/sci/ - Science & Math

>>15561606
>getting filtered by calc 2
pleb

>>15561606
1st is trivial.
2nd is necessary to exactly solve a 3rd degree equation, even when all three solutions are real.
https://www.google.com/search?hl=en&q=Y%3Dx%C2%B3%2D5x%2D3

>>15561606
1 = 3/3
0.3333... = 1/3
0.3333... x 3 = 0.9999.... = 1/3 x 3 = 1

A third is 0.3344444444 you faggots.

>>15561766
Tfw decimals are stupid concepts invented by the big ol Jew and 2023 noone noticed.

>>15561606
1 = 0,1 + 0,9
1 = 0,01 + 0,99
Etc.
1 = 0,000...0001 + 0,999...9999

So saying that 0,9999... = 1 is the same as saying that 0,000...0001 = 0

1/3 s 0.33333... repeating to the infinity. We write 0.3 with a line on top.

Here another demonstration of why 0.99999...=1 :

Let's a=0.99999...
10a=9.999999
10a=9+a
9a=9
a=1
0.99999...=1

[math] \displaystyle
0. \bar{0}1
= \lim_{n \to \infty} 0. \underbrace{0 \dots 0}_{n ~ \text{times}}1
= \lim_{n \to \infty}
\left [
\left (
\sum_{k=1}^n \dfrac{0}{10^k}
\right )
+ \dfrac{1}{10^{n+1}}
\right ]
=0
[/math]

>>15561771
Almost true, but it's not saying
0.000 00... ...00 001= 0,
It go to infinity so it's saying
0.000 000 000 000 000 000 00...=0

>>15561641
So 0.6, ..., 0.7, ..., 0.8 , ... ,1?

>>15561905
You're embarrassing yourself, stay in school kid.

>>15561606
0.999... = i

>>15561942
i^2 =/= 0.999...

1+2+3+4+... = i^2/12

>>15561641
I'm learning this right now. Preddy cool

>>15561606
>first
learn long division and then calculate me what 1/3 is both in base ten and in base twelve
>second
lmao, literally filtered by beyblades and nomenclature
have as good a life as you can get, mein goober

>>15561771
0.999...≠0.999...999, the first doesn't end and the second does

>>15562016
and i^4=0.999...=1

>>15562046
i mean in a sense due to ramanujan, cesaro or abel summation , but next time try to bracket things properly, that could be interpreted as i^(2/12) when the proper one is (i^2)/12

>>15562046

>>15561771
Fine me a number between 0.999... and 1 and we'll talk.

>>15562101
0.999...1

>>15562066
It makes sense due to the prolongation of the Riemann's fonction zeta for negative numbers. Indeed ζ(-1)=1+2+3+4+5+6+... with the first formula but -1/12 with the prolongation

>>15562066
> that could be interpreted as i^(2/12)
Operator precedence says no.

>>15561759
>0.3333... = 1/3
wrong

>>15562664
1/3=0.33333...

>>15562690
no its not
1/3 ⪰ 0,333...
2/3 ⪰ 0,666...
3/3 = 1

Its really weird to see ⪰, ⪯ and even stuff like ± to be used in equations while you have no idea what they do
Even saw some maps with ⪰ and ⪯ used while there was no prior calculations involved kek

>>15562118
So in other words, 1+2+3+... != -1/12.
Show me a system of arithmetic where the partial sums of 1+2+3+... approach -1/12, and then we'll talk.

>>15561606
>square root (-0.9999...) = ?

>>15562699
If 1/3 is greater than 0.3333... then there should be a positive number d that satisfies 1/3 - 0.3333... = d
Show me that number.

>>15563124
You can show me instead since you are the one who says that there should be a number like that

>>15563124
Also
>If 1/3 is greater than 0,333..
Failing to understand the difference between > and ⪰

>>15561606
Mathematics, like most things in life, need to be pragmatic to be able to work with it. It's like having a wife, which I obviously don't have.

>>15561606
i prefer the notation i learnt at school, it makes things much easier
0,(9) = 1
0,999...=/=1

>>15561606
at least you know the proper definition of the complex unit

>>15563301
>at least you know the proper definition of the complex unit

OP did not post the proper definition of the complex unit.

What OP posted was an equation that has two solutions: i and -i. OP's equation doesn't define i, it simply shows a property that both i and -i have.

A more precise definition is: i = √(-1), where the √ symbol specifies only the positive square root.

>>15562664
1/3 = 3/10 + 1/30
= 0.3 + 1/30
= 0.33 + 1/300
= 0.333 + 1/3000
:
= 0.333... + 1/inf
= 0.333... + 0
= 0.333...

>>15563216
[math]
0.(9)=0.999...=0. \bar{9}=1
[/math]

>>15563388
>1/3 = 3/10 + 1/30
Because your calculator says so ?

>>15563413
>filtered by adding fractions
lol

>>15563420
3/10 + 1/30 = 10/30
Now why did you even start to complicate things when the definition was already LITERALLY 10 times easier

You are filtered atleast 10 steps earlier so shut the fuck up

>>15563438
>i have no argument

>>15563446
1/∞ = 0
not math

>>15561643
1 is wrong

>>15561759
2nd line is wrong

that repeating decimal NEVER equals 1/3

>>15561606
Retard mathlet here, if i^2 = -1 then does i^4 = 1?

>>15561606
0.9999... cannot go on forever, it has to become one at some point

>>15563471
by definition, inf is larger than any number.
1/inf=0

>>15563517
yes, [math] \text{i}^{4n}=1 [/math]

>>15563487
>>15563575
brainlets should not dabble in mathematics

>>15563591
>inf is larger than any number.
Infinity is not a number but operation because it's endlessly reoccurring, you cannot do operation on a operation because it's like dividing a plus sign

>>15563653
but sometimes operators can have operations done on them https://www.codecogs.com/library/maths/calculus/differential/the-d-operator.php

>>15563314
you have that literally completely backwards there chief.
using a square root as the definition introduces the ambiguity. using the square eliminates it.

