Convergent series

based

But they are at the same point, there's no distance between them.

no of course

even though the distance between the two points is infinitely small, they are still fundamentally separate

But they're not.

Congratulations, you misunderstand convergence. Slap yourself for being retarded and then go back to Calc 2.

Reads like something from Professor John Gabriel's chef-d'œuvre
drive.google.com/file/d/0B-mOEooW03iLd3gyMGlWNk02amM/view

>not using the cauchy sequence method of proving 0.9 = 1

eople from Wikipedia do not agree with you and with him.

here u go OP
[math]
.99999... = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{9}{10^{i}} = 1
[/math]

Thank you

It's the time of the year when all the calc 2 students are learning seq/series and keep making threads about them

They are representations, there is no distance between them in the first place. Distance is an illusion caused by belief in a number line. Don't fall for dogmas children.

false

thats literally the definition of the decimal expansion of a real number

en.wikipedia.org/wiki/Decimal_representation

I propose to officially exclude John Gabriel from the scientific community. He is engaged in pseudoscience.

People from wikipedia don't have PhDs in vector spaces

I support your decision.

People from Wikipedia are smarter than pseudo-scientist-loser.

Source?

... for you.

John Gabriel is not allowed to wear the title of professor.

Read the definition of the limits.

1 = 0,999... + 0,000... ...001

Oh so NOW we can count infinite zeroes?

it's an infinity between two known points

This n = 0.000...001 is false.

Suppose n = 0.000...001.
n + 2 ends in an odd digit.
[math]\sqrt{n+2} ends in an odd digit \implies \sqrt{2} ends in an even digit. [\math]
Clearly contradictory to math done over two thousand years ago and math done today.
n = 0.000...001 does not end in an odd digit.
n is undefined.

This is just one of many contradictory results which will always arise when inconsistent assumptions are taken.

This n = 0.000...001 is false.

Suppose n = 0.000...001.
n + 2 ends in an odd digit.
[math]\sqrt{n+2} ends in an odd digit \implies \sqrt{2} ends in an even digit. [/math]
Clearly contradictory to math done over two thousand years ago and math done today.
n = 0.000...001 does not end in an odd digit.
n is undefined.

This is just one of many contradictory results which will always arise when inconsistent assumptions are taken.

I don't get this logic t b h

Please enlighten me, what is the last digit of sqrt(2)?

Two numbers are NOT the same if there is SOME finite difference between them.
Please show this finite difference.
...
Can not be done, therefore they are the same number.

see

FINITE number. That is not a finite number.

finite = what has an end

It is odd and even whatever it happens to be. A better why of looking at it is to say it does not terminate.

No you didn't, there are still =/= signs where there should be = signs.

Saying that .9 repeating is not the same as 1 is as retarded as claiming that .5 and 1/2 are different numbers because "HUR DUR 5 IS NOT THA SAM AS 1 AND 2 HAHA DUR I SO SMRT!!!!1!ONE!!111!!"

.9 repeating and 1 are different representations of the same number. There is no reasonable argument to the contrary. You are wrong. Get over it. Accept that math is not your calling in life.

Neither do you if you think 1 =/= .999...

finite does NOT equal what has an end!
one third is a finite number with an infinite number of digits needed to represent it.

Wrong. The ellipses at the end of the 9 means that there are an INFINITE number of nines. That 0.000...001 retardation means that there are a FINITE number of zeros between the decimal point and 1.
You fail, go back to whatever shithole in Europe you came from.

That's not what was fucking written. It's not infinite just because you're dumb enough to think it is.

what's the smallest possible number

0.9+0.09+... denotes a LIMIT
It's MEANINGLESS otherwise. You can't add an infinite infinite amount of numbers.
This means, that by definition 0.9+0.09+...= limit(partial sums) = 1

>You can't add an infinite infinite amount of numbers.
wrong

Also, let me add into it.
You can not only add an infinite amount of numbers into this, but as you are only adding nines into it, you know for sure that the end of it is 9
So, just imagine you are "fast-forwarding", warping, into the end of this number, you can only see nines.
so 0,999... 999.

if we just swapped to using base .999— instead of base 10 we wouldn’t have this issue

Again, you can't know for sure it is a 9. That is still an assumption.

Suppose:
[math] n=\lim_{x\to\infty} \dfrac{10^x-1}{10^x}=0.99...9\\
m=\lim_{x\to\infty} \dfrac{10^x-2}{10^x}=0.99...8\\
s=\lim_{x\to\infty} \dfrac{n-m}{10^x}=0.00...1\\
[/math]
s is undefined because it defines the terminating digit of an irrational number as even or odd. m or n(or both) is an inconsistent definition.

Correction:
s = n - m = 0.00...1
(Although in either instance s is the same)

your supposition is exactly what I'm saying it is
is it that impossible? why can't it ever be like this?

