. I mean n can tend to infinity. [editline]27th March 2012[/editline] is it because I can't reach a stage where infinity comes from n + 1? It doesn't seem very solid[/QUOTE] You can use the above link from JohnnyMo1 to demonstrate this in IR: ε real, ε > 0, l1/x-0l<ε <=> 1/x<ε, as we are looking for the limit in +oo <=> 1/ε<x, as ε>0, with 1/ε>0 So, for each real ε > 0 there exists a real δ = 1/ε, δ > 0 such that for all x with x > δ, we have |1/x − 0| < ε, so lim1/x(x->+oo)=0

[QUOTE=Bahumhat;35320671]x = .999 12x = 11.988 12x - x = 11.988 - x 11x = 10.989 x = .999 Do I win?[/QUOTE] Congratulations, you just figured out that .999 is the same as .999. Now do the operation for .9...

[QUOTE=sealclubber;35320677]Induction is a proof for all [i]natural numbers[/i]. Infinity is not in the natural numbers. For example you might prove that 1 + .... + 1 is finite by induction for any [I]finite number[/I] of 1's, but this does not mean 1 + 1 + .... is finite![/QUOTE] But the reason induction works is because, if the first proposition is right and one proposition being implies the other being right. We can work , number by number, until we reach that proposition. 1 is real and n +1 is also real. 3 is real because 1 +1 = 2 which is real and 2 +1 = 3 there,fore, it's real. We can reach infinity by adding ones infinitely, why doesn't it hold there, as well?

Because 1/n never reaches 0. It's its [I]limit[/I]. Therefore you can't prove 1/n=0 because [I]1/n=/=0 on IR[/I] Beside, induction only only works with naturals, not reals.

[QUOTE=Bahumhat;35320671]x = .999 12x = 11.988 12x - x = 11.988 - x 11x = 10.989 x = .999 Do I win?[/QUOTE] How do you think 0.999 = 0.9... and more importantly how do you feel qualified to try and lecture others about maths while thinking something like that?

[QUOTE=_Axel;35321280]Because 1/n never reaches 0.[/QUOTE] Never reaches and reaches at infinity is the same thing.

[QUOTE=MountainWatcher;35321265]But the reason induction works is because, if the first proposition is right and one proposition being implies the other being right. We can work , number by number, until we reach that proposition. 1 is real and n +1 is also real. 3 is real because 1 +1 = 2 which is real and 2 +1 = 3 there,fore, it's real. We can reach infinity by adding ones infinitely, why doesn't it hold there, as well?[/QUOTE] Because you cannot "add to infinity". Infinity is not a number. [i]No matter how many times you do it[/i], you've still only done it finitely many times.

Theorem. A cat has nine tails. Proof. No cat has eight tails. Since one cat has one more tail than no cat, it must have nine tails. Enjoy.

This thread.... [img]http://i249.photobucket.com/albums/gg221/mooman1080/happillyputonsunglassesthenleave.gif[/img]

[QUOTE=Bradyns;35322002]Theorem. A cat has nine tails. Proof. No cat has eight tails. Since one cat has one more tail than no cat, it must have nine tails. Enjoy.[/QUOTE] Not really a math theorem. More like a nifty manipulation of the english language.

[QUOTE=sealclubber;35321681]Because you cannot "add to infinity". Infinity is not a number. [i]No matter how many times you do it[/i], you've still only done it finitely many times.[/QUOTE] Why can't I define the series of 1 + 1's an infinite series, like 0.9 + 0.09 ...? And it'd add to be infinite, wouldn't it?

[QUOTE=MountainWatcher;35322854]Why can't I define the series of 1 + 1's an infinite series, like 0.9 + 0.09 ...? And it'd add to be infinite, wouldn't it?[/QUOTE] Because an infinite series only makes sense when it is convergent. An infinite sum of 1's, however, doesn't converge to any number--we say it is divergent.

[QUOTE=Anglor;35316781]It's really amusing seeing how certain you are about this. Who are you to decide wether mathematical definitions are redundant or not?[/QUOTE] Hey this is off topic but I love your avatar. All Hour Cymbals is one of my top 50 favourite albums.

Fractions are our friends but decimals are dicks.

[QUOTE=Ninja Duck;35324471]Fractions are our friends but decimals are dicks.[/QUOTE] Fractions are way easier to understand than decimals, especially repeating ones. [QUOTE=Bahumhat;35320671]x = .999 12x = 11.988 12x - x = 11.988 - x 11x = 10.989 x = .999 Do I win?[/QUOTE] I'll try to make it really clear for those who still don't get it: 1/3 + 1/3 + 1/3 = 3/3 (AKA 1) 1/3 = 0.3.. 0.3.. + 0.3.. + 0.3.. = 0.9.. (AKA 1) [B]0.9 =/= 0.9..[/B]

ITT: We're trying to do math, and we're failing miserably.

