[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 4 KB, 125x125, xaxaxaxaxaxaxaxaxaxaxaxaxaxaxaxaxa.jpg [View same] [iqdb] [saucenao] [google]
11437743 No.11437743 [Reply] [Original]

1=0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...+0.00000000000000000000000000000000000000000000000000000000000000000000000000000...1

>> No.11437762

>>11437743
wrong
1=3/3
1/3=0.3°
0.3°x3=0.9°
1/3x3=1
0.9°=1

>> No.11437763

>>11437762
>1/3=0.3°
that’s where you’re wrong

>> No.11437767

>>11437743
brainlets will disagree

>> No.11437789

>0.00000000000000000000000000000000000000000000000000000000000000000000000000000...1
What do you mean by this? Formal definition please.

>> No.11437793

>>11437763
and the earth is flat.
take your meds shizo

>> No.11437798
File: 152 KB, 1018x776, whatisconvergence.jpg [View same] [iqdb] [saucenao] [google]
11437798

>> No.11437801

>>11437743
0.000000000000000000...00000000001 = 0
0.9999999999999999999...9999999999 = 1
1 = 1 + 0
you are correct

>> No.11437825

>>11437789
Infinite zeroes with a one (after) infinity obviously.

>> No.11437829

>>11437801
wrong, completely misunderstood the point.

>> No.11437835

>>11437825
0.0000....00001 can also be expressed as one over infinity, which isn't a viable mathematical expression at all.
Using limits we could do 1/(lim->infinity) which is definitively equal to zero. So the expression either doesn't exist or just means 1 = 1+0

>> No.11437931

>>11437829
i am sure that writing out that many 0's and that many 9's is much more high IQ than just writing them like 0.000... and 0.999... so therefore you must be right because that's so high IQ

>> No.11437966

>>11437931
oh boy here we have the fag who cant go into a conversation without talking about iq.

>> No.11437978

>>11437966
oh boy here we have the newfag who can't sense basic /sci/ irony

>> No.11437995

>>11437978
oh boy!

>> No.11438298

0.0000.....01 is not a number. there is no number where there is an infinite number of 0s yet it also ends in a 1.

>> No.11438376

>>11437801
/thread

>> No.11438414

>>11438298
Yes there is. Normally you w

>> No.11438743

>>11437743
1/3=0.333...
2/3=0.666...
3/3=0.999...=1

>> No.11438768
File: 171 KB, 1018x776, 1583276313055.jpg [View same] [iqdb] [saucenao] [google]
11438768

>> No.11438816

>>11437763
1/3 = 3/10 + 1/30
= 0.3 + 1/30
= 0.33 + 1/300
= 0.333 + 1/3000
:
= 0.3... + 1/inf
= 0.3... + 0
= 0.3...

>> No.11438834

1=0.999...
2=0.999...+0.999...
N=N(0.999...)
N+1=(N+1)(0.999...)
N+1=(N(0.999...)+1)(0.999...)
N=N(0.999...)^2+0.999...-1
N-N=N(0.999...)^2+0.999...-1-N
0=N(0.999...)^2+0.999...-(1+N)

0.999...=((-1)[+/-]sqrt(1^2+4*(1+N)*1))/2N
For N=1
0.999...=(-1+3)/2)
0.999...=(2/2)
0.999...=1

>> No.11438894

> it's another undergrad "revelation" thread
Wew

>> No.11438966

>>11437801
>see this number that factually isn't 0?
>that's 0

>> No.11438977

>>11438414
no there isn't you retard

0.000... implies that 0 continues without end
yet 0.00...1 means that 0 continues without end yet at the same time ends with 0. fucking stupid

>> No.11438979

>>11438834
circular logic

>> No.11438981

>>11438977
ends with 1*

>> No.11439052

>>11438977
Not a very strong case when one of the bases of calculus is the concept of an infinitesimal and a limit. Defining arbitrarily small numbers is already something that is done regularly in math. Perhaps the history of the 0.333..., namely the fact that it comes from a recursive division in 1/3, is important for some mathematician's pet project and throwing that away with an = could have implications. If you don't use approximately equal to for 1/3 and 0.999..., you are a terminal brainlet.

>> No.11439073
File: 246 KB, 728x728, 1560337741586.jpg [View same] [iqdb] [saucenao] [google]
11439073

>>11439052
>number that has an infinite number of trailing 0s
>yet at the same time ends in 1
>infinite
>ends in 1

shut the fuck up you dumb retard.
this has nothing to do with calculus or infinitesimal but everything to do with your LACK of understanding of mathematics.

pic related is you

>> No.11439094

>>11439073
Your spastic outbursts only further evidence your sheer incompetence. Don't worry though, if you're not cut out for STEM you can always fall back to being a barista.