My idea is that it is inconsistent or contradictory. Not sure if that is what you were saying or not.

I'm saying the very basic axioms are wrong on this and 0,000...1 is a real number. Also 0,999...9 is a number.

this

fuck all you stupids

.9 repeating = 1

.000...001 doesn't fucking exist, it isn't even well defined in the hyper reals

.3 repeating = 1

and yes, = equals, not some gay fucking arrow

If you chose to define the number that way, then we have to say
0.333...3 = 1/3
3(1/3) = 1 = 0.999...9
1-0.999...9 = 0.000...1 = 0.000...0

Technically, under this definition having n+0.000...0 OR n+0.000...1 is the next largest number after n. The additive identity element, "0", becomes very muddy in this choice.

but it's not 0,333...3
it's something bigger than that, but smaller than an ending 4.
it can't be represented like that, but it can by saying 1/3

You would be defining a different set of numbers using different axioms.

a different set than the current in use or the one we are talking? I don't see why it can't be like that, it's just that some numbers cannot be represented, 0,999... not being one of them

there don't exist Real numbers that differ at an infinitely far away decimal. All numbers can be represented since the Real numbers are complete

en.wikipedia.org/wiki/Completeness_of_the_real_numbers

yeah, I'm believing Wolfram|Alpha over a random Veeky Forums shitposter
wolframalpha.com/input/?i=(0.333...) -1/3

okay, faggots, if you think you are so smart, and think those are two different real number, there must be numbers between them
give me a decimal representation of any number you want between 0.9999.... and 1.000...
1.999....5 is not a valid candidate because you can't do that in real numbers, that is, say there's an infinite sequence of numbers which somehow ends in a different number
0.9999..... is not 0.999....9, because the sequence doesn't end
fuck off

Are you ready to be wrong?

1-epsilon equals .9 repeating

Therefore 1 cant be .9 repeating or epsilon is 0. It isnt.

Reals are Archimedean, so there are no infinitely small not infinitely big numbers, that is numbers smaller or bigger than any other real number

>false

limits are not defined as "attainable"

is 0,000... = 0?

0,000... ...001 = 0

Maybe you should look up "Bilinear forms"

Y'all fucking brainlets. Clearly OP is working with the non-Hausdorff space [math]\mathbb{R}\cup\{\overline{1}\}[/math].

you mean [math]\mathbb{R}\cup\{\overline{.999...}\}[/math]?

youtube.com/watch?v=2TCDiK7GpNM
geometric series -- best video on youtube ever (apart from the on e.) I'm not even kidding.

Convergence absolutely does not have the same value or result as equation and anyone who writes an infinite sum is "equal" to something should be shot in the back of the head when they least expect it.

Also

The DEFINITION of equality is the convergence you show...

>Convergence absolutely does not have the same value or result as equation
Convergence is the DEFINITION of EQUALITY.

No.

Uneducated retard.
Go learn something, you clearly have no clue what you are talking about.

Have a look when two real numbers are equal.

SPOILER: Since a real number is just the equivalence class of cauchy sequences of rational numbers, they are equal iff they are in the same cauchy series, by definitions that is iff their difference converges to zero.

>PhDs in vector space
That's a good one lul

he's bas(is)ed

You are legitimately retarded if you couldn't follow the math presented in

Fun fact: Infinity is never used anywhere in calculus
[math] \forall \epsilon>0,\ \exists M>0:n>M \implies\ |(\sum^n_{k=1}\frac{9}{10^k})-1|

Epsilon is 0, 1 - .(9) = 0.(0) =0

that filename lmao

moron
en.wikipedia.org/wiki/Laplace_transform#Formal_definition

We were told that the = regarding a limit has its own definition. The real value it approaches OR a improper limit (uneigentlicher Grenzwert), I guess undefined could be the result too, if there is a 1/0 or something in there?

The line about 0.9999... being unequal to 1 is kind of wrong, because the limit sum is basically included in that.

[eqn]F(s)=\int^\infty_0f(t)e^{-st}dt=\lim_{a \rightarrow \infty }\int^a_0f(t)e^{-st}dt \iff \\ \forall \epsilon>0,\ \exists M>0:a>M \implies \bigg |\int^a_0f(t)e^{-st}dt-F(s) \bigg |

1 = (9/10 + 1/10) = 0.9 + 10/100
= 0.9 + (9/100 + 1/100) = 0.99 + 10/1000
= 0.99 + (9/1000 + 1/1000) = 0.999 + 10/10000
= 0.999 + (9/10000 + 1/10000) = 0.9999 + 10/100000

and so on

each line = 1
exactly one, not approaching it.

line #1 is exactly 1
line #10 is exactly 1
line #98327498236483689 is exactly 1

At infinity, it still is exactly 1

>angry concave brainlet.png