Why does 0.999... = 1

...

>there's no natural number between 1 and 2, therefore 1=2
nice argument brainlet

but what if there is actually not 0.333.... in the glass

How is it clearly false?

the "minus" operation can be loosely translated as "what is difference between such in such".
Answer me, numerically, what is the difference between 1.000 and 0.999..?

There are real and rational numbers between them.

Then you're a greedy faggot who doesn't want to share precious water.

1 is a whole 0.999... is not

What's missing?

if 0.999... =1 then does 0.888... = 0.89?

the number n such that n + 0.999.. = 1

Try subtracting them.

The idea of real numbers is to have a set of number such that to each point on the line there would correspond exactly one point. So we can think of real numbers as points on the line.
Now let's try to figure out how to represent those numbers in decimal form. Mind you, we identify numbers with points on the line. So take one point on the line and call it 0, now take another point that is to right from our point 0 and call it 1.
Now we will try to assign a decimal to each point between points 0 and 1.
Divide the interval between 0 and 1 into 10 equal parts, and label them from left to right with 0,1,2,3,...,9.
Now for each of those interval do the same thing. So for example, you take the interval with the label 1 divide it into ten equal parts and label each of them with numbers from 0 to 9.
At this stage you divided the interval into 100 equal parts and you can identify each of those parts by two numbers that range from 0 to 9. And for example the 100th interval here is identified by the pair (9,9).
Now for each of the 100 interval you repeat the process of division in to ten parts and labeling them with numbers from 0 to 9.
Now you clearly see you can continue this process as long as you want, and at the n-th step of this process you will have divided the interval from 0 to 1 into 10^n parts and you can identify each part with an n-tuple of numbers from 0 to 9.
Now here comes the trick, if you would take any point P on the interval from 0 to 1 at each step of the construction there is only 1 interval in which that point is contained.

It's called zero and is expressed this: 0

(cont.)
Moreover, for each other point P', no matter how close to P, there is a step in the construction and an interval from that step such that P is in that interval and P' is not.
From this remarks you see that each point on the interval is defined by an infinite sequence of interval, where each step n of the sequence is the unique interval from the n-th step of the construction that contains P. But each of those intervals has an unique label - a n-th tuple of numbers from 0 to 9. Now we will identify each number on the interval as the sequence of those n-tuples. Instead of the sequence of the n tuples w can think of each points ans an infinite string of numbers fro 0 to 9. Now we will take take infinite string of numbers, put "0." and call this new string of characters is a decimal representation of the real number P. Now my question to you is, what is the decimal representation of the ponit 1?

I guess the question here is, why is 1/3 == .333....? Are they literally the exact same?

Try the division

n has to be larger than zero

Then you'd be making a false statement

>How do you go from a number that's not one to a number that is?

"The father of Ivanka Trump"
and
"The current president of the USA"
is the same person.

0.999...
and
1
are the same number.
And so is
1/2 + 1/2
and
1/4 + 3/4

2/6
and
1/3
and
0.333...
are the same numbers too

if 0.999... = 1 then 0.999... + 0.999... should = 2

but it's not possible to add together infinite numbers because you'd be adding for an infinity long time.

>if 0.999... = 1 then 0.999... + 0.999... should = 2
Indeed it is.

>but it's not possible to add together infinite numbers because you'd be adding for an infinity long time.
You can't do computable arithmetic on these representations of numbers, but that doesn't mean the algebra doesn't go through.

Except not TO one

That logic isn't sound in the case of equality. Its only sound in the case of equivalency. The real proof is as follows:

0.99~ = 0.33~ * 3
0.33~ = 1/3
1/3 * 3 = 1
Therefore 1 = 0.33~ * 3
OR 1 = 0.99~

The proof's pretty straightforward. I think someone posted it here last time someone started a threat like this.

let x = 0.99....
10x = 9.999
10x - x = 9
9x = 9
x=1

Alternatively,
1/3 = 0.33...
0.33 * 3 = 0.99...
1/3 * 3 = 3/3 = 1
0.99... = 1

Can you justify that?

