Math

[Professor dickwrinkle No.11041691]

Hello students, I bet you can’t prove me wrong when I say: 0.999... isn’t equal to 1.

Pro tip, don’t try because you will fail and look stupid compared to enormous intellect.

[Anonymous No.11041702]

We all know they are not the same. Stop looking for attention.

[Professor dickwrinkle No.11041720]

>>11041702
Ok sorry, plz forgive me

[Anonymous No.11041727]

>>11041691
>>11041702
Retards. Not even going to bother explaining why you're wrong. Just Google it.

[Anonymous No.11041733]

>>11041727
Please go sit in a real analysis class before commenting on this post. Let me guess, you are referring to the 10x - x proof? Wow dude what a mathematician you are. You can recite other peoples shit basic flawed proofs and say you know math because you know no better. Clap clap brainlet.

[Professor dickwrinkle No.11041860]

>>11041727
I can sense your knowledge in math is based on YouTube videos from numberphile and such.

[Anonymous No.11041862]

>>11041733
Please finish high school before posting on this board.

[Anonymous No.11041865]

>>11041691
it is and everyone who thinks otherwise apparently doesn't know what a real number is

[Anonymous No.11041871]

>>11041691
you must be 18 to post here

[Anonymous No.11041876]

.

[Anonymous No.11041886]

>>11041876
lol do you actually think 1/3 exactly equals 0.333...? Why do you think it goes one forever

[Anonymous No.11041892]

>>11041886
>do you actually think 1/3 exactly equals 0.333...?
Demonstrate that it doesn't, using math, or shut the fuck up you dim witted cunt.
"Pretending" to be retarded is no substitute for having a personality.

[Anonymous No.11041895]

>>11041691
x = 0.999...
10*x=10*0.999...
10*x = 9.999...
x = 0.999... (as we stated earlier)
10*x-x = 9.999...-0.999... (take the difference)
9*x = 9
x = 9 / 9
x = 1

If x equals both 0.999... and 1, the transitive property asserts that 1 must equal 0.999...

[Anonymous No.11041896]

>>11041691
0.999...=1-(1/10^n)
10*0.999...=10*(1-(1/10^n))=10-10/10^n=10-1/10^n-1
If its true that 0.999... equals 1 then 9+0.999... should be equal to 10*0.999...

but 9+1-1/10^n=10-1/10^n and is clearly diffrent from 10-1/10^n-1

So that's that and now we don't need this thread in our lives anymore.

[Anonymous No.11041897]

>>11041892
I answered your question with "math" as you so elegantly put it >>11041895

[Anonymous No.11041899]

>>11041895
wrong, check this answer:
>>11041896

[Anonymous No.11041900]

>>11041896
Exactly my point, you just did it a little differently. Its basic Algebra, I dont understand why people dont see this.

[Anonymous No.11041902]

>>11041899
were both correct you just did it differently

[Anonymous No.11041907]

>>11041892
I guess you whant me to show you wrong because you can't show me wrong.

I'll try and hope you will understand.

The reason that there is infinite amounts of threes is because no amount of numbers can express 1/3 in base ten. the closest we get is 0.333...
because 0.333...<1/3<0.333...4
The problem is that the number is between 0.333... and 0.333...4

[Anonymous No.11041908]

>>11041899
your answer is simply too abstract and does not hold with concrete numbers.

[Anonymous No.11041910]

>>11041900
no im saying you are wrong.
i mean 0.999... is 1-1/10^n
so 9*(1-1/10^n)=9-9/10^n
clearly not 9 which could be written 9-0

[Anonymous No.11041914]

>>11041899
1-0.999...=?

1.000000000000...
.999999999999...
-_________________
= .000000000000...

0.000...1, but youll never reach the point of one because 0 is ever repeating... so:

0.000... still holds, thus the difference between point nine repeating and one is zero, meaning there is absolutely no differentiation between the two values, making them equal.