>>15563653
>confusing the ramp with the mountain
lnf just is what it is, it's not a growing living thing or a diesel engine chugging along

>>15563653
ahem.... reeeeemann sphere

>>15562866
Yes, 1+2+3+4+... is divergent, going to plus infinity. I said that people sometimes say that it's -1/12 because when we use Riemann's zeta function, zeta(-1)=1+2+3+4+..., and when we use a prolongation:
zeta(-k)=((-1)^k)×B{k+1}/k+1 when B{k+1} is the k+1 th number of Bernoulli.

Compute for k=1, you got zeta(-1)=-1×(1/6)/2=-1/12

But also zeta(-1)=1+2+3+4+5+... with the original definition.

If we use partial sum it's divergent and it makes sense that keeping adding number bigger and bigher for infinity gives us infinity. But if you want to give a value (a number) at the sum of the natural integers, it will be -1/12 (by using Riemann's zeta function prolongation)

>>15563684
>but sometimes
Dont add more stuff, get this one thing right
If you divide 1 infinite times and you say that it equals some definite thing, lets call it 0 for now.
That means the operation had a stop
Which means it was not infinite

>>15563727
You have to understand what division means
1/1
=
"0.5, 0.5" (product 1) (you divided one once)

1/2
=
"0.5, 0.5"
"0.25, 0.25" (product 0.5) (you divided one twice)

1/3
⪰
"0.333.."
"0.333.."
"0.333.." (you divided one into thirds) (as close as possible not reaching it)

1/4
=
"0.5, 0.5"
"0.25, 0.25"
"0.125, 0.125" (product 0.25) (you divided one twice, twice)

etc.

Now lets do 1/0
=
"1"

>>15561906
Ok redditard genius, what comes before and after 0.9999.....

>>15563471
Wrong [math] 1/x [/math] is discontinuous and the limit at 0 from the left and the right aren't equal

>>15563820
>what comes before and after 0.9999.....
This sentence don't mean anything. Say me what come just before and just after pi, or any number...

If you didn't meant "just before", 0.5 come before 0.9999... and 1.5 come after

>>15563820
Ok redditard genius, what comes before and after 5?

>>15563847
your mom at my house

>>15563199
>Failing to understand the difference between > and ⪰
I am saying it is a strict = and not ⪰. While technically ⪰ can be used as well (the "greater" part will be trivial) the other anon insisted that = is incorrect to use while it definitely is correct until a number d is found that satisfies 1/3 - 0.3333... = d.

>>15563856
Thats how it works when you use greater/lesser equals
1/3 ⪰ 0,333..
1/3 - 0,333... ⪰ 0

Its a different system but with that you dont need to find a new number x

>>15563826
Well it's a fucking good thing that x isn't near zero then.
lrn2read

>>15561641
>calc
>posts grade school shit
huh?

>>15563124
epsilon greater than 0, chose epsilon small

Does 1.999... = 2?
What about √i?

>>15563999
>Does 1.999... = 2?
yes, due to continued fraction every number has two representations(infinite really if we count zeroes to the left like in 1=01=001=0001=... and so on)
>What about √i?
https://www.youtube.com/watch?v=Z49hXoN4KWg
algebraic closure is neat
https://en.wikipedia.org/wiki/Algebraic_closure

>>15563999
(1+i)/sqrt(2)

>>15563999
yes, 1.999...=2
for a a positive integer, a,9999...=a+0,99999...=a+1

for the square root of i, you are looking for numbers that squared gives i. There is 2 solutions, (√2)/2+(√2)i/2
and -(√2)/2-(√2)i/2
Here the explaination:
https://m.youtube.com/watch?v=Z49hXoN4KWg

>>15564010
I understood it. This guy is good

>>15563314
No, a more rigorose demonstration is exp(i*x)=cos(x)+i*sin(x)

>>15564094
Hmm... Its better than the other.
Indeed, we can't write "i=√(-1)"
(sqrt function is not define for negatives, we can get 1=-1 using this definition)

The original definition is i2=-1, that's simplier than Euler's formula and thats not less rigourous

>>15564140
i^2, my latex lagged

>>15563314
even if x^2=4 admit 2 solutions (-2 and 2), if I say 2^2=4 it's not false

>>15563915
I 0.333... always adds another 3 to the end and your epsilon won't be good anymore. There is no one single epsilon to beat 0.333...

>>15562062
>The first doesnt end.
>The second ends at infinity.

Yes, he said it bad. If that end that can't be at infinty. It's like if I say that world is eternal but will end...

>>15564686
He said it back but he is true
0.9999... =/= 0.999...999

>>15564709
>>15564712
https://www.britannica.com/science/infinity-mathematics
>infinity, the concept of something that is unlimited, endless, without bound.

Let's do some logical sophism :
"ends at infinity" = "ends at something that never ends" = "doesn't end"

Ha!

>>15561641
Infinity is a suggestion and therefore a flawed concept. Infinity cannot exist. It disrupts the definition of exist. These terms cannot be used in definition of each other. It is self insistent tautological reasoning. It unifies with no other concepts. It has no extension. No intension. It is a device for logical leaps outside of explicit statements.

>>15563754
here's your (You)

>>15564727
True

>>15563851
fucking kek'd

0^0=1
can you /sci/?

>>15564948
0

>>15564948
>>15564983
No

0.01^0.01 = 0.999079390...
0.0001^0.0001 = 0.999079390...
Etc.
0^0 = 0,9999...

>>15565127
Error :
0.01^0.01 = 0.9549926...