>> No.11439112

>>11439094
>>infinite
>>ends in 1
kys

>> No.11439135

>>11439112
Finite but arbitrarily small. Ends in 1. Maintains a sign and a significand. Ellipses are easier than writing out an unimaginably large number of zeros. It's a perfectly valid way of writing the truncation error in 1-0.999... the only thing you really have to note in order to avoid contradictions when performing algebra are the relative positions of the last digit. 1-.999...n = 0.000...n 1, where n is the number of digits you choose to expand to. lim n->infinity approaches zero. Being aware of this gets rid of most contradictions. i.e. 0.000...1 = x, x*10 = 0.000...10 = 0.000...1, 10x = x, 10 = 1. Stops working when you note that you shifted the index of the terminal digit by dropping the 0.

It's a kludge, obviously, but no more imprecise than representing a fraction as a repeating decimal in the first place, and such a representation is a helpful aid if you're trying to keep track of those particular properties of the construction. 0 has no sign. 0.000...1 does.

>> No.11439188

>>11438298
[math] \lim_{x \to \infty} f(x), ~g(x) ~~where~~ f(x)=2*10^{-x}, ~~g(x)=10^{-x} [/math]

[math] f(x)-g(x)=0.00...00001 [/math]

>> No.11439192

>>11439188
retard

>> No.11439202

>>11439192
explain how it's wrong, cretin

>> No.11439209

>>11439202
open up a book on analysis and see for yourself, retard. We've had this thread hundreds of times already.
What you wrote is meaningless. It literally doesn't have any meaning.

>> No.11439271

>>11439209
I am creating a number that always ends in 1.

Prove that it doesn't, little brainlet.

>> No.11439299

>>11439135
>>11439112

>> No.11439308

>>11439271
sure but it won't have an infinite number of trailing zeros.

>prove that there is no number with infinite number of trailing zeros that also ends with 1

infinite numbers don't end therefore it cannot end with 1.

>> No.11439323

>>11439188
[math]f(x) - g(x) = 2*10^{-x} - 10^{-x} = 10^{-x}*(2-1) = 10^{-x}[/math]
your point?

>> No.11439327

>>11439308
g(x)/f(x) will always equal 0.5. Even if x is infinite. This is only possible if the last digit is enumerated.

>> No.11439332

>>11439323
I did it like that to discourage brainlets like you thinking that simplifying the equation invalidates the first result.

>> No.11439334

>>11437743
Love the fact that /sci/ still argues about this even if OP is such a bad troll.
>sage goes in all fields

>> No.11439338

>>11439332
so [math]10^{-x} = 0.00\dots000001[/math] ?

>> No.11439341

>>11439338
Yep. Problem?

>> No.11439348

>>11439341
it's not true

>> No.11439349

>>11439348
then explain
>>11439327

>> No.11439354

>>11439349
it's also not true (the statement doesn't even make sense to begin with)

>> No.11439360

>>11439354
It's provable via induction you midwit undergrad

>> No.11439365

>>11439360
what is provable via induction ? give the exact statement

>> No.11439370

>>11439365
g(x)/f(x)=0.5 for any x

>> No.11439372

>>11439370
that's true. so what ?

>> No.11439379

>>11439372
So, given x tends to infinity, g(x)/f(x)=0.5.

Doesn't matter how many zeros proceed the ending.

>> No.11439381

>>11439372
>>11439379
precede*

>> No.11439388

>>11439379
that's also true. so what ?

>> No.11439392

does anyone else smell poo?

>> No.11439407

>>11439388
Good. So I'm right. This >>11438298 is wrong.

>> No.11439443

>>11437825

if the zeros are infinite then the 1 never comes because the zeros never stop. Your understanding of inifnity is flawed.

>> No.11439458

>>11439370
>>11439372
>>11439379
>>11439388
>>11439407
let g(x)=x^2, f(x)=x.
then lim g(x)/f(x) != 0.5 as x -> inf

That was really difficult to disprove.

>> No.11439477

>>11439458
you missed the part where I defined g(x) and f(x) you dribbling brainlet >>11439188

>> No.11439482

>>11439188
in this situation f(x) - g(x) is exactly equal to 0

>> No.11439483

>>11439477
The retarded definition where 2*0 in the numerator is not 0? Sorry, forgot about that retarded abnomination.

>> No.11439492

>>11439483
if g(x) and f(x) are defined as such, g(x)/f(x)=0.5 for all x

>> No.11439508

>>11439492
nope. You are just trolling or very stupid, because the behaviour for the function (that you defined) is not what you are describing. In my country children learn about lim and this when they are about 13 to 14 years old.

>> No.11439516

>>11439508
Kek, you are very dumb my very stupid friend, for you see nothing I mentioned is unconventional. I just presented in this thread for halfwits like you to out themselves.

>> No.11439542

>>11439516
Are you even trying? (2*10^-x)/(10^-x) as x -> inf is 2. I even told you that every 13 to 14 year old school kid will know this. You funny invention of 0.000...0001 does not make it better.

>> No.11439578

>>11439542
So you agree that in order to compute the ratio as x tends to infinity, the final digit must be enumerated.

>> No.11439606

>>11439578
What final digit?

>> No.11439649

>>11439578
no. g(x)/f(x) or the limit of g(x)/f(x) tells you nothing about whether "final digit must be enumerated" (whatever that means).

take f(x) = 2pi and g(x) = pi. then g(x)/f(x) = 0.5 for all x and so does the limit, but all of f(x), g(x) and f(x) - g(x) have infinite decimal expansion.