[Anonymous No.11041916]

>>11041914
Well you won't reach it if you tried to write it out in real life yes. but that doesnt mean the number doesn't exist.

You can also write it as: 1-(1-1/10^n)=1-1+1/10^n=1/10^n n=how big of a number you want.

[Anonymous No.11041919]

>>11041910
Next proof of no difference: any given two distinct real numbers will always stand to have infinite values between them, so on a number line of reals, there is never a logical next number.

between 24 and 25... could be 24.5, or 24.588

I challenge you to name one real sequentially after 0.999.. REPEATING and the value 1. The answer is you cannot, thus if we are to call them real numbers, and they do not follow this rule, then they must be the same number.

[Anonymous No.11041924]

>>11041919
0.000001

(just kidding)

Lol that number doesn't exist because 0.999... infinitly close, but that doesn't mean that they are the same.

[Anonymous No.11041926]

>>11041691
If I have an infinite number of dice and roll all of them an infinite number of times, will there exist a die that rolled a 6 every single time?

[Anonymous No.11041927]

>>11041907
>The problem is that the number is between 0.333... and 0.333...4
0.333...4 isn't a real number. You can't have an infinite number of threes followed by a four: nothing can come after the infinite string of 3s precisely because the string is infinite. Another way of saying this is that the sequences 0.33, 0.333, 0.3333 ... and 0.34, 0.334, 0.3334... represent that same real number.
But of course, you already know this.

[Anonymous No.11041928]

>>11041910
0.999... is 1-1/10^n
This is a false assumption.

[Anonymous No.11041929]

>>11041924
Actually it does. Unless you are dealing with hyper real numbers, which its theory suggests there is a finite sequential next, however that far evades the scope of all things modern math especially since computers, structures, human minds, animals, and all things natural and living are dealt with using real numbers. You cannot deflect three valid proofs by saying 'muh abstract maths'

[Anonymous No.11041930]

>>11041929
no it doesn't because if we where only talking about positive numbers and negative in a number line, you realise we can't say 1/4=0 just because we can't express rational numbers in this number line

[Anonymous No.11041933]

>>11041886
It's not going anywhere retarded. It's a number, it doesn't have legs, it can't move.
It's easy to see it must be an infinite expansion. Just do the division algorithm:
10 (3
-9 0.3
10
-9 0.33
10

Notice that now it's just back to 10/3 and so it's just going to repeat

[Anonymous No.11041934]

>>11041930
But rational numbers and positive whole numbers still fall within the same scope of general rules even if they themselves follow individualized restrictions. Hyper real numbers which allow you to disprove my 3rd proof (my other two still stand dumb ass) are within a completely different scope to the point of basic linear combinations no longer being valid.

[Anonymous No.11041935]

>>11041927
Maybe it is, maybe it isn't but you 0.333... is missing +1/(3*10^n) for it to actually equal 1/3.

You didn't object to 0.333...<1/3 so i guess you understood that part.

[Anonymous No.11041936]

>>11041930
its not a different group of numbers as your elementary example showcases, but a different numerical system all together, thus comparing them is tardish.

[Anonymous No.11041937]

>>11041933
Oh sorry maybe you don't understand english, The infinite string of 3's goes on forever.

[Anonymous No.11041939]

>>11041934
You missed everything I said, not all rational numbers exist in a positive and negative number line so they are not possible to express without ending up with things like 1/4=0.

[Anonymous No.11041942]

>>11041896
>10-10/10^n=10-1/10^n-1
wut

[Anonymous No.11041946]

>>11041939
Alright, you're correct. THAT proof no longer stands. My other ones are still valid, however. 0.999...=1

[Anonymous No.11041947]

>>11041942
Example:10/100=1/10
10-10/10^2=
=10-10/(10*10)=10-1/10^1

[Anonymous No.11041951]

>>11041946
Which one?