>>15565127
What makes you think you can use decimals to prove something that happens before decimals

0^0
Left is the starting product 0
Operation is ^0 which clearly states that you do nothing to your starting product
its 0

>>15565145
Because of this :
>>15561771
>0,000...0001 = 0

Also I'd correct it as :
0,000...0001 = 0+

>>15561771
>So saying that 0,9999... = 1 is the same as saying that 0,000...0001 = 0
not true. There will never be 1 in 0.(0) because those zeroes are infinite so it's literally 0.0000000000000000000 = 0

>>15565198
>Because of this :
>Didn't explain anything related to the question

>>15565145
He used the limit when x goes to 0 if the function x^x.
But we can also use the limit of 0^x when x goes to 0, we got 0.
And if we use the limit of x^0 when x goes to 0, we will got 1.

That's why 0^0 is indeterminate.

This debat is quit stupid, it's like if we debate to know if infinity/infinty is 1 or infinity or 0 or whatever

>>15564833
Anyone without knowledge of mathematics.

>>15563487
Ah ok, and why? Your God don't like it?

>>15565681
But im asking how can determine something with decimals that is happening before decimals

When you divide integers enough you reach this new system which is decimals, now you are at the new system and it doesnt work the same way as the old system but works differently like:
1/2 = 0,5
and
0,1/0,2 = 0,5

You cannot do anything with decimals when figuring 0^0 because you can see that the systems works differently and are not linear

>>15561606
Go away to another place where no use that. O wait, nothing.

>>15561606
0.999...=1 is not a concept, that can proof.
i^2=-1 is a concept i=(0,1) a point in R^2

>>15561766
0,3344444444 x 3 = 1,0033333332

1,0033333332 = 1

>>15565770
math] \displaystyle
\frac{1}{3} = 0.\overline{3}= 0.1_3
[/math]
[math] \displaystyle
3 \cdot 0.\overline{3} = 0.\overline{9} = 0.1_3 + 0.1_3 + 0.1_3 = 1_3 = 1
[/math]

>>15565770
[math] \displaystyle
\frac{1}{3} = 0.\overline{3}= 0.1_3
[/math]
[math] \displaystyle
3 \cdot 0.\overline{3} = 0.\overline{9} = 0.1_3 + 0.1_3 + 0.1_3 = 1_3 = 1
[/math]

>>15563199
1 ⪰ 1

>>15565850
>copemath
pathetic

>>15562062
Any nth decimal of the number of the left is equal the nth decimal of the number of the right. At no point are the numbers any different therefore they are equal.
Also 0.999...999 is a nonsensical notation. It doesn't mean anything.

>>15561606
I don't know why the premise of two different notations for the same number filters you retards so badly. It's as though your conception of a number is the symbolic representation itself and you can't separate the two. How do you even cope with the existence of roman numerals?

>>15563732
That's not what it means. Taking a limit of 1/x as x approaches infinity gives you the exact value of the limit (which is 0). Saying 1/inf is more of a thought shortcut but the limit has an actual specific value. The process didn't "stop". We just have the tools required to investigate what happens after an infinite number of "iterations".

>>15561771
>0,000...0001
> 0,999...9999
These aren't things.

>>15563886
>1/3 ⪰ 0,333..
1/3 = 0.333...

>>15565764
>you can see that the systems works differently and are not linear
Elaborate

>>15565942
1/2 has downwards pull to 0,5 and
0,1/0,2 has upwards pull to 0,5
Same quantities but the other one works within decimal system and other does not

You are figuring 0^0 which is in downwards pull and you are "proving" it with upwards pull when the systems are not linear to eachother
Which means you cannot prove it with that system because decimals comes after the thing you are proving

Remember that you only have decimals because you get it from 1/2 first so its a system thats after

>>15565955
You are aware how fractions work, right?

>>15565996
go on

>>15561606
>Rotating 90 degrees twice in the same direction is the same as rotating 180 degrees
How is this stupid?

>>15561606
Good thread, let's add some shit :

....99999999 + 1 = 1.......000000000000 but this last 1 would never appear, so :

....99999999 + 1 = ...0
....99999999 = -1
Then :
....11111111 x 9 = -1
....11111111 = -1/9

...1111111 x 9/12= -1/9 x 9/12
...11111111 x 3/4 = -1/12
...(10^3 x 3/4) + (10^2 x 3/4) + (10^1 x 3/4) + (10^0 x 3/4) = -1/12
...(75) + (7.5) + (0.75) = -1/12

...3333333333333.25 = -1/12

>>15566018
P-adic numbers ?

>>15566000
You want me to explain fractions to you? You know you have to be at least 18 to post here, right?

>>15566018
>Good thread
Bad thread but it's at least somewhat entertaining considering the usual quality of posts on /sci/ lately.
>....99999999
If you are using made up notation you should explain what it means first, otherwise it's meaningless to anyone but you.

>>15566081
>explain fractions
nah, more like explain your hallucinations

>>15566089
It's not my hallucinations that make you unable to understand elementary school concepts like fractions.

>>15566082
>If you are using made up notation you should explain what it means first, otherwise it's meaningless to anyone but you.


https://mathcircle.berkeley.edu/sites/default/files/handouts/2020/BMC_Adv_p-adic_numbers_notes.pdf


>>15566074
10-adic in this case.

>>15566098
your hallucinations are not elementary school concepts

>real numbers
>not actually real

>>15566154
And ?

>>15566154
> /sci/ people
> not actually sientists

>>15565762
go ahead and try to make them equal

no amount of additions will allow it

>>15566734
>no amount
no finite amount

>>15566734
>infinite sums are fake and gay
Don't worry, you can still live a fairly normal life with just elementary school education.

>>15565161
if it where a sum it'd be the empty product of a addition which is indeed 0, but it is not, its the empty product of multiplication which is 1

>>15565770
learn long division and the work out what 1/3 is in base ten and in base twelve, then ask yourself why you got what you got

>>15565924
i think to them its some weird stuff used in football and nothing else, given that they skipped the use of L by itself i wonder how many americans know that it's 50

>>15567476
I don't know man, roman numerals are all over american public buildings, monuments, etc. But I guess people don't pay attention to anything but their TV.