Okay, so, we can think about this as "distances between two numbers."
Proposition
If a - b != 0, a and b are different numbers
if a - b = 0, a and b are equivalent

2-3 = -1, so 2 and 3 are different numbers
2-2 = 0, so 2 and 2 are the same number

1 - 0.99... = 0.00... so 1 and 0.99... are the same number

but 0.333... is an infinite nnumber clearly there is only finite amount of water in the glass.

How can I be the same person as my wife's son's father?

...

0.999... does not exist

prove me wrong

It's defined as the geometric series 0.9 + 0.09 + 0.009 + ..., i.e. the sum from i=0 to infinity of 0.9 * 0.1^i. Elementary analysis clearly shows that this limit does indeed exist.

how can something requiring """infinity""" exist

Go study some analysis and find out. We epsilon-delta now.

ok i actually will do this thank you

You're not. Some other guy is.

>SCIENTISTS SAY THEY CAN ONLY SAY WITH 99.999...% CERTAINTY THAT YOU CAN NEVER GO BACK IN TIME

Figured it out ur welcome every1

1/3 = 0.333.... so
3/3 = 0.999....
3/3 = 1
0.9999..... = 1

>1/3 = 0.333.... so
prove it

The summary (lightly simplified, but not much) of it is this:

Consider the "infinite sum" 0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 + ...; in other words, 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... . I claim that these infinitely many values together sum to 1.

What that means is the following: consider the sequence of *partial sums* of this infinite sum: the first number, the sum of the first two numbers, the sum of the first three numbers, etc. This yields 0.5, 0.75, 0.875, 0.9375, 0.96875 ...; AKA 1/2, 3/4, 7/8, 15/16, 31/32, ... . This is an infinite sequence of numbers that get ever closer to 1, but never reach it.

Now it so happens that the numbers in this sequence, while all being less than 1, get *as close to 1* as you like. All numbers from 15/16 (= 0.9375) onwards are between 0.9 and 1; all numbers from 127/128 (= 0.9921875) onwards are between 0.99 and 1. More generally, for *any distance D larger than zero* that you can name -- here I named 0.1 and 0.01 -- there is a *point in the sequence* such that *all points afterwards* are no further than D removed from 1. All points in the sequence after 15/16 are AT MOST 0.1 removed from 1 (i.e. between 0.9 and 1.1); all points after 127/128 are at most 0.01 removed (between 0.99 and 1.01).

Based on this, in calculus, we say that the infinite sum has the value 1. For you can make *finite approximations of the infinite sum* that are *as close to 1 as you would like*. This is called a limit, and the formal statement would be that the limit as i goes to infinity of (1 - 1/(2^i)) is 1. Moreover, formally, the SUM for i from 0 to infinity of 1/(2^i) is 1.

0.999..., AKA 0.9 + 0.09 + 0.009 + ..., AKA the sum for i from 0 to infinity of 0.9 * 0.1^i, has a value of 1 by much the same reasoning.

See en.wikipedia.org/wiki/Limit_of_a_sequence and en.wikipedia.org/wiki/Series_(mathematics) for further introductory reading.

So this is the stock response, but I've never quite understood it. If you won't permit .333... as 1/3's decimal representation, then what is?

"""facts""" like 1+2+3+... = -1/12 invalidate the whole concept of infinite series

That's not a limit though

>"""facts""" like 1+2+3+... = -1/12 invalidate the whole concept of infinite series
1+2+3+... = -1/12 is NOT true in the theory of infinite series. 1+2+3+... is just infinite, just as you'd expect it to be.

I mean it's literally the decimal representation of 1/3. Just divide 1 by 3.

I appreciate this picture, these threads are just boring. 1 Million "3*1/3=3/3=0.3...=1"-posts.
But yellow 2 is false, there are many models with infinitely small numbers (although not [math] \mathbb{R} [/math]).