[Anonymous No.11041959]

>>11041951

>>11041914 and >>11041895 are both still valid.
basic algebraic combinations as well as the fact that the differentiation between these two reals is 0

[Anonymous No.11041968]

>>11041959
This one: >>11041895
Is in valid as shown here >>11041896

[Anonymous No.11041979]

>>11041968
Im done, people already disproved your proof.

[Anonymous No.11041980]

>>11041895
>10*x=10*0.999...
Where is the ending zero? When you multiply anything by 10, the product ends in a zero.

[Anonymous No.11041988]

>>11041959
This one: >>11041914
Is invalid because there is a zero and it might sound weird but 0.000...1 thats is what it is.
1/10^n

[Anonymous No.11041994]

>>11041980
I can't perfectly explain it since i'm not a real mathematician believe it or not, but imagine it like this:
0.999...99*10=
9.999...90

[Anonymous No.11041995]

>>11041914
Prove that 0.999... is a valid number and that you can perform arithmetic with it. You’re just making assumptions.

[Anonymous No.11041996]

>>11041994
Then that would ruin the proof

[Anonymous No.11042001]

>>11041914
Subtraction begins from the right of the numbers, not the left.

[Anonymous No.11042002]

>>11041994
Where would the 0 come from?

[Anonymous No.11042004]

>>11041924
Two numbers are by definition equal if their difference is zero. The difference between these numbers does exist, and it is zero, hence they are the same. To see that such a number exists, just look at the limit of the sequence 0.1, 0.01, 0.001, etc. and observe that this converges to 0.

[Anonymous No.11042005]

>>11041691
>what is a geometric series?
You're a retard and shouldn't be allowed to reproduce.

[Anonymous No.11042006]

>>11042002
from my anus

[Anonymous No.11042008]

>>11041935
>You didn't object to 0.333...<1/3 so i guess you understood that part.
I DO object to that. The difference is a sequence that is equal to zero.
>+1/(3*10^n)
That is equal to zero in the n -> infty limit.

[Anonymous No.11042012]

>>11042004
You didn’t properly subtract those numbers those. You subtract by taking the rightmost values and subtracting those first. But there is no rightmost value in 0.999... so the subtraction can’t take place

[Professor dickwrinkle No.11042020]

>>11042005
Nigga this isn’t even a fucking geometric aerie you ducking retard

[Anonymous No.11042028]

>>11041947
I mean, what you're saying makes sense, but why when I preform the algebra, I get
10 - 10/(10^n) = 10 - 10^(1 - n)?

[Anonymous No.11042051]

>>11041995
[math] 0.999...= \sum_{i=1}^{ \infty } \frac{9}{ 10^{i} } [/math]. Let [math] S_{n} = \lim_{ n \to \infty } \sum_{i=1}^{n} \frac{9}{ 10^{i} } [/math]. If [math] \epsilon > 0 [/math], then there is some [math] k \in \mathbb{N} [/math] such that [math] k > \frac{9}{ \epsilon } [/math]. Thus, we have [math] \frac{9}{ 10^k } < \epsilon [/math]. It's obvious that [math] \frac{9}{ 10^{k} } < \frac{9}{ 10^{k+1} } [/math]. So [math] S_{k+1} - S_{k} = \frac{9}{ 10^{k+1} } < \frac{9}{ 10^{k} } < \epsilon [/math]. Therefore, the sequences is a Cauchy sequence and 0.999... is a real number.

(hoping I didn't fuck up the latex)

[Anonymous No.11042054]

>>11042051
Well, I didn't fuck the latex but I did fuck the proof [math] S_{n} = \sum_{i=1}^{n} \frac{9}{ 10^{i} } [/math]. There, all is well now.