>>15566018
>...99999+1=...00000 so ...99999=-1
yeah, k-adics(k since we are using a non-prime as the base)
in fact and non-jokingly ...999.999...=0

>>15567472
0.999...=1 is not a concept anywhere. You know that if catch that representation 0.999... is a infinity serie.

>>15567482
ok, yeah your right, im actually being to harsh on them, its just these threads man, they warp your notion of what those that do not know math are like

>>15567467
sure

>>15567491
>You know that if catch that representation
what country are you from?, and my condolences on the lackluster mathematics curriculum you guys have over there

>>15567500
Hahaha you're pathetic. Go away a your university of shit, child.

>>15567500
>>15567502
Sorry, surely you go to a highschool.

>>15565924
As a guy who didn't get the .999... = 1 thing when I first learned it, it was because I didn't actually understand infinite series. I would see shit like [math]\sum_{n=1}^{\infty} \frac{1}{2^n} = 1[/math] and not even blink an eye, because it just seemed obvious that, at some point, you'd hit 1. So I basically substituted the idea of an infinite series with that of a large but finite series.
But the .999.... thing hit a wall because it's an infinite series which very obviously will never equal one after any finite number of terms.

If the difference between one number and the next one is infinitely small, that means the number is the same, no?

>>15567813
what is infinitely small? probably you meant "can be shown to be arbitrarily small". then yes, those are the same numbers

Haven't seen that thread in a while. I'm 99% convinced that it's bait every time it's posted.

>>15561606
Let [math]x = 0.\bar{9}[/math].
Multiplying both sides by ten, we have [math]10x = 9.\bar{9}[/math],
by which we can get the difference [math]9x = 9[/math].
And from this we clearly see that [math]x = 1[/math]

>>15561606
Not being able to divide by zero when it can be treated as multiplying by 0.

>>15567489
>in fact and non-jokingly ...999.999...=0
Nice.
Also let's jump the rabbit hole :

0 = e^(i.Pi) + 1 = ...999.999...

>>15568131
its nearly 100% bait.

This thread is so tiring to read. If you no absolutely nothing about math beyond school why even try discussing math? To embarrass yourself?

>>15569116
know*

>>15569116
Come on !
A thread with convergent series and k-adic numbers is theoremicaly fun.

>>15567842
the opposite of infinitely big

>>15568854
>Multiplying both sides by ten, we have 10x=9.9¯
>by which we can get the difference 9x=9.
How did you get the difference ?

>>15561606
L'hopitals rule which is just to avoid teaching taylor series.
lol. lmao even.

why are people itt writing fractions and exponents without latex

[math] 0.999... + 0.999... = 2 [/math]

>>15570132
Bro, people write 0.999... as
0.9999899,.......
in here

[math]\frac{1}{3} > 0.3 \\ \frac{1}{3} > 0.33 \\ \frac{1}{3} > 0.333 \\ \frac{1}{3} > 0.3333 \\ \frac{1}{3} > 0.33333 [/math]
at which real time moment of incremented additional 3's does the equation evaluate as equal to satisfy an assumption otherwise?

>>15570805
at the infinity

>>15570805
>let's pretend finite is infinite

>>15561771
yes.
0.00...01 can be mathematically proven to equal 0.
0.999... can be mathematically proven to be 1.

>>15569644
not him, but he got it through this little flat nigga over here "-" ,using it like so:
10x-x=9.9...-0.9...=9x=9

>>15570559
ye, 0.9...+0.9...=1.9...=2

>>15563851
9.999... /10 response

>>15570805
>>15571021

0.999... is infinite
1 is finite

That is a huge pretend alright

>>15571778
he didnt say 9.999.. though
he said [math]9.\bar{9}[/math]

Show the operation using that

>>15561906
1 = 0.99999999.... = 0.9899999999... = 0.97999999.... = 0?

>>15571974
[math]9.9...=9.\bar{9}[/math], its just another notation for repeating decimals, pic related, god do i pray that no children is left under your tutelage

>>15572022
Then use that notation to do the calculation

>>15571964
>0.999... is infinite
god i wish subwits like you never learned basic arithmetic

>>15572053
?
tell me what "..." means then other than infinite

>>15572087
>>15572022

>>15572128
>ignores the question
many such cases

>>15563314
Neither your definition nor OPs definition are capable of differentiating i and -i: the complex numbers cannot be ordered so the word "positive" is essentially meaningless in this context. If you want to make this differentiation, you should do what Ruden does and define the complex numbers as ordered pairs of reals (a,b) with multiplication given by (a,b)*(c,d) = (ac-bd, ad+bc), and declare that you're going to write (a,b) as a +bi. But this is such overkill, I don't even think Alfhores does it.

>>15563999
[math] \sqrt{i} = i^{1/2} = e^{\log(i)*\frac{1}{2}} = e^{\frac{1}{2} (\pi i /2 + 2n\pi i)} = e^{(\pi / 4 + n\pi) i} = \pm \frac{1}{\sqrt{2}}(1+i) [/math] (where the sign of the result depends on the parity of n).
Complex exponentials aren't the most intuitive at first but they're really cool.

>>15562070
How exactly do people just “see” equations like this? Are they so smart that they’re able to connect dots that no other person would ever even comprehend?

>>15561606
The top one is pseudo-math because imagine if you have a thimble that is divided by 10 in iterative processes, you will have a thimble at the end result. This resulting thimble is smaller than any real number and thus is an example of a non-Archimedean number. Now a sperg will tell you this thimble is not any real-number-distance from 0 and thus equals 0. The first claim is correct, the second however is an abstractive convention that is inferrential but not deductive from the first point. I am happy to ask questions on this.
Think Bill Clinton here (pic related).