[QUOTE=Zoran;35325555]ITT: We're trying to do math, and we're failing miserably.[/QUOTE] More like some people can't understand the proof on the right of the image posted numerous times and want to say Euler and other mathematicians are wrong with no proof of their own

[QUOTE=Zoran;35325555]ITT: We're trying to do math, and we're failing miserably.[/QUOTE] Speak for yourself.

[QUOTE=JohnnyMo1;35325829]Speak for yourself.[/QUOTE] We is first person plural. First person implies that I'm part of we. I never implied I was any better than anybody else.

[QUOTE=JohnnyMo1;35317077]I hate to break this to you but she is right. Do you need me to type a proof up for you?[/QUOTE] [QUOTE=Glorbo;35317737][img]http://filesmelt.com/dl/0.999_1_.jpg[/img] [B]PLEASE, TELL ME EXACTLY WHAT THE FUCK IS WRONG WITH THE PROOF ON THE RIGHT.[/B] Don't tell me "IT'S NOT LOGICAL" or "IT DOESN'T MAKE SENSE". Tell me what is mathematically wrong with the proof given to the right. Tell me what illegal operation was supposedly given here that disproves this equation, please. Because until you do, all of your bullshit is null and void.[/QUOTE] Fourth time that image has been posted in this thread, Wikipedia links to proofs and various other proofs were provided, and yet these idiots still deny it. Please, I beg you, close the thread already.

No because 2/3rds = 0.667 GG WP i won.

No dude, 0.9.. never equals 1 same way 0.0...1 never equals 0 because infinitely small numbers exist. 0.9... is an asymptote of 1. And nobody would be arguing if some nerd wouldn't try to refute a common mistake that people are making. What an asshole. Also I don't understand what a reciprocal is but I'm definitely good enough at math to tell people that their proof is wrong, without actually giving my own. I can also only read up to 3 last posts. :downs: Did I miss anything?

[QUOTE=Glorbo;35321503]Never reaches and reaches at infinity is the same thing.[/QUOTE] Uh, no ? Take ƒ(x)=1/x-exp(1-x) for example. ƒ(1)=1/1-exp(1-1)=1-exp(0)=1-1=0, yet; lim 1-x = -oo x->+oo lim exp(X)=0 X->-oo So, by composition lim exp(1-x)=0 x->+oo Also, lim 1/x = 0 x->+oo So, by sum, lim ƒ(x)=0 and f(1) equals 0, so never reaches 0 and reaches 0 at infinity doesn't mean the same thing. x->+oo

If x = 0.999999(infinitely) and y = x - 1 then y = -0.000000000000...etc y will NEVER reach ...00000001.

I don't see your point. You're not really proving anything as infinitesimals don't exist in IR.

[QUOTE=_Axel;35327939]I don't see your point. You're not really proving anything as infinitesimals don't exist in IR.[/QUOTE] Exactly. 0.0.... is equal to 0. That means .9... is equal to 1.

Yep, since 0,999... can be seen as the limit of the following (a[I]n[/I])sequence: a[I]n[/I]=1-1/10^n (as 1-1/10^2=1-0.01=0.99, 1-10^3=1-0.001=0.999, etc...) We have lim10^n=+oo x->+oo & lim1/N=0 N->+oo So by composition, lim1/10^n=0 n->+oo So lim(a[I]n[/I])=1 n->+oo So 0,999... as we define it by the limit of the (a[I]n[/I]) sequence is 1. As a result 0,999...=1

[QUOTE=_Axel;35328075]Yep, since 0,999... can be seen as the limit of the following sequence: (a[I]n[/I])=1-1/10^n (as 1-1/10^2=1-0.01=0.99, 1-10^3=1-0.001=0.999, etc...) We have lim10^n=+oo x->+oo & lim1/N=0 N->+oo So by composition, lim1/10^n=0 n->+oo So lim(a[I]n[/I])=1 n->+oo So 0,999... as we define it by the limit of the (a[I]n[/I]) sequence is 1. As a result 0,999...=1[/QUOTE] I don't want to sound like a total noob, but what type of math is that anyways? I'm at quadratic algebra, so I'm pretty uneducated as far as math goes.

That's differential calculus. I have no idea what quadratic algebra is, though. In the French equivalent of 12th grade, we rarely talk about math types anyway. In my main math class we mainly do calculus, whereas in speciality math class we study arithmetic and very basic complex analysis.