[Professor dickwrinkle No.11042061]

>>11042028
I don’t know what mistake you made but I as I see it you get rid of the division which shouldn’t happen.
Here’s the way I do it and we can compare them:

10/(10^n)=10/(10*10*10*10...(for n amount of times))
Divide both upper and the lower part by ten so you keep the same value of it
(10/10)/(10*10*10*.../10)=1/(10^n-1)
Since we don’t know what n exactly is all we can write is n-1

[Anonymous No.11042066]

>>11042061
>I don’t know what mistake you made
>

[Professor dickwrinkle No.11042068]

>>11042008
Let’s say 0.000...1 is 0
Then 1/9999...=0
That means 0*9999...=1
And 1/0=9999...

[Professor dickwrinkle No.11042072]

>>11042066
Oh they are the same thing, if we replace n with 2 we can see that the results are the same. Try 10-10^1-2
And then 10-1/(10^2-1)

[Anonymous No.11042076]

>>11042072
Imagine having no education at all but thinking you can argue about math

[Anonymous No.11042080]

>>11042072
I am just wondering how they are necessarily the same if one can be proven algebraically and the other can only be shown by using an example.

[Professor dickwrinkle No.11042098]

>>11042080
The computer you use is just making the awnser shorter and less complicated.

[Professor dickwrinkle No.11042103]

>>11042076
Thanks

[Anonymous No.11042116]

>>11041896
Ignoring the fact that you probably fucked up the algebra and your proof is unintelligible because of a lack of latex or at least parenthesis. You're assuming your result to be true from the very first line and the shuffling the terms around with algebra. This is not how mathematical proofs work. You have to first prove 0.999...=1-(1/10^n) (Hint: It does not for any value of n).

[Professor dickwrinkle No.11042125]

>>11042116
If n is infinite it works so gotcha.
Btw maybe i can’t write the shit perfectly I mean I’m not a real mathematician as I said. But it stands and if you want I can show a video telling it more “correct”.

[Anonymous No.11042133]

>>11042116
>You have to first prove 0.999...=1-(1/10^n) (Hint: It does not
Correct. That's because 9/10+9/100+9/1000+...=1

[Anonymous No.11042135]

>>11042125
If n is infinite (or rather, tends to infinity), then 1/10^n=0 and you have proved 1=0.999...

[Professor dickwrinkle No.11042137]

>>11042133
Wrong because the nines will always be added after each other and they never change each other’s value. So the very first nine. 0.9 will never change by definition of the number.

[Professor dickwrinkle No.11042142]

>>11042135
Well it’s obviously not zero because then you are implying you can divide by zero.

[Anonymous No.11042144]

>>11042137
Nigger, I already BTFO you in the other thread. You don't know what a power series is. You don't know what a geometric series is. You don't know what a limit is. What ground do you stand on?
The limit of the the sequence of partial sums (which is how infinit sums are fucking defined) of 9/10+9/100+9/1000+... is exactly equal to 1.

[Anonymous No.11042147]

>>11042142
What in the fuck, no you arent.

[Anonymous No.11042152]

>>11042125
But that's a limit. lim(10^(-n)) with n approaching infinite equals 0.
>I'm not a real mathmatician.
LaTeX is not only for real mathmaticians by any stretch. But nevermind that, you cannot be expected to learn it just to post here. Plugging it into a math program of a free Office (OpenOffice / LibreOffice) package requires you to only understand the templates and screenshot it and it would work just as swell.

[Anonymous No.11042153]

>>11041860
My knowledge in maths is based on studying maths in Cambridge.
You're retarded.

[Anonymous No.11042154]

>>11042020
>this isnt a geometric series
Yes it is

[Professor dickwrinkle No.11042156]

>>11042147
Well yes if 1/10^infinite=0 then 1/0=10^infinite

[Anonymous No.11042159]

>>11042156
Good thing we are never dividing by zero, and merely subtracting 0.

