Why 0.999... is not 1. here

lim n->inf

1/10^n=0

9/10^n=0

1/10^n+

9/10^n+9/10^(n-1)+9/10^(n-2)+•••

=0

but sum is "1"

therefore,

lim n->inf, 1/10^n is not 0

so 0.999... is not 1

isnt it?

Other urls found in this thread:

en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_from_Cauchy_sequences
twitter.com/NSFWRedditVideo

If 0.999 = 1 doeth that meanst that there be no conservation of energies whilst people whomst applyeth limits calculateth with wholest numbers?

Kek I see what you're doing.

No because you're making a sum from n=infinity to 0, which for our purposes is the same as the sum from m=0 to infinity where m=-n, of the same problem. You're just trying to do a proof by induction by saying the limit for 9/10^(n-p) as n->inf where p=0, equals zero, and it's true for p=1, etc. That doesn't really work here, because you're really just reversing the number line, as in, if p=n-2, then p is within epsilon of n.

I guess I didn't really explain that well, but that's the best you're going to get from me.

the series will eventually add
9/(10^-1) + 9/(10^-2) + 9/(10^-3) ...
you are not describing the series that represents 0.9999......

This is not technically correct. Reference pic

[math] \displaystyle
1 = \frac {3}{3} = 3 \cdot \frac {1}{3} = 3 \cdot 0. \bar{3} = 0. \bar{9}
[/math]

that reference is complete trash though

want an intuitive way to understand it OP? given two real numbers x and y, x = y if and only if there doesn't exist z such that x < z < y

To say that 0=\=0.999 implies that there is some difference 0.000...1
But by definition in the 0.000... the ... suffix means there 0s go on forever. There cannot be a trailing 1 because the 0s never terminate. Not just that but it's fundamentally wrong because it's merely a 'mistake' in our decimal system, just like long division of 1/3 leads to 0.333...,
No matter how much you divide there's always that remainder. Multiply it back by 3 however and that mythical 0.000...1 fills in the difference. But this is a trick because in reality 1 and 0.999... are the same thing, in the same way that 10x-x = 9.999...-0.999...

That reference is not trash

why is this question even relevant

you will never have to use .999999... in your entire life

I mean, you will never have .99999..... of an item. You have one or something, or you don;t.

>even number of water molecules in that water bottle

Brainlet

whoever made it doesn't even know what induction is

Proving that
9.999... - 0.999... = 9
does require induction.

You latex you fucking retard.

Secondly 0.999...=1 is true by the definition of the real numbers. Your proof is non-nonsensical and idiotic.

That pic is complete bullshit and completely wrong.

any of these "proofs" are assuming the hypothesis to be true and then showing that from the truth of the hypothesis the hypothesis follows.
It makes absolutely no sense.

No.

>any of these "proofs" are assuming the hypothesis to be true

Actually they aren't, but you can go ahead and pretend that they are if you want.

You're fucking retarded.

Allow me to PROVE this easily.

0.99999... is equal to:

[eqn] \sum_{n=1}^{∞} (\frac{9}{10^n}) [/eqn]

And the partial sums are:

[eqn] \frac{9}{10},\frac{99}{100},\frac{999}{1000},... [/eqn]

So the sum is this limit:

[eqn] \lim_{x \to ∞ } \frac{10^x -1}{10^x} [/eqn]

Using L'hopital we get:

[eqn] \lim_{x \to ∞ } \frac{ln(10)10^x}{ln(10)10^x} = \lim_{x \to ∞ } 1 = 1 [/eqn]

The answer is 1.

>No.
ok then, which one of the equal signs is wrong?
#1, #2, #3 or #4? ()

Not him, but your argument is flawed.

You have to prove

[eqn]\frac {1}{3} = 0. \bar{3}[/eqn]

Let me go through them from left to right okay?

1.
x=0.999... already implies that the equivalence class of the Cauchy series representing 0.999... is converging.
By the definition of equality in the real number is is obvious to see that 1 is also a member of that equivalence class, so 1=0.999... .
2.
The claim that 0.333...+0.333... is actually defined implies the equality that 1=0.999... for the same reason as stated above.
3.
The convergence theorem used already implies that R is complete which for the same reason as stated in 1. means that 1=0.999... .