>>15561606
Axiom of choice. It is not the issue with modern mathematics, rather the paradox is how we define infinity wrong. We need different distributions of sets that behave differently (we do this partially with 'measurable sets' but we need to go further set-theoretically). I believe so much as picking a single element from an infinite set is awry and should be a classification that we inhibit unless otherwise permitted. The paradoxes arise from essentially an infinite picking of infinite elements and this is all stemming from that we can pick elements at all not that we are doing so infinitely (though this is a good showcasing of why picking elements from infinite sets can be dangerous logically).
The axiom of determinacy is not so much stupid since it exemplifies how different infinite sets can behave differently.

Use a base 3 numeral system.
1/3 in base 3 is 1/10 = 0.1

>>15572025
118 KB PNG

>>15571974#
9.9...=9.9¯" role="presentation" style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14.6667px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">Let's a=0.9¯
10a=9.9¯
10a=9+a
9a=9
a=1
0.9¯=1

>>15572154
"..." in 0.3333... mean a repetition of a character an infinite times, that don't mean that the number is infinite but that its decimal writing is not finished. In 0.999... we can't say that the decimal writing is not finished, because we can also write it as 1

0.\bar{9}

>>15572744
nah, its enough with the base being divisible by 3, like 6 or 12

>>15572025
Idk how to use LaTeX, but here the proof that 0.9¯=1
Let's a=0.9¯
10a=9.9¯
10a=9+a
9a=9
a=1

>>15561606
Decimal system is kike poison to make math prove bullshit concepts that lead nowhere to keep mongoloid anons like those in this thread arguing ad nauseam.

>>15573128
you were born after your father pissed in your mother

>>15571964
1 = 1.000...

>>15573130
Not how reproduction works, math rabbi.

>>15573139
non-retard reproduction, yes

>>15573143
You're typing in a pretty retarded manner I must say. I wouldn't trust a retard such as yourself about the topic of reproduction, let alone axioms.

>>15573147
>t. retard
lol

>>15569644
>>15571974
>>15572025
I'm the one who you replied to, originally ( >>15568854 )
Do read Basic Mathematics and practice bc you are too unfamiliar with the basic arithmetic tools and a little dose of algebra and notations. Boy are you stupid!

>>15573352
projection, the post

>>15571964
Everything in the '''real numbers''' is a limit. The number 6 is actually a 'cauchy sequence' that 'converges to 6'.
0.999... (infinite 9's) is 1 because you can't fit in any value between 1 and 0.999... so it's just 1.

Outside of university/academia nobody uses real numbers they use approximations with floating point numbers so 0.999... can exist because you can always add another .9 in the finite world.

>>15573369
Do you really think he is able to comprehend ideas like a Cauchy sequence? [math] 0.\bar{9} \neq 1[/math] people are either retarded or trolling. I hope they are the latter.

>>15572822
When you multiply the digit 9 by 10 you get 90.
You have to keep the same decimal precision because of this fact
10 x 0,9 = 9,0
10 x 0,09 = 0,90
10 x 0,009 = 0,090
It doesn't matter how many steps you go, you are going to have 0 at the end of it becaue anything multiplied by 10 does that
>>15573369
>because you can't fit in any value between 1 and 0.999... so it's just 1.
Sounds to me that this statement is just a big assumption.
>>15573444
0.999... = 1 people are the ones whos trolling, see >>15562699

>>15573527
>the digit
Ignore this, was typing something else first

>>15573527
Lol
>It doesn't matter how many steps you go, you are going to have 0 at the end

But there is no end...

Bro, we talk about an infinity of steps, who say you that it will work like for finite steps ?

0.9999... with ∞ the number of 9
10×0.999=9.999999... with ∞-1 numbers of 9
infinity is so big than is this case we can consider ∞=∞-1

Infinty is a hard concept that we took millenium to fully understand it. So don't worry if you don't understand it

>>15573527
>0.999... = 1 people are the ones whos trolling

Sorry but I'm not trolling, we gave rigourous proofs that 0.9999...=1
If you don't think it, give me a proof that 0.9999... is different of 1, and then we will talk

>>15573714
I did right here >>15572628
nobody decided to look at it for whatever reason.
The proofs that you guys gave (like >>15572822) all rely on algebra, an algebraic system that breaks in-part if you include infinitesimals like I did.

>>15573729
Bro, that's your explaination here >>15572628 that is pseudo mathematic. I don't really understand the example with the thumble and how it's related to this problem. That will be an infinite number of little chunks, ∞×0 is sometimes egal to 1 (I didn't said the opposite). Probably you are true but I'm too dumb to understand it. I just ask you where is the mistake on my demonstration here >>15561781 . I also ask you a proof with calculs,without using an example, that 0.9999... is different of 1

Wikipedia gave me answers all my life, I don't think this page is a troll:

https://en.m.wikipedia.org/wiki/0.999......

>>15573714
>give me a proof that 0.9999... is different of 1, and then we will talk
I would say because decimal system works with 10 numbers ( 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 )
And you cannot equally divide them into thirds ( 1/3 ) because you always have 1 number left over
1, 2, 3
4, 5, 6
7, 8, 9
0
So 1/3 is not equal to 0.333... in the first place

>>15573850
Bro another demonstration don't start by 1/3:

a=0.99999...
10a=9.99999...
10a=9+a
9a=9
a=1

>>15573803
My issue with >>15561781 is the subtraction step of 10a=9+a to 9a=9. This is a nonobvious step since subtraction is troublesome with infinitesimals. A clear insight to why this subtraction can go awry is lets say 0.9999 = 1 - e, where e is an infinitesimal. Then lets suppose e is ''equivalent'' to 10e. Then 10a=9+a » 10 - e = 9 + 1 - e » 9 = 9. This resulting scenario is showing how you are basically saying 1=1 and making an algebraic error to get there from 0.999999…

>>15573696
>we can consider ∞=∞-1
https://youtu.be/SrU9YDoXE88?t=23m10s

>>15573527
>It doesn't matter how many steps you go, you are going to have 0 at the end
By definition there's an infinite number of 9s in the decimal expansion. Explain how you can have a 0 after an infinite sequence? Where is that 0? What nth digit of 9.999... is 0?