[Anonymous No.11042164]

>>11041691
[math]0.999\dots = 0.9 + 0.09 + 0.009 +\dots = \sum_{k=1}^{\infty}\tfrac{9}{10^k} = \lim_{n \to \infty}\sum_{k=1}^n\tfrac{9}{10^k} = \lim_{n\to\infty}9\tfrac{\tfrac{1}{10}-\tfrac{1}{10^{n+1}}}{1-\tfrac{1}{10}} =9\tfrac{\tfrac{1}{10}}{\tfrac{9}{10}} = 1[/math]
>inb4 limit not real lol

[Anonymous No.11042169]

>>11041947
But they're equal as n -> infty

[Anonymous No.11042172]

>>11042051
good attempt, but seems like you don't actually understand the cauchy criterion and there are some typos in it as well.
>>11041691
>>11041702
>>11041733
>>11041860
>>11041907
>>11041928
>>11042061
>>11042137
Read and understand this proof to get smart:
By the Cauchy sequence definition of real numbers, a real number is an equivalence class of rational sequences a:N->Q such, whose n'th member is a_n, with the equivalence relation being a~b <=> |a_n-b_n|-> 0 as n-> infinity. This says that for every rational x>0, there is a natural number N such that for all n>N, |a_n - b_n|<x. Now that this is established, we need to define decimal expansions.
For an infinitely long decimal expression, we define the real number that it represents by taking a sequence of increasing finite truncations of the decimal expression. In particular, we write 0.9999... to denote the real number represented by the sequence (0, 0.9, 0.99, 0.999, and so on). We inject rational numbers into reals by letting the real number q' by (q,q,q,q,....) for a rational number q. Then, in order to prove that the real numbers 1 and 0.9999... are equal, it suffices to show that they lie in the same equivalence class (because gain, real numbers are equivalence classes, not elements in those classes themselves). The n'th term in the canonical sequence for 0.999... is 1-(10^-n). The n'th term in the sequence for 1 is 1. Hence it suffices to prove that |1-(10^-n) - 1 | = 10^(-n) -> 0 as n->infinity. Looking back at the definition of a limit, let x>0 be any positive rational number. We want to find an N such that for all n>N, 10^-n < x, or 10^n > 1/x. But that is simple, just take any N such that 10^N > 1/x. Hence, the limit is proved and 0.9999... = 1
Faggots.

[Anonymous No.11042173]

>>11042156
Using inifite in arithmetic does lead to some stupid shit. That's why we have limits and [math] \lim_{n \to \infty } \frac{1}{10^{n} } = 0 [/math], which implies 1=0.999....

[Anonymous No.11042177]

How can OP ever recover

[Anonymous No.11042207]

man this thread would be so much bettter if OP understood what a series, or even a limit was.

[Anonymous No.11042208]

>>11042172
I wasn't trying to prove 0.999...=1. Just that 0.999... is a real number and that's what was done. And, since the numbers of terms between x_j and x_i are finite, then the statements [math] \forall \epsilon > 0 ( \exists n \in \mathbb{N} ( \forall i,j \geq n ( | x_{j} - x_{i} | < \epsilon ))) [/math] and [math] \forall \epsilon > 0 ( \exists n \in \mathbb{N} ( \forall k \geq n ( x_{k} - x_{k+1} < \epsilon ))) [/math] are equivalent when we assume [math] \{ x_{n} \} [/math] is an strictly increasing always positive sequence. And yes, there's probably plenty of typos because I'm phone-posting.

[Anonymous No.11042212]

>>11042208
FOR FUCK'S SAKE. [math] x_{k} - x_{k+1} [/math] should be [math] x_{k+1} - x_{k} [/math] but you get what I'm saying.

[Anonymous No.11042213]

>>11042012
>But there is no rightmost value in 0.999... so the subtraction can’t take place
Look up "subtraction of real numbers".

