Bug in mathematics

This is the millionth time we've had this thread, and you're probably a troll, but just in case.
Ask yourself what exactly a decimal expansion should mean. If I write 0.374, that means 3/10 + 7/100 + 4/1000 right? For infinite decimal expansion, I need to do an infinite sum. For example, 1/3 is 0.333... because the following equation is true,
[math]\sum_{n=1}^\infty \frac{3}{10^n} = \frac{1}{3}[/math]
This is the sort of thing you learn when you do calculus. So now that we have a sensible definition for a decimal expansion, we can easily prove, with calculus, that 0.999... is 1. It's just because we have the following equality
[math]
\sum_{n=1}^\infty \frac{9}{10^n} = 1
[/math]

If you want 0.999... to be something else, you'll need to give me a different definition of what a decimal expansion is supposed to be that makes it work. If you want 0.999... to be "infinitely close to 1, but not equal to 1", that won't work in the real numbers, because you can find a real number between any two distinct numbers. So you would need to change a natural definition and leave the real numbers to make it work. It's not really something that makes sense to do.

But OP is saying it's a bug. Not that the math doesn't work out.

We all understand that the math checks out. What we're saying is that it's incorrect despite the correct math.

That just because the math is true doesn't mean it actually is.

Sorry man, 1 and 0.99999... are different numbers. They are literally different numbers. One of them has a 1 before the decimal. The other only has value after the decimal -- no matter how infinite the expansion is, you cannot cross that unless another value is added to it.

This problem has nothing to do with the math. We get the math. The math checks out.

If you really believe this then you don't really get the math, sorry

This is a problem that is beyond math, man. Some things just aren't actually explained by math even when it works out.

This simple 1 = .999... problem is also a metaphor for how myopic you people can be. Even when the math works, it doesn't really work in the real world.

>in the real world
this is math, the real world is irrelevant, we dont care that numbers cant get infinitely accurate in the real world. Thats why we use fractions to approximate pi

1=.999...

For any n; for any k;
[math]\sum_{n=1}^{\infty} \big[ A: \frac{1}{2^n} \big] < \big[ B: \frac{2}{3^n} \big] < \big[ C: \frac{3}{4^n} \big] < \big( D_{1 \rightarrow \infty} \sum_{k=4}^{\infty} \frac{k}{(k+1)^n} \big) \neq 1 [/math]

Repeating decimals betray math because they aren't written sensibly. Repeating decimals implies A, B, C and D.1~infinity would each read as 0.999..., yet we can prove these are all unequal values infinitely many times for infinite tests of n and k.

[math]0.\bar{9}[/math] has no intelligible meaning as written and is missing information. It's like trying to identify the difference between 8 and 9 by only judging the upper half of their symbols. All we see is o . 8 < 9 but o = o

mathlets are retards and must manually breathe.

[math].\bar{9}[/math] has a defined value

[math].\bar{9} = \sum_{n=1}^{\infty} \frac{9}{10^n} [/math]

not missing any information at all, the infinite sum comes straight out of the meaning of the decimal representation

Maths is defined because we say so. It's hilarious and makes it sounds like you should be able to break it for no reason, but it all holds together somehow. Make it up, prove it, magic.

No because 1/3 is not 0.33333333333 repeating. That's just an approximation of it. So 0.99999 is not 1, it's just an approximation of it.

Thanks for proving my point

I'll prove it with a simple geometric series. So you have a series that goes 9/10+9/100+9/1000... this is equal to 9(1/10)+9(1/10)^2+9(1/10)^3+...
Now comes the formula a/1-r our a=9/10 and our r=1/10 so we get (9(1/10)/(1-1/10))= 9/10*10/9=1 now stop making this retarded post.

call [math] a = .333... , 10a = 3.333..., 10a-a=3.333...-.333..., 9a=3, [/math] so [math] a=\frac{1}{3}[/math]

Pic related is how I feel when I encounter posts like that of OP written by curious people with no knowledge of analysis.

algebra using approximations is still an approximated answer cunt.

n = 0.999...
10n = 9.999...
10n-n = 9.9999.... - .99999
9n = 9
n = 1
.9999.... = 1

the real world doesn't have infinity in it so we could never get .999....

n = x
10n = 10x
10n - n = 10x - x
9n = 9x
n = x

im proving that theyre equal asshat, i assume the existence of .3 repeating and show that its equal to 1/3.

no fucking approximation involved retard, learn some basic fucking math instead of throwing a tantrum because you cant wrap your thicc skull around infinite decimals

r=0.999...
10r=9.999...
10r-r=9r=9
therefore
r=9/9=1

that problem mainly exists in base 10, in base 12
1/3 equals 0.4

This is a wonderful explanation. If someone truly doesn't get this then they should try writing down these equations and messing about with them until they do.