For
The definition of multiplication and equality of real numbers (Cauchy sequences in Q) already implies that 1=0.999 for the same reason stated in 1.

Wrong. Assuming the convergence of the series already implies that 0.999... =1.


The only valid reasoning for proving that 0.999... = 1 is that they represent the same equivalence class.
Which is true because:
[math]x_n \in 0.999...[/math], where [math]x_n=\frac{10^n-1}{10^n}[/math]
and [math]y_n \in 1[/math], where [math]y_n=1[/math] obviously implies that:
[math]|x_n-y_n| \to 0[/math]

but, assuming 1/3=0.333...

it would be correct, ok?

>assuming 1/3=0.333...
That already implies the hypothesis...

just answer the question

>non-nonsensical
So it makes complete sense?

That is the answer...

it's a simple yes/no question

either #3 is correct or not

Yes, but you haven't shown anything at all. You can always """"prove"""" anything by assuming the hypothesis.

is that a yes?

Yes. Can you read?

Makes sense, since the hypothesis is an a priori truth.

>a priori truth
No.
It is a result of the definition of the real numbers. Showing that 0.333...=1/3 is exactly the same as showing 1=0.99... .

Good, so now we can forget the 0.999 thing and concentrate on the 1/3 thing.
They are equivalent, no?
Proove one and you have proven the other. ok?

Yes. I already wrote the proof here:

1/3 has the advantage that by doing the division manually, you get the sequence 0.333... and no one has ever shown where the pattern breaks.

So there you have it, no fancy sums, a simple syntax-based proof.

No. That makes no sense. How do you actually show that 1/3 equals 0.333...?
I don't get what you mean by saying "doing the division manually".

>doing the division manually
pen and paper, old school style
you keep having to subtract 9 from 10

What are you accomplishing with that?

Are you seriously telling me that it is "good enough" to calculate the first few terms and then claim "the pattern holds"?

This is absurd and completely idiotic.

we can agree to disagree

you can by all means blow my proof apart
by showing where the pattern breaks

>we can agree to disagree
No, you are wrong.

>my proof
It isn't a proof. It is a conjecture. Your claim is "the pattern never breaks" but you gave no arguments or any kind of proof why this should be true.

If you make the claim "the pattern holds" you have to show it and seriously doubt that you are able to do that.

I don't have no fancy learnin degree, but even I know Op is a complete faggot

>you are able to do that

and you are not able to do the opposite

I stick to my claim.

I feel you have the burden of evidence since every calculator, wolfram alpha etc. seem to support that the pattern holds.

>since every calculator, wolfram alpha etc. seem to support that the pattern holds.
I dont know whether I should laugh or cry.

Do you know about the Riemann hypothesis one of the most famous unsolved problems in mathematics? All calculation have shown it to be true.
But it still remains unsolved.

Mathematics doesn't work by saying "the pattern holds" and you are only demonstrating that you have not one clue about mathematics by saying such outrageous things.

>I feel you have the burden of evidence
THAT IS NEVER THE CASE. The burden ALWAYS lies with the person who makes the claim.

>and you are not able to do the opposite
Of course not I ALREADY PROVED IT.
I know that it is. But you haven't provided ANYTHING.
You haven't written anything that could even resemble mathematical reasoning.

>Wrong. Assuming the convergence of the series already implies that 0.999... =1
What the fuck are you talking about? I know there is a missing step where I don't prove the series converges but....

Assuming something converges is not the same as assuming it converges to 1, it could be 2, or 3.

Seriously what the fuck are you on?

>The only valid reasoning for proving that 0.999... = 1
Wrong. I can think of thousands of different ways to prove it.

>I know there is a missing step where I don't prove the series converges
Yes. And that missing step already includes the proof for 0.999...=1. (R is complete)

>different ways
There are certainly different way but if you don't apply the definitions of the real numbers or their equality it is not a valid proof.

>Riemann hypothesis
vs 1/3
who's laughing now

>The burden ALWAYS lies with the person who makes the claim.
you are the one claiming the pattern breaks

>you are the one claiming the pattern breaks
I dont I ACTUALLY PROVED THAT IT DOESNT, which is something you dont seem to be able to do.

>vs 1/3
>who's laughing now
I dont know who is laughing, but your reasoning would mean that you just solved the RH.