>> No.11439657

>>11437743
It's true though. Cope.

>> No.11439691

>>11439649
You can only deduct that by simplifying the expression, where pi is cancelled. Mathematics must remain consistent for all expansions.

So to calculate pi/2pi as a/b, then you must calculate pi itself. HOWEVER pi cannot be calculated to its extent thus you have contradicted yourself by assuming cross viability.

>> No.11439701

>>11438816
>1/inf
not a number

>> No.11439712

>>11439701
it's zero, retard

>> No.11440013

>>11439712
It approaches zero retard

>> No.11440106

>>11437743
infinite 0 with a one at the end don't exist because infinite 0 have no end
but i agree that 0.99999... is just infinite close to 1, and not 1 itself. not that it would make any difference...

>> No.11440113

>>11439691
>he never calculated pi to the end
brainlet detected

>> No.11440213

>>11440013
infinity isn't a diesel engine chugging along

>> No.11440748

bump

>> No.11440753

Room temp IQ ITT

>> No.11440756
File: 125 KB, 317x341, 1534497174036.png [View same] [iqdb] [saucenao] [google]
11440756

>it's an anon learns the difference between countable and uncountable infinity episode

>> No.11440761

>>11437743
Physicists are fucking retarded

>> No.11440767

>>11440756
explain, midwit

>> No.11440796

>>11437743
Highwit pretending to be retarded
>>11437801
Highwit
>>11438966
Midwit
>>11438834
>>11437789
Brainlets

>> No.11440801

>>11440796
Extreme deficiencies.

>>11439691
Immense proficiencies.

>> No.11440811

>>11440801
IQ range estimate is 65-95.

>> No.11440855

>>11438834
>>11439542
Snuff porn actress
>>11437762
>>11439657
Gimp
>>11437801
>>11439135
>>11440106
Bratty goddess
>>11439691
>>11440801
Got a purdy mouth on 'em

>> No.11440863

>>11440796
Cringe
>>11440801
Based
>>11440855
Basinge

>> No.11440875

>>11440767
Bu it's more fun for me to not explain and leave you wondering what I meant

>> No.11441555

>>11437763
Let Pn be the statement that
sum [k = 0 to n] (a*10^-k) * b
=sum [k = 0 to n] (a*b*10^-k).
P1 is the statement that
(a + 0.1*a)*b = a*b + 0.1*b
By distributivity or whatever, this is true for all a, b. Thus P1 is true. Assume Pn is true, that is
sum [k = 0 to n] (a*10^-k) * b
=sum [k = 0 to n] (a*b*10^-k). Note that
sum [k = 0 to n+1] (a*10^-k) * b
=sum [k = 0 to n] (a*b*10^-k)+a*b*10^-(n+1)
=sum [k = 0 to n+1] (a*b*10^-k).
Thus Pn+1 is true, and therefore Pn is true for all n.
Thus Pn is true as n approaches infinity.

>> No.11441565

>>11437743
well of course, anything plus 0 is itself

>> No.11441571

0.00000... 1 has 1 at the omegath decimal place retards. Just because there are infinite zeroes doesn't mean a one can't come after. There are infinitely many numbers in [0,1) yet you still believe 1 is a number. Learn some math dumbos.

>> No.11441623

>>11441571
>>11439112

>> No.11441893

>>11441623
>infinite numbers between 0 and 1
>1 comes after

>> No.11441903

>>11441571
>omegath decimal place
no such thing

>> No.11441909

>>11441571
0.000...01 is not allowed because its set representation is a recursively defined set that is forbidden by zfc

>> No.11441975

>>11441903
Because you said so?

>> No.11441979

>>11441909
I'm talking about numbers here, not sets

>> No.11441993

>>11441571
Finally, some high IQ has arrived.

>> No.11442011

>>11441909
Yikes. [0,1) is also recursively defined you dumb rat

>> No.11442042

>>11441975
because the generally accepted definition of real numbers said so

>> No.11442051

>>11442042
argumentum ad populum :D

>> No.11442060

>>11442051
>argumentum ad populum :D
not really

>> No.11442248

bump

>> No.11442280

The argument essentially boils down to a single question in which we fundamentally disagree. Does 1/inf = 0? The answer is no. Lim h->inf (1+1/inf)^inf=e. Lim h->inf (1)^h = 1. The only reason retards think 1^inf is indeterminate is because they think 1/inf = 0.

>> No.11442282

>>11441893
What's the last digit in .9 repeating? 9. There are infinite 9's before it the 9 comes after.

>> No.11442292

Bump

>> No.11442640

bump

>> No.11442694

bump

>> No.11442755

bumperino pupernino le xd herp derp ROFLCOPTER Lulz

>> No.11442987

If we only look at whole numbers, and assume that decimals and fractions of a number don't exist for the purposes of this example, why isn't 9=10, or 3=4?

>> No.11443283

>still using actual numbers

WHAT IS THIS THIRD GRADE SHIT LMFAOOOOO

>> No.11443294

>>11437762
1/3=0.3.....4≠0.3...