>That logic isn't sound in the case of equality. Its only sound in the case of equivalency.
And that is exactly the point. One way to construct [math]\mathbb{R}[/math] is by equivalence classes of cauchy series of Rationals.
Unfortunately, your proof (as all the others in this thread) is not valid, it just shifts the "problem" from 0.9...=1 to 0.3...=1/3 (yes I know, I'm able to divide, but it's about representation).

So, open the book and give a proof.

Lets say theres some infinitesmal k between 1 and .999...

1 = .999... +k

We can divide both sides by any number, lets divide by 3 for familiarity and demonstration's sake.

.333... = .333... + (.333...)k => .333... = .333... +k

Well, thats odd, both sides are not equal. Huh, it turns out that the infinitesmal doesnt make sense for any other iteration of 1 = .999... +k. If thats the case, then 1 =/= .999... +k.

Thus 1 = .999...

>0.999... = 1
>0.999... = 1 - (1/infinity)
>0.999... - (1/infinity) = 0.999
>[0.999... - 2(1/infinity) = 0.999
>1 = 0.999 - (infinity)(1/infinity) = 1-1
>1=0

I maed a meem xD

>brainlets still getting baited by trolls who contest this

If none of the proofs satisfy you please provide one, you vapid twit

if 0.999... = 1

what does 0.1/10^(infinity) +0 .999... = ?

What we adjusted our model to include an infinitely small number? Let's call it Horse.

Horse would have these properties:

Horse + Horse = 2 Horse
Horse * X = X Horse
However because the Horse is infinitely small, any X Horse is also equal to Horse.

So Horse = 2 Horse
Divide by Horse
1 = 2

That doesn't make much sense until you realize there is no Horse. Or should I say, zero is also a Horse? It looks like a Horse, it walks like a Horse, gosh darn-it it must be Horse.

Sky's the limit.

0.1/10^infinity = Horse

i hate you Veeky Forums

0 + 1 = 1 ? am I missing something?

First of all, unless you're working in an extended number system, infinity isn't a number. It's a possible cardinality for a set, but that doesn't make it a number. You can't just raise something to the power of infinity.

What you CAN do is add 0.999... to the LIMIT of 0.1/10^(x) as x APPROACHES infinity (or, to use more appropriate language, grows without bound). The result is, of course, 0.999..., because the limit of 0.1/10^(x) as x grows without bound is exactly 0.

So, if your point is 0.999... can't equal 1 because adding the limit of 0.1/10^(x) as x diverges would also give you 1, I agree with your premise but not your conclusion, because the limit of 0.1/10^(x) as x diverges is exactly 0, and it's absolutely expected behavior for adding 0 to a number to give you the same number back.

It's what this user said. If you understand how to do the sum of a geometric series, .999... can be represened by the series of 9*(1/10)^n starting at n = 1.

Sure, you can argue that it will never EQUAL one, and there is no proof that would adequtely address that. What every proof does show is that the difference between .999... and 1 is absolutely and completely negligible in every possible sense; a condition that is sufficient to say that they are "equal".

So they're not equal but we say they're equal because it's very close?

Why didn't someone just post this at the beginning of the thread?

why do people try so hard to convince others that 0.999... = 1 and not just admit that it doesn't but it's very close so we say it is.

It's unbelievable how hard people try to shill 0.999... = 1, you'd think they were being paid.

>So they're not equal but we say they're equal because it's very close?
It's not so much that it's "very" close. It's INFINITELY close, and any two numbers that are infinitely close are equal.

kek

Suppose [math]0.999 \dots \neq 1[/math]. Using [math]d \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}[/math], [math]d(x, y)=|x-y|[/math], we get a metric space. Metric spaces are Hausdorff, so there are disjoint nbds [math]U \ni 0.999 \dots, V \ni 1[/math].

On the other hand, let [math]W[/math] be any nbd of [math]0.999 \dots[/math]. Now, since [math]d(0.999 \dots, 1)< \varepsilon[/math] for all [math]\varepsilon > 0[/math], there is an open ball [math]B(0.999 \dots, r) \subset W[/math], but then [math]1 \in W[/math], contradicting the Hausdorff property.