>I ACTUALLY PROVED THAT IT DOESNT,
no you didn't, your proof was based on fancy sums, not syntax patterns.
Two different things altogether.

>And that missing step already includes the proof for 0.999...=1

Well.. yeah... you can't prove something without a missing step so duh....

All I'm saying is just because you assume something converges when it really doesn't, doesn't mean that it will always converge to 1.

Take 1 + 2 + 3 + ... = -1/12

-1/12 does not equal 1.

>Assuming the convergence of the series already implies that 0.999... =1.

No, it could be imply it equals 1, or 2, or -1/12, or......

>no you didn't
You are retarded I did.

>your proof was based on fancy sums
No sum was included in my proof.

> not syntax patterns.
So you mean my proof was based on more then "it think this is true"?

>Two different things altogether.
Yes. ONE OF THEM IS VALID MATHEMATICS (mine) and your ""proof"" (WHICH IS NOT MATHEMATICS) is just gibberish demonstrating your lack of knowledge in mathematics.

Do you even have a high school diploma? you certainly have no background in mathematics and I HIGHLY doubt that you knwo what a real number even is.

>Syntax patterns isn't math.

Gödel might disagree

>All I'm saying is just because you assume something converges when it really doesn't, doesn't mean that it will always converge to 1.
Of course not.
BUT THE EXISTENCE OF A LIMIT MEANS THAT YOU ARE ASSUMING THAT 1=0.999... . THAT IS JSUT THE DEFINITION OF THE REAL NUMBERS.

>1 + 2 + 3 + ... = -1/12
ahahah that is a Veeky Forums joke not an actual true equation.

>No, it could be imply it equals 1, or 2, or -1/12, or......
LOL. Read about completion of metric spaces and the definition of reals.

Gödel proved nothing by saying "the patter holds".

All his proofs are rigorous mathematics and have very little to do with "patterns" neither do his results.

Dude, I know that you have no clue about mathematics but you dont make yourself look smarter by namedropping people who's work you dont understand even in the slightest.

Okay then, I don't know why you're arguing then

>Read about completion of metric spaces and the definition of reals.
What should I read about them and why?

>What should I read about them and why?
So you know what a "real number" actually is. Which you dont seem to do.

Else you would understand why your proof doesn't work.

>So you know what a "real number" actually is.

Can you just fucking tell me what a real number is without being a dick?

>Can you just fucking tell me what a real number is without being a dick?
Not within the time span of a few hours.

en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_from_Cauchy_sequences
Here is a brief Wikipedia article about different methods. One of the most used is the construction from Cauchy sequences in Q which makes perfectly clear why 0.999... and 1 are the same real number.

Soo, found the reason why the pattern would break yet?
Because your ad hominem mouth-foaming is getting kind of old.

>Not within the time span of a few hours.

"If you can't explain it simply, you don't understand it well enough."

- Albert Einstein

>Soo, found the reason why the pattern would break yet?
Again. I ALREADY PROVED THAT IT HOLDS.

I don't really understand why you keep using the same arguments again and again.

Seriously read a bit about mathematics you know very little about the subject and this is not an "adhominem" (nice argument btw.) just a mater of fact.

Airtight proof for brainlets:

[eqn] S = 0.9999… [/eqn]
Add 0.111… to both sides
[eqn] S + 0.111… = 0.9999… + 0.111… [/eqn]
Now let’s calculate the right side: 0.9999… + 0.111… is equal to:
[eqn] \sum_{n=1}^{∞} (\frac{9}{10^n}) + \sum_{n=1}^{∞} (\frac{1}{10^n}) [/eqn]
Which is equal to:
[eqn] \sum_{n=1}^{∞} \frac{10}{10^n}[/eqn]
Which equals:
[eqn] 1 + .1 + .01 + .001 + … = 1.111… [/eqn]
So now we have this equation:
[eqn] S + 0.111… = 1.111… [/eqn]
Subtract 0.1111... from both sides:
[eqn] S = 1 [/eqn]
Sub for S:
[eqn] 0.999… = 1 [/eqn]

>"If you can't explain it simply, you don't understand it well enough."
I certainly can't give you a completely rigorous explanation in a few minutes.

The rough Idea looks like this:
You look at the rational numbers and notice that you even when you have a series of rational numbers where the terms get very close they dont necessarily converge against any rational number.