1/3 = 3/10 + 1/30 = 0.3 + 1/30
= 0.3 + 3/100 + 1/300 = 0.33 + 1/300
= 0.33 + 3/1000 + 1/3000 = 0.333 + 1/3000
= 0.333 + 3/10000 + 1/30000 = 0.3333 + 1/30000

and so on

each line = 1/3
exactly, not approaching it.

line #1 is exactly 1/3
line #10 is exactly 1/3
line #98327498236483689 is exactly 1/3

At infinity, it still is exactly 1/3

>If you want 0.999... to be "infinitely close to 1, but not equal to 1", that won't work in the real numbers, because you can find a real number between any two distinct numbers.

If anyone wants clarification on what user is talking about here, the property he is referencing here is called density

mathonline.wikidot.com/the-density-of-the-rational-irrational-numbers
we can prove that in between any two unequal real numbers there exists infinitely many rational numbers. So if you show that two numbers dont have any numbers inbetween them, that is sufficient to prove that they are the same number

technically the bottom one is incorrect. you mean to take the limit of it as n --> inf

the limit at a point is never technically reached

>you mean to take the limit of it as n --> inf
No. Try writing that down on a bit of paper.

Squaring the circle milesmathis.com/square.html

Protip: The bug is in infinitesimals, not math itself.

Ask yourself: "what does the '...' means?" There lies your problem.
1 = 0.9 + 1/10
= 0.99 + 1/100
= 0.999 + 1/1000
etc. At each line you have equality with 1, so 1 = 0.999...

Could someone please help me out, then?
Because if what you say is true, and it makes a lot of sense, then what most of calculus is based on (that really small positive number that's smaller than any other positive number, I think it might be different from an infinitesimal) wouldn't exist. That, or it'd have to be zero, as the moment you declare such a quantity, you could find another quantity smaller than it.
Could someone tell me if this is wrong?

the more you add nines the more close to 1 you get. since there's an infinity of them you can basically say that the difference between 0.9 with no end and 1 is almost unpercievable.
It always depends on what you're doing.

are you 8 years old ? this is just logic

>what most of calculus is based on (that really small positive number that's smaller than any other positive number
This isn't what calculus is based on. Calculus is based on limits. The "really small positive number" is just a metaphor to give intuition, something people tell brainlets so they don't have to worry about proper rigor and can go on to become engineers.

Engineers dont work in aspects of infinity or infinitesmals. Calculus doesnt produce engineers, it produces mathlets.

There is only one proveably correct answer in this thread and it's

One is God. Zero is absence of God.

>your iq

it's the difference of day and night

Nice bait bro

You can check. Pic.

Do people not understand limits?

>There is only one proveably correct answer in this thread and it's

except for the fact that the post you linked is actually incorrect in a number of ways, and other posts are much more helpful while also being correct

If if 1/3+1/3+1/3=1 where did the 0.000 vanish?

into the second if

The only problem is people failing to understand basic fucking maths: decimal representation of numbers is not unique.

The decimal expansion of a number is a representation of the that dedekind cut, not the number itself learn analysis REEE

No, calculus is actually incorrect and needs to be reformed, as math in general does not provide enough information in a repeating decimal, which was proven by the post.

>m-muh limits
use a finite number limit or fuck off cause you don't actually understand the definition of infinity. You literally don't need to understand it either, ya just need to fuck off.