>>15573863
Do 9 x 10 even once

>>15573895
It has to be the last one
10 times N is 10N
You dont see it ?

>>15573884
There are no infinitesimals in reals. Under reals 1 = 0.999... since there are no real numbers between 1 and 0.999...

>>15573913
See what? You're trying to say that there's a zero after the infinite decimal expansion and I'm asking you to clarify what that means.

>>15573915
That is my point! I am not using reals. See: >>15572628
I am pointing out that if you need to define your set as containing no infinitesimals to prove 0.999…=1 then you are missing the full picture.

>>15573919
9 x 10 = 90
Do it to every number in the decimals
Every number is now 90

>>15573925
So where is that 0? Point to the specific digit which is 0.

>>15573886
Yet another person who doesn't even understand what vsauce's math videos are about nor the video itself. Pop math/sci and it's consequences, etc.
Learn some set theory before you bring the topic of ordinals here or you're going to embarrass yourself.

>>15573947
I cannot say where it is other than the end

You start to work from left to right and you always get 90 before you start working on the next digit where the next 90 overwrites the previous 0, having 90 again you work onwards

>>15573955
There's no such thing as an end of an infinite sequence so by your own admission that 0 is nowhere in the sequence.

Excellent bait anon. Make mathfags suffer more.

>>15573962

[eqn]0.999...= 0.9 + 0.09 + 0.009 + \dotsm=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{9}{10^k}=1[/eqn]

>>15573975
>I cannot say where it is other than the end
You already conceded.

>>15573982
Well he didn't take that as an answer first

>>15564833
chatgpt answer

>>15574083
nah, chatgpt isn't that dumb

>>15574093
oh is it?

>>15574106
pic rel sounds reasonable. I'm gonna stick with that

Good thread, let's add somme shit :

ocean - droplet = ocean
1 - (0.000...1) = 0,999... = 1

Am I good at math ?

>>15573884
>This is a nonobvious step since subtraction is troublesome with infinitesimals.
the reals have no infinitesimals, you are thinking of hyperreals you freakish baboon
https://en.wikipedia.org/wiki/Hyperreal_number

>>15573925
that zero appears because there are infinite 0 to the right of the 9, so 9.0...*10=90.0..., can you perceive from this what 0.9... is?

>>15574296
nah, you absolute shit, 0.9...*anything* is a non-sense statement

>>15573886
Ok, My sentence was wrong. But I didn't talk about transfinite numbers, I just said that remove 1 at the infinity, it stay infinity

>>15574683
You Big Bad Baboon, I never said nor OP never said anything about reals. I mentioned several times in the thread how we need to consider more than the reals.

>>15562049
>I'm learning this right now
Good luck with high school anon

0.999... = 1.000...

>>15570844
>>15571021
at no real time moment of incremented additional 3's does the equation evaluate as equal.
infinity and the infinite are not qualities of the past tense, they are qualities of the future tense. You can never have, nor can you have ever had, reached infinity.

The fractional [math]\frac{1}{3}[/math] simply doesn't equate to a decimal representation, no more than pi does. Pi goes on and on, and so does [math]\frac{1}{3}[/math].

any useful arithmetic utilizing pi will only use pi to some decimal level of accuracy, like possibly 7 or 15 decimal places.
but
[math]\pi = 3.14159[/math] is a false statement, while
[math]\pi > 3.14159[/math] is a true statement.

similarly,
[math]\frac{1}{3} = 0.33333[/math] is a false statement, while
[math]\frac{1}{3} > 0.33333[/math] is a true statement.

if you want to stretch, there are real provided options to stretch without needing to confuse existing definitions.
such as:
[math]\pi \approx 3.14159... [/math]
and
[math]\frac{1}{3} \approx 0.33333...[/math]

approximation is not equality, for everything that it implies.

0.19 = 0.2

9 = 10

Fuck off now kys

THERES SOMETHING WRONG WITH BASE 10

I am making the sacrifice against base 10 and iq

>>15575358
Owwwh

There's nothing wrong with base 10.
There's something wrong with ignoring it, though.
December isn't a name like John or Sally.
It literally translates to "tenth month"

Try to recall the last time you celebrated December as the 10th month, or is all you can remember the retarded fake knowledge spoonfed to you while you recklessly accepted december as the 12th month?
September = Septagon = 7
Octoboer = Octagon = 8
November = Nonagon = 9
December = Decagon = 10

>>15575352
Did you saw wikipedia ?
https://en.m.wikipedia.org/wiki/0.999......


What they say:

>Some students interpret "0.999..." (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity".

>Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value, since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999..." as meaning the sequence rather than its limit.

Infinity don't works like classical numbers.

In your opinion, the picture on this post is false ? because we need an infinite number of iterations (of adding numbers) for that be equal and you said that we can't reach infinity. so it's not equal but approximatly equal ? Explain me

>>15573901
?

>>15573884
You are true, I checked and it's not the more rigourous proof that 0.9999...=1

>>15561641
Why is there a sigma when that should be a limit
The summation symbol is absolutely incorrect

>>15575549
It's correct because it dum to infinity, that mean that we took the limit of the sum to n when n goes to infinity

>>15575554
*it summ to infinity

>>15561606
anything involving infinity is nonsense and word games

>>15575559
This

EVERYONE IGNORE THIS THREAD

IT CONTAINS PEOPLE WITH BELIEF IN BASE 10 MYSTICISM

KEEP OUT

>>15575559
>>15575561
t. retarded mathlets

>>15561641
>0.999...= 1 because 9/10 equals 0.999...
Nice circular fucking logic, still no evidence though, lmao

>>15575358
>THERES SOMETHING WRONG WITH BASE 10
Yeah, let's try base 2 instead
[math]\sum_{n=0}^{\infty} \frac{1}{2^n} = 0.1 + 0.01 + 0.001 + 0.0001 + ..... = 0.111... = 1[/math]

1/9 = 0.111...
+
8/9 = 0.888...
=
9/9 = 0.999...