Now you want to expand the real numbers to get even these "missing" numbers, you do that by taking all these sequences where the terms get arbitrarily close together and call that the "real numbers".

Now you need to define things such as equality, looking at 2 such sequences you say they are equal if the difference between their terms goes to zero.

Wrong, for the reasons outlined here. You are proving the statement by using the completeness of R itself is prove enough that 1=0.999... .

dude I said the proof was airtight, you think this is a game?

No but seriously, point out 1 line in that proof that is wrong. You can't.

>No but seriously, point out 1 line in that proof that is wrong.
Every time you add real numbers you are using a definition which already implies that 0.99...=1.

Nothing you do is "wrong" but you are also never showing anything.

The proof has to done by using the definition of the real numbers and when they are equal.

>Every time you add real numbers you are using a definition which already implies that 0.99...=1.

No I'm not, you don't see that until the end. You can add any two numbers you want. Addition is a very well understood function, you can't prove addition because you need addition to prove addition.

>Addition is a very well understood function,
Do you understand what adding real numbers means?

I am certain that you don't, it is about equivalence classes of Cauchy series in Q.

Actually you can't define addition unless you already know what it is.

If I gave you "1 # 1" you would no idea how to use #. Math is circular logic, as you know, and cannot be broken down into a list of axioms.

>Math is circular logic, as you know, and cannot be broken down into a list of axioms.
Are you serious?

Look into ZFC, math is traditionally axiomatic and all of modern mathematics is completely axiomatic. I am afraid that you have no clue what you are talking about.

Once you have set theory you can define addition first on N then Z then Q then R then C.

Well technically you would have to redefine mathematics if you wanted to do such a thing, and at that point we have redefined the math we started with and we end up not learning anything about what we originally wanted to know: addition.

Besides, it doesn't matter anyway:

Just because you can prove B from A, does not mean that you must have A to go from C to B.

>Well technically you would have to redefine mathematics if you wanted to do such a thing, and at that point we have redefined the math we started with and we end up not learning anything about what we originally wanted to know: addition.
Wrong. This was already done, mathematics is ONLY defined through these axioms. There is no other addition.
It is and has always been defined through the use of axioms. This concept is over 2000 years old when Euclid wrote his "elements" where he also started with axioms and defined everything from there.

>Besides, it doesn't matter anyway:
It does you don't know the definitions and so you don't have the required knowledge to judge whether your proof is correct.


>Just because you can prove B from A, does not mean that you must have A to go from C to B.
But you are assuming A and then you go on to show A.

>But you are assuming A and then you go on to show A.

In your proof, you assume A and use it to prove B. So you are actually assuming A anyway.

In my proof, I assume A, go from A to C, then C to B.

In ANY proof, you must assume A first. Even in your proof. So there is nothing wrong with either of our proofs.

Just because you can conclude B from A, and you start from A, doesn't mean that you are assuming B. And if you say it does, then congratulations! You can't prove anything!

If the simplest logic like that escapes you... ask your school for a refund.

>So there is nothing wrong with either of our proofs.
Wrong. You are arguing from a position where what you are showing is already true. You are basically representing a real number as a series and then evaluate the limit of the series. While doing that you actually use that the limit exists and is unique. That step is something you leave unproven, but it contains everything important.
While making the leap back from the series to the real numbers you NEED to already know that 1.11...=0.999...+0.111... .
You never explain why series converges, or why it is unique or even why it is a real number at all.
Do you even know WHY?

You are only hiding the point where a real proof is needed.

Either your proof is invalid, because you never proved anything at all or you are assuming the hypothesis.

have you even taken precalculus? did you miss the lecture on limits or something?

This is going to be fun:

According to that logic, you can't prove that 0.999...=1 using the definition of real numbers.

Because the definition of real numbers requires that 0.999...=1

So you are already assuming the conclusion you are trying to prove.

>you can't prove that 0.999...=1using the definition of the real numbers
You can.
It requires 1-2 lines and is fairly straightforward.

>Because the definition of real numbers requires that 0.999...=1
No. But you are using that exact fact when you go from the series to the limit and you provide no proof for it.

>So you are already assuming the conclusion you are trying to prove.
Again no. I was talking only about your proof where you are actually never showing that the series converges.
But the convergence of the series is EXACTLY what you need to show.