Clarifying, Equality in terms of real numbers means that they are the _same_ number, in our case, that means they have the same dedekind cut representation
[math]1 = \{x\in\mathbb{Q}: x < 1\}[/math]
whereas,
[math]0.\overline{9} = \bigcup_{i=0}\{x\in\mathbb: x

>calculus is actually incorrect
no it isnt
picture related

>as math in general does not provide enough information in a repeating decimal
you have the fucking definition of the number coded inside of the decimal representation. It is a specific fucking number regardless of how it agrees or disagrees on your views of what math should "really be like"

It's even simpler than that, because we don't NEED cuts,

Rationals is the result of taking the cartesian of the integers with themselves and quotienting it by a specific equivalence relation and you prove it algebraically. we don't even need LIMITS

>t.underage

[eqn]\mathbb{Q} := \mathbb{Z}\times\mathbb{Z}/{{}\sim{}} [/eqn]
Where, [math]{{\langle a, b\rangle}\sim{\langle \alpha, \beta\rangle}} \Leftrightarrow a*\beta = b*\alpha [/math].

It is trivial to prove this set, equipped with induced operations from [math]\mathbb{Z}[/math] is a Field.

[math]1/3 = \lbrack\!\lbrack1,3\rbrack\!\rbrack[/math]
[math]3 = 3/1 = \lbrack\!\lbrack3,1\rbrack\!\rbrack[/math]

[math]3/1 * 1/3 = \lbrack\!\lbrack3,3\rbrack\!\rbrack[/math]

but [math]\lbrack\!\lbrack3,3\rbrack\!\rbrack = \lbrack\!\lbrack1,1\rbrack\!\rbrack[/math], because 1*3 = 3*1 in the integers.

In short 3*(1/3) = 1, come on, it's not that fucking hard

I'm not saying you're wrong, but let me tell you my opinion: There is another way to construct IR from IQ through equivalence classes of limits of cauchy series. I like it more, I think it is more precise. Instead of "the same", "is in the same equivalence class" suffices. It is immediately clear, that the completion depends on the metric and it already is a foundation for doing analysis in the constructed completion of IQ.

Lets just slow it down to matrix mode so you can follow each step, step by step
Lets just start with two sums.
[math]A: \sum_{n=1}^{\infty} \frac{1}{2^n} | B: \sum_{n=1}^{\infty} \frac{9}{10^n}[/math]
Got it?
For ANY and EVERY test of n passed to both A and B, even an INFINITE amount of unique tests performed, the partial sums of B will ALWAYS be greater than the partial sums of A.

A and B absolutely under no circumstances equal the same value.

If you think they do equal the same value, you are literally retarded and just need to fuck off because you don't understand the definition of infinity and you've been more than capable of getting this far in life without understanding it, so whether you understand it or not is not even a valid response. Disagreement with this is just benign retardation cause it implies you are agreeing with some other definition, yet you do not understand and cannot invoke that definition.

This is not a conversation or a debate. This is a grade and a judgement.

Aye, they are essentially equivalent, but I like the metric emergent from the topology and thus from the order...

I guess if you're more analysis-minded, Cauchy classes are okay, but I feel they are yucky.
To each their own, and still,

If you consider real numbers classes of Cauchy sequences, then it is even more trivial to prove that 1 = 0.999.., I mean just prove it.

your constructed IQ has failed

If you define something to be the end result of an "infinite" process, (it's not, it's just notation) you can't simply do finite tests, senpai.

I'm not even the same guy but please, do enlighten me with > the definition of infinity

if you want to talk partial sums you came to the right guy.

for any partial sum of B, lets say you take the partial sum up to b places, then there exists some number a such that [math]\sum_{n=1}^{a} \frac{1}{2^n} >= \sum_{n=1}^{b}\frac{9}{10^n}[/math]

therefore they are equal

hm, no, A can be just bigger, you need both ways doode
[math]\heartsuit[/math]

1/x is always bigger than 1/x^2 when x>1
so, at x=1 they are different
riiiiiight. where's my nobel

Do you people never get tired?

i dont usually partake in these threads, i think people cycle through so that its new people most of the time this thread happens.

But i think thats mainly true for the side of the argument who knows that .999... = 1

B can evaluate to 1 by rounding at any decimal place. We can do one iteration of 9/10^n, get 0.9, and with the second iteration get 0.99 where the hundredths 9 will round up -> 0.9 + 0.1 = 1

For A, it is different. To get a number that rounds up on the hundredths, we must go through 1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 0.96875. We can then take the 0.06 to round up 0.9 + 0.1 = 1.