[Anonymous No.11042214]

>>11042208
Don't mean to sound rude, but this is actually false : "And, since the numbers of terms between x_j and x_i are finite, then the statements <...> are equivalent if we assume x_n is strictly increasing and positive." Take for example the sequence of finite harmonic sums 1 + 1/2 +1/3 +.... This is clearly not a cauchy sequence. It's strictly increasing and always positive. Yet, it satisfies your second definition. Hence your definition is much weaker.
Your heart was in the right place though. Peace ;)

[Anonymous No.11042232]

https://www.youtube.com/watch?v=--HdatJwbQY
https://www.youtube.com/watch?v=xSlS2xE8rH8

All you guys BTFO'd GO CRY TO YOUR BITCH MOM NIGGER LOVING KEK, PLZ SUCK MY DICK AHAHAHHAHA

[Anonymous No.11042238]

the problem is that many people think of limits as something which is obtained after actually performing infinitely many steps, a result of an infinite process etc. limit is just a number with a certain property: the sequence gets arbitrarily close to this number. not infinitely close, not after infinite number of steps, but arbitrarily close. pick any precision and the sequence eventually (after finite number of steps) gets there. people are right that "the sequence approaches something but never actually reaches it", but point is that the symbol 0.999.. etc. does not denote the sequence, it denotes the actual limit. if you reject 0.999... = 1, then you're rejecting that 0.999.. is a number in the first place and the discussion stops being meaningful.

[Anonymous No.11042240]

>>11042172
I understand that my post was probably the first time a lot of people here saw what real numbers actually were. I will take this opportunity to teach you some more. We will continue with the Cauchy definition of real numbers. How do we define the field operations?
Simple. From now on, we will refer to real numbers by arbitrary elements in their equivalence classes. This means you have to prove that the definitions are well defined independent of the choice of element in the equivalence class, but that can be done without too much effort.
The most basic operation is addition. We define (a_n)+(b_n) be (a_n + b_n) (essentially sum each entry in the sequence. so (1,1,1,1,1,1,...) + (1, 1/2, 1/4, 1/8,...) will be (1, 1.5, 1.25, ....). We need to prove that this defines a unique real number independent of choice of sequences (a_n) and (b_n) in their respective equivalence classes. This is easy: assume (a_n) ~ (a'_n) and (b_n) ~ (b'_n). We need to show that (a_n + b_n)~(a'_n + b'_n), or equivalently that |a_n + b_n - a'_n + b'_n | -> 0 as n-> infinity. But |a_n + b_n - a'_n + b'_n | = | (a_n - a'_n) + (b_n - b'_n)| <= |a_n - a'_n| + |b_n - b'_n| (here we used the triangle inequality |a+b| <= |a| + |b|). Hence since |a_n - a'_n| tends to 0, as does |b_n - b'_n| , their sum does too, hence (a_n + b_n)~(a'_n + b'_n) and addition of real numbers is well defined. This is actually not the complete proof yet: we need to establish that addition actually defines a real number by showing that (a_n + b_n) is also a Cauchy sequence, but that proof is similar and left as an exercise to the reader :)
Now that you've seen how addition is defined, try and yourself define: addition, multiplication and division. Not just tell us how to do it, but show that your definition actually works, as I did.

[Anonymous No.11042250]

>>11042240
>>11042172
if after reading these posts, you still think that 0.999... != 1, consider hanging out with noose-chan

[Anonymous No.11042252]

>>11042250
FUCK YOU MADA FOCKI SUCKI BTH TIETS FACKU

[Anonymous No.11042255]

>>11042252
Sorry, I don't speak retard.

[Anonymous No.11042276]

>>11042255
HEY TKE UT BICK YOU MOMA IS A BITH FUCK YOU IL DUUUUUHHH (druwling like alut) AHAHHAHAA

[Anonymous No.11042292]

The supremum of the sequence of {0.9, 0.99, 0.999, ...} is 1

[Anonymous No.11042346]

>>11042214
>Take for example the sequence of finite harmonic sums 1 + 1/2 +1/3 +....
Well, you are correct. Thanks, anon.

[Anonymous No.11042617]

>>11042207
but OP is a mongoloid Indian faggot who thinks he understands math because he understands the definition of an axiom, and therefor refuses to accept them.