>But you are using that exact fact when you go from the series to the limit and you provide no proof for it

No, I'm using real numbers, but I'm not using 0.999...=1 lol

If you are saying that anytime you use real numbers you are assuming 0.999...=1 then, according to that logic, you can't use real numbers to prove 0.999...=1.

>when you go from the series to the limit

What limit?

>What limit?
When you make the sum into a real number
>1+.1+.01+.001+…=1.111…
You never proved that this is actually true and NO it isn't obvious, but a very short proof.
Which is exactly the same proof as 0.99..=1.
That you actually don't realize that you are using limits explains a lot.

>If you are saying that anytime you use real numbers you are assuming 0.999...=1 then, according to that logic, you can't use real numbers to prove 0.999...=1.
Again, the proof that 0.99..=1 is at most 2 lines and you only need the definition of equality for real numbers.
You are using:
>1+.1+.01+.001+…=1.111…
Which you never showed, but the proof for that is the exact same as for proving 0.99..=1.

Oh boy, let's see, how would I explain this to someone like you....

Your job is to prove that 1+.1+.01+.001+…=1.111… without using real numbers.

>That you actually don't realize that you are using limits explains a lot.
>asking you to clarify what limit you are talking about means I don't know what limits are
Fascinating.

>Your job is to prove that 1+.1+.01+.001+…=1.111… without using real numbers.
You can use real numbers all you want.
But you only need their definition and the definition for equality.

This is really straightforward if you know when 2 real numbers are equal, I am sure that you can do it. That is also the whole proof of 0.99...=1.

wtf I actually hate .999...=1 defenders now?!

induction does not help there you moron

Set S to 0.999...

10S = 9.999...
S= 0.999...

10S-S=9S
9S=9
S=9/9
S=1

>9S=9
How did you even get that?
Why would 10S-S=9?

Who gives a fuck
For all practical purposes in calculation it's 1.

This is why mathematicians piss me off with their general retardedness. In literally 100% of real world calculations you have a healthy error margin chasing all your numbers around. Obviously it doesn't matter if it's 0.9999.... or 1 in the calculation, it will come to practically the same result. Computers have algorithms that parse off the ends of long decimals for calculation sake anyway

Proofs are great, but autism is not. Stop thinking like a clod and actually produce something useful

What are you even on about?

Do you seriously think that mathematicians are discussing with other mathematicians whether 0.999..=1?
Showing that this is true is about 1-2 lines, just applying the definition of equality for real numbers.

>Obviously it doesn't matter if it's 0.9999.... or 1 in the calculation, it will come to practically the same result.
It REALLY matters 1 is easily representable on a computer 0.999... is not.

>Computers have algorithms that parse off the ends of long decimals for calculation sake anyway
They don't. The way numbers are stored in a computer limits their accuracy not some algorithm.

To all who think these are different numbers, please describe the number BETWEEN them.

0.999 = 0. (3 nines)
0.999*10 = 9.99 = 9. (2 nines)

so
S = 0. (infinity nines)
10*S = 9. (infinity-1 nines)
9. (infinity-1 nines) - 0. (infinity nines) =/= 9

That's like saying 9.99 - 0.999 = 9, which is not true

>That's like saying 9.99 - 0.999 = 9, which is not true
It is completely true, but didn't prove it.

Also infinity is not a number you can't add to it nor subtract from it, infinity - 1 is not defined.

But you're not adding 1/10 and 9/10. You're adding two limits that approach 0.

>1/10^n+ 9/10^n+9/10^(n-1)+9/10^(n-2)+•••

Why did you write this?^
Limits aren't infinite series. They're limits. You don't sum all the values of n up to the point to which the limit approaches. You determine the value to which the function appears to approach as it reaches the limit.

Here's one: the surreal number [math]1-(0|y \in \cup_{n\in \mathbb{N}}S_n:y>0)[/math],
where [math]S_n=\{(x,y)|x,y\subseteq\cup_{i

Switching to a different theory of mathematics doesn't magically create new real numbers though...

nifty

It's a somewhat unintuitive artifact of our decimal system, bipolar weirdos treat it like yet another conspiracy

I don't know why you blame mathematicians for this, when they consider any values of constants or variables arbitrary, and prefer to replace them altogether with letters.