>d(0.999…,1)0

you can't just assert this when you're trying to prove that .999...=1

Yes I do and I will ask a question please answer me. I have been thinking about it since yesterday night.

1)Take n number of things out of which p are identical to one another and other q are identical to one another.
Total number of arrangement of n number of things among themselves without repitition is
nPn = n!
2)But total number of unique arrangement is
nPn / (p! * q!)

>How do i conceptually and logically infer 2 from 1?
Please explain.
Don't give example to prove this.
Best is to give pure mathematical path of infering 2 from 1.

pls halp

Base case
1 + 0.99... = 2
Induction Hypothesis
x_n + 0.99... = x_n + 1
Induction Step
x_(n+1) + 0.99... = x_n + 1 + 0.99... = x_n + 2

where's my prize

>why do people try so hard to convince others that 0.999... = 1 and not just admit that it doesn't but it's very close so we say it is.
nice strawman faggot

if two numbers x and y are distinct, then there is a number between the two of them

now tell me which number is between x=0.999... and y=1 ?

(1+.999...)/2

i like this

0.99... + ((1 - 0.99...) / 2)

ha I got you there is no way you'll think of a flaw in my argument

0.999..., except a 0.999... that's bigger than the one you wrote. That's which number.

re.2: Yes they do.
Example: dx

let y = .0000...00009 + your number, that's closer to 1 than your x
thus your x was not .999...

In case anyone wants the real reason why it's because the sequence of numbers

{.9, .99, .999, .9999, ... }

converges to 1 in R. Here's what this means exactly. Let a_n be the nth term in the sequence. For any number e>0 we can find an integer N such that |a_n - 1| < e whenever n>N. This is clear since we can always make |a_n - 1| smaller by taking larger n.

This limit is unique. Suppose there was some other number than 1 it converged to, say C. Then |C - 1| = |(C - a_n) + (a_n-1)| N and n>N' where N' is the number associated with convergence to C. Since e is arbitrary, the inequality |C - 1| < 2e must always hold for every positive e. Therefore C=1. Since the limit of the sequence is unique and real numbers are defined by convergent sequences, we identify .99... with its limit 1.

if two numbers differ by an arbitrarily small amount they're the same number

you shouldn't end your arguments with ellipses...
it makes you look like an edge lord........

Your intuition wants there to be an infinitely small number, an infinitesimal between the two. Infinitesimals aren't well defined in the real number system, a whole lot of maths break if you try to shove them in. There are number systems that use infinitesimals though. Accept that in R there are multiple decimal expansions for numbers.

False.

Let m be a variable that changes.
Let its infinitesimal unit of change dm be less than 0.0...01.
Let x be 0.999... + dm.
My x is now closer to 1 than 0.999... is.

>not knowing ellipses is one way to denote an infinitely repeating decimal
back to /pol/ please

now let x = 0.999... + dm + dm
still closer.

thus your x=0.999... + dm did not equal the true 0.999...

>>not knowing all mathematicians are edgelords
no indeed i did not

>actually 1.000... - 0.999... = 0.0...01 so it is actually true because 0.0...01 = 0

i'm afraid you got meme'd, my friend. good luck in your introductory real analysis class!

Thanks but I'm already done with it. Have to keep things simple for those not in the know.

How do you mean?

think of 0.9999.... as 1- lim(1/n) as n --> infinity

Take any arbitrarily small ε. If it is between 0 and 1-0.999...9 for some number [math]n[/math] of 9's, then [math]|1-(0.999 \dots 9 + 9\cdot 10^{-(n+1)})|< \varepsilon[/math].

...

So you are telling me
0.(9) exists but 0.(9)8 cannot, right?
so much for a continious number line
Cuz you know, we could pretty much deduce all numbers are equal by that logic
1 = 0.(9) = 0.(9)8 = 0.(9)7 = ...