>>15575636
>n=0
lolno

>>15575638
yeah, fucked up that bit

>>15575632

[math]
\boxed{0 < p < 1} \\
1 = p + (1-p) ~~~~~~ \overset{1}{ \overbrace{[=====p=====|==(1-p)==]}} \\
\text{divide p using x} ~~~~~~ \overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====x=====|==(p-x)==}]} ~~ + ~~ (1-p)}} \\
\\
\text{solve x and (p-x), when length ratios must be the same} \\
\dfrac{x}{p-x}= \dfrac{p}{1-p} \Rightarrow x- xp = p^2 - xp \Rightarrow \underline{x=p^2} \Rightarrow \underline{(p-x)=p(1-p)} \\
\overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====p^2=====|==p(1-p)==}]} ~~ + ~~ (1-p)}} \\
\\
\overset{1}{ \overbrace{ \underset{p^2}{[ \underbrace{=====p^3=====|==p^2(1-p)==}]} ~~+ p(1-p)+(1-p)}} \\
\overset{1}{ \overbrace{ \underset{p^3}{[ \underbrace{=====p^4=====|==p^3(1-p)==}]} ~~+ p^2(1-p)+p(1-p)+(1-p)}} \\
(1-p)+p(1-p)+p^2(1-p)+p^3(1-p)+ \cdots =1 ~~~~ \left | ~ \times \frac{1}{1-p} \right . \\
1+p+p^2+p^3+ \cdots = \dfrac{1}{1-p}
[/math]

>>15575651
>1=p+(1-p)
>1=1
You just owned yourself retard lmao

>>15575659
wait, does 1 not equal 1?

>>15575660
It does, what it does not equal is 0.999...

>>15575662
so why did you claim that saying 1=1 is "owning himself"?

>>15575665
Because I'm asking for evidence that 0.999...= 1, so far there still isn't any

>>15575672
he showed you evidence
part of understanding that evidence is accepting that 1 = 1

>>15575674
No all it proves is that 1=1

>>15575651
[math] \displaystyle
p=0.1 \\
\dfrac{1}{1-0.1}=\frac{10}{9} = 1 + \frac{1}{9} \\
\sum_{j=0}^\infty 0.1^j= 1 + \sum_{j=1}^\infty 0.1^j \\
9+1=9+9\sum_{j=1}^\infty 0.1^j \\
1=9\sum_{j=1}^\infty 0.1^j = 0.999...\\
\dfrac{1}{3} = 3 \sum_{j=1}^ \infty 0.1^j = 0.333...
[/math]

>>15575679
>i
>used intrinsically
>suggestive

>>15575690
j also works

>>15573884
here no substractions, where is the "mistake" ?
>>15575632
No, we he just didn't showed the calcul:

Let's a=0.9999...
a=0.9999...
a=0.9+0.09+0.009+0.0009+...
a=9(0.1+0.01+0.001+0.0001+...)
a=9S
with:
S=0.1+0.01+0.001+0.0001+...
0.1S= 0.01+0.001+0.0001+...

S-0.1S=0.1
0.9S=0.1
S=(0.1)/(0.9)
S=1/9


a=9S
a=9×1/9
a=1
0.9999...=1

>>15575561
Literally every integer base has this problem. There's no "base 10 mysticism".

>>15575632
>evidence
This is math, anon.

>>15563124
I could show you that number exactly on a base 12 numerical system. Just because the mathematical basis set is flawed doesnt mean numbers the basis set cant describe properly don't exist.

>>15575637
and 8783654/9999999=0.(8783654)...,
what might you gather from this?

>>15575690
he could have put an emoji if he so desired, the answer would not change

>>15576710
>can't show his tiny d
lol

>>15576710
There are still numbers that can't be represented properly in base 12, so why stop at base 12, why not just use base infinity?

>>15561606
0.999... = 1 is arguably not even a "math concept", it's just a notational concept. It's like saying putting the number on top as the numerator and the number on bottom as the denominator is a "math concept" or a vertical and horizontal line crossed over each other meaning "addition" is a "math concept."

>>15577362
>represented properly
the fuck you mean with "represented properly" you absolute meta-schizo?

>>15577383
I mean representing the whole value with actual numbers without just cutting it off at some arbitrary point like we currently do with irrational numbers.

>>15561606
[eqn]\sum_{n=0}^{\infty}\left(x^n\right)=\frac{1}{1-x}\\\mathrm{let}\ x=\frac{1}{10}\\
0.999...=\frac{9}{10}*\sum_{n=0}^{\infty}\left(x^n\right)=\frac{9}{1-\frac{1}{10}}=1[/eqn]Will generatingfunctionchads ever stop winning?

>>15577386
0.333... is not irrational.

>>15577386
So we must to write 1.000000000... instead of 1 ?

>>15574296
No, your assumption is off, if the ocean is 1, a drop is actually 0.00000000000000000000000004055 give or take, which is not 0.0...01

>>15561606
If 1 = 0.999... is that still true in higher bases?
In hexadecimal would 0.999... still be 1 or would 0.FFF... = 1 instead?