This still does not make A and B equal. A is equal to "1@n=2", and B is equal to "1@n=5".

Seriously, how literate are you? I'm not about to be mean, I just want to know if you're a school kid, an undergrad etc. Because, this is unreal. If this is bait, it's just unimpressive, but if you really think you're making sense it is worrying.

who cares if the n is different for finite choices, the whole point is that we take the limit to infinity

>doing infinite tests is doing finite tests

And in both cases, A and B do not sum 1x they sum a repeatomg decimal 9, and this is where math needs reformation because B is proveable greater than A so they can't both simply equal [math]0.\bar{9}[/math]

what?
What I'm saying is, just because something holds for EVERY case before the limit, it doesn't mean that it holds FOR the limit

eg.
All ordinals less than aleph_0 are finite, aleph_0 isn't, but aleph_0 is the limit of all ordinals less than it.

Infinity is a bitch, swallow it

More literate than you. Please, use a constant definition of infinity that isn't what I've described that does not also contradict itself. Could you come up with one?

b is not provably greater than a, just because its against common sense doesnt mean its right.

B and A are equal, since they both equal 1

I'm a 3rd year undergrad of pure mathematics

>constant definition
Infinity doesn't only come in one flavor, you know that? right? Not all two infinite sets are bijectable, you know that, right?

>what I've described
which?

[math]\sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} +\frac{1}{8} + ... = a= \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{4} +...) = \frac{1}{2}(1+a) = a [/math]
then solve for a

[math]\sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} +\frac{9}{1000} + ... = b = \frac{9}{10}(1 + \frac{9}{10} + \frac{9}{100} +...) = \frac{9}{10}(1+b) = b [/math]
then solve for b

theyre both equal to 1, precisely because they are infinite

i did b wrong because the latex preview box is ass to use, but you can fix it yourself

If you want a very silly definition,
Infinite, that which is not bijectable to a finite ordinal.

Finite ordinal, member of the smallest inductive set.

Smallest inductive set, The intersection of all inductive sets provided by the axiom of infinity, which states that there is at least An inductive set.

Inductive set, a set that contains the empty set and that, for every set X it contains, it contains {X}UX.

Empty set, the set that no element belongs to;

Finite, that which is bijectable to a finite ordinal

It doesnt hold for every case before infinity. It holds for an infinite case of infinite tests, which is what setting the limit to infinity means. If it were capable of naturally summing 1 at any point, any extended infinite work would take that 1 and blow it up towards infinity. You can only get 1 by:
>doing finite work
and
>having a decimal cutoff limit for rounding
and in such case, the decimal accuracy limit and the finite work limit would be parts of the answer itself, that it does not suffice to just say they both equal "1", but rather more like [math]A = 1_{n=5}^{d.=1} | B = 1_{n=2}^{d.=1}[/math], or more loosely as [math]A = 0.\bar{9}_{\frac{1}{2}} | B = 0.\bar{9}_{\frac{9}{10}}[/math] or even [math] A = 0.\bar{9}_{\underline{0.5}} | B = 0.\bar{9}_{\underline{0.9}}[/math] towards the efforts of not fucking up the working definition of infinity so we dont literally just say A = B like actual retards, so we don't say 2 = 1 like mongoloids.

>It holds for an infinite case of infinite tests, which is what setting the limit to infinity means.
It just isn't what it means.

in this simple case, the limit has to do with metric, it has to do with [math]\forall[/math].

Saying that x is the limit of a sequence s is saying

[math]\forall\epsilon>0:\exists n\geq 0: |s(n) -x|

[math]
\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} +\frac{1}{8} + ... = a= \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{4} +...) = \frac{1}{2}(1+a) = a
\\\\
\displaystyle
\sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} +\frac{9}{1000} + ... = b = \frac{9}{10}(1 + \frac{9}{10} + \frac{9}{100} +...) = \frac{9}{10}(9+b) = b

[/math]

>It doesnt hold for every case before infinity. It holds for an infinite case of infinite tests, which is what setting the limit to infinity means
No, it isn't.

no one gives Weierstrass the credit for making the concept rigorous.