>>15578685
math] \displaystyle
\begin{align*}
1 = \left (\frac{15}{16} + \frac{1}{16} \right )
&= \text{0x0.F} + \frac{1}{16} \\
= \text{0x0.F} + \left ( \frac{15}{256} + \frac{1}{256} \right )
&= \text{0x0.FF} + \frac{1}{256}\\
= \text{0x0.FF} + \left ( \frac{15}{4096} + \frac{1}{4096} \right )
&= \text{0x0.FFF} + \frac{1}{4096} \\
= \text{0x0.FFF} +\left ( \frac{15}{65536} + \frac{1}{65536} \right )
&= \text{0x0.FFFF} + \frac{1}{65536} \\
&\vdots
\end{align*}
\\ \displaystyle
\Rightarrow \text{0x}0.\overline{\text{F}} = 1
[/math]

>>15578685
[math] \displaystyle
\begin{align*}
1 = \left (\frac{15}{16} + \frac{1}{16} \right )
&= \text{0x0.F} + \frac{1}{16} \\
= \text{0x0.F} + \left ( \frac{15}{256} + \frac{1}{256} \right )
&= \text{0x0.FF} + \frac{1}{256}\\
= \text{0x0.FF} + \left ( \frac{15}{4096} + \frac{1}{4096} \right )
&= \text{0x0.FFF} + \frac{1}{4096} \\
= \text{0x0.FFF} +\left ( \frac{15}{65536} + \frac{1}{65536} \right )
&= \text{0x0.FFFF} + \frac{1}{65536} \\
&\vdots
\end{align*}
\\ \displaystyle
\Rightarrow \text{0x}0.\overline{\text{F}} = 1
[/math]

>>15578685
in base n, where m=(n-1), 0.m...=1

>>15578685
>>15579005
yet it is objective that
0.F > 0.9
0.FF > 0.99
0.FFF > 0.999
as
[math]\frac{15}{16} > \frac{9}{10} \\ \frac{255}{256} > \frac{99}{100} \\ \frac{4095}{4096} > \frac{999}{1000} [/math]

surely it must also be true that 0.FFF... > 0.999...
which means neither equal 1, and neither equal each other.

>>15579508
nice

To these people 9 = 10

Fuck off kys now

>>15579508
Is 2(1+2+3+...) larger than (1+2+3+...)

>>15579508
Yet it is objective that each line =1.
Exactly 1, not approaching it.

>>15579530
For every step the other is twice as large

>>15579554
For every faggot is one just like you

>>15579557
i know, a straight one

>>15579566
baited.

>>15579538
>objective
i don't believe you know what that word means.

>>15579530
>is [math] -\frac{1}{6} [/math] larger than [math]-\frac{1}{12}[/math]
?????????????????????

>>15579580
talking about >>15579005 oc.
bless your heart

>>15579554
so (1+2+3+4+5+6+...) is smaller than (2+4+6+....)?
really?

>>15579629
Yes for every step

>>15561606
>/his/tards trying to do math
stay in your lane and let the STEMchads work

>>15579508
a = 15/16 + 15/256 + 15/4096 + 15/65536 + ... = 0.ffff...
b = 9/10 + 9/100 + 9/1000 + 9/10000 + ... = 0.999...
c = (b1 + b2) + (b3 + b4) + ... = b = 0.999...
c = (9/10 + 9/100) + (9/1000 + 9/10000) + ... = b = 0.999...
c = 99/100 + 99/10000 + ... = b = 0.999...

a = 0.f + 0.0f + 0.00f + ... = 0.ffff...
c = 0.99 + 0.0099 + 0.000099 + ... = 0.9999...

0.f < 0.99
0.ff < 0.9999
0.fff < 0.999999
0.ffff < 0.99999999
0.ffff... < 0.9999...
which contradicts 0.fff... > 0.999...

>>15579632
>finite
lol ok, have a great day

>>15579629
>Short answer
Yes
>Long answer
Ye...

>>15579632
And? How is that relevant? For any finite number of "steps" of one expression there is a finite number of "steps" where the 2nd expression is greater. And the number of steps is infinite. Which means that they must necessarily be equal.
0.1 + 0.01 + 0.001 + ... < 0.9 + 0.09 + 0.009... because there's no finite number of steps of the summation 0.1 + 0.01 + 0.001 + ... such that it's greater than 0.9 (one finite step). You cannot group 0.1 + 0.01 + 0.001 + ... in any way to make it larger than 0.9 + 0.09 + 0.009....

>>15579651
In fact you cannot group 0.1 + 0.01 + 0.001 + ...
in any way to even make it larger than 0.12

>>15579005
>>15579332
So would quaternary be 0.333... = 1 then?
But then is 0.333... not 1/3?
I am an admitted mathlet so not trying to prove anything I'm just curious.

>>15579651
What if i do this

(+1+2+...) = (+2+4+...)

>>15579687
If it's not base 10 you have to specify that otherwise it's just confusing.
3 base 4 is not equal 3 base 10

>>15579689
(ω+1+2+...) = (ω+2+4+...)

>>15579689
>>15579694
Both diverge to infinity.

>>15579693
>3 base 4 is not equal 3 base 10
How so?

>>15579705
typo probably
likely meant 0.3

>>15579687
>>15579705
>>15579710
Whoops. I meant 0.333... base 4 and 0.333... base 10. 3 base 4 and 3 base 10 are in fact equal.

[math]\left ( 0.333... \right )_4=\sum_{n=1}^{\infty}\frac{3}{4^n} = \frac{3}{4} +\frac{3}{16} +\frac{3}{64} + \cdots = \left ( 0.75 \right )_{10} + \left ( 0.1875 \right )_{10} + \left ( 0.046875 \right )_{10} + \cdots \\

\left ( 0.333... \right )_{10} < \left ( 0.75 \right )_{10} \\
\left ( 0.333... \right )_{10} < \left ( 0.333... \right )_4[/math]

>>15579710
>>15579719
Ah I see.

>>15579687
0.3333333333_10 = 0.1111111111_4
https://www.rapidtables.com/convert/number/base-converter.html

x = 0.999...
10x = 9.999...
9x = 9
x = 1
simple as

>>15579702
but it is the same infinity
(I'm just asking)
(If I remember, Eulet showed that the Harmonic series diverge to log(log(∞)) or something like that)

[math] \displaystyle
\begin{align*}
\sum_{n=1}^{ \infty} \frac{1}{n} &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}
+ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots \\
&> 1 + \frac{1}{2} + \left (\frac{1}{4} + \frac{1}{4} \right ) +
\left ( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right ) + \cdots \\
&= 1 + \lim_{n \rightarrow \infty} n \cdot \frac{1}{2} = \infty
\end{align*}
[/math]

>>15580332
Yes