Wrong.
[eqn] \forall \epsilon > 0 \exists N \geq 0: \forall M\geq N, |s(M) - x| < \epsilon [/eqn]

OMG you're right, I'm retarded sorry

Ignore , this is the right version >you'd think after Calculus V I'd know this shit

[math]
\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} +\frac{1}{8} + ... = a= \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{4} +...) = \frac{1}{2}(1+a) = a \\\\ \displaystyle \sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} +\frac{9}{1000} + ... = b = \frac{9}{10}(9 + \frac{9}{10} + \frac{9}{100} +...) = \frac{9}{10}(1+\frac{b}{9}) = b
[/math]

...

sauce?

you surely don't trust ((((Books)))) to ((((teach)))) you ((((maths))))?

>But how?
1/3 = 0.333...^-
0.333...^- * 3 = 0.999...^+

Kenneth Ross' Elementary Analysis. I like this book a lot, it's very readable. That proof was the most convincing to me that 0.999...=1.

[math]
\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} +\frac{1}{8} + ... = a= \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{4} +...) = \frac{1}{2}(1+a) = a
\\\\
\displaystyle
\sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} +\frac{9}{1000} + ... = b = \frac{1}{10}(9 + \frac{9}{10} + \frac{9}{100} +...) = \frac{1}{10}(9+b) = b
[/math]

what the fuck, user? you have to be 18+ to post on Veeky Forums

thanks user

18=17.999...

In this case?
In any case the limit is treated as a natural number. Whether n is 1 to 100 or 1 to 100000 or 1 to infinity, the limit is a treated as natural. This is the first fuckup because infinity is not a number.
Treating it as a number in the equation of classical calculus to sum the results [math]A|B = 0.\bar{9}[/math] leads to another fuckup of redefining infinity. In the working sum cases, infinity meant do "a (presumably concatenated) infinite amount of work". In the answer repeating decimal case, infinity then means "an equivalent finite number" in [math]0.\bar{9}[/math]

Classical calculus abuses infinity like a whore without a pimp, constantly reassigning it any arbitrary definition that just so happens to produce the requested result and completely regardless to whether the current definition contradicts another definition so the definition of infinity in the equation is different than the definition of infinity extracted as a resulting sum from the very same equation.
It is fucking schizophrenic paradoxical bullshit. Infinity means uncountably large. If you wish to set a limit to infinity but still receive some kind of sensible result right now by instantiating hidden rules of decimal accuracy to produce a result like [math]0.\bar{9}[/math], where the overline says there is an uncountably large amount of 9's, we have to add the same concatenating feature of infinity used in the sums, because:
[math] x = 1\rightarrow\infty \sum_{n=1}^{\infty} \frac{x}{(x+1)^n} < \frac{(x+1)}{(x+2)^n}[/math] defines an infinite amount of unequal values that classical calculus claims they all equal [math]0.\bar{9} = 1[/math].

Calculus' use of infinity is a brainlet paradigm.

we ARE NOT using infinity, and why is a limit treated as natural number WAHT

what is A|B, A divides B? what is this notation

Limits are not about infinite processes, that is a naïve interpretation of calculus,

the thing is, 1/3 + 1/3 + 1/3 = 1 because of Algebra even if calculus was invalid!

Mathematicians do not need to use infinity for limits, because we have the fucking definition of limits namely the epsilon and delta def. and the equivalent for sequences def

Nowhere infinity and infinite checks appear, you only need the set of natural numbers which comes from an Axiom, the Axiom of the inductive set

>[math]\frac{1}{3} × 3 = 1[/math]
A valid statement even though what you were implying was
>[math]0.\bar{3} × 3 = 0.\bar{9} = 1[/math]
Which is absolutely not a valid statement
The classical problem needs reformation to actually evaluate as
[math]0.\bar{3}_{\frac{1}{3}} × 3 = 0.\bar{9}_{\frac{3}{3}} = 0.\bar{9}_{1} = 0.\bar{9}_{\stackrel{1}{\leftarrow}} = 1[/math] for logical arithmetic. And you're right, 1/3 + 1/3 + 1/3 has nothing to do with calculus but thats not the main issue. The main issue is classical representation of repeating decimals is inadequate.