/sci/ - Science & Math » Thread #11653094

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Anonymous Sat 09 May 12:20:11 2020 No.11653094 View Reply Original Report

Quoted By: >>11653101 >>11653119 >>11653165 >>11653167 >>11653194 >>11653262 >>11653381

I ) 1 is clearly a natural number
II ) 0.999... is clearly a real number
therefore 0.999... =|= 1

Anonymous

Anonymous Sat 09 May 2020 12:24:12 No.11653101 Report

Quoted By: >>11653113

>>11653094
>II ) 0.999... is clearly a real number
wrong

Anonymous

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Anonymous Sat 09 May 2020 12:32:49 No.11653108 Report

Quoted By: >>11653124

I love the method of 'proof by "clearly"'.

Anonymous

Anonymous Sat 09 May 2020 12:35:09 No.11653113 Report

Quoted By:

>>11653101
Define 0.999... with the peano axioms then

Anonymous

Anonymous Sat 09 May 2020 12:40:21 No.11653119 Report

Quoted By: >>11653124 >>11653288

>>11653094
>is clearly
your fields medal is in the mail

Anonymous

Anonymous Sat 09 May 2020 12:43:14 No.11653124 Report

Quoted By: >>11653128

>>11653119
>>11653108
>of course you need to prove that an infinite sum of fractions is not a natural number
kek seething brainlets

Anonymous

Anonymous Sat 09 May 2020 12:46:24 No.11653128 Report

Quoted By:

>>11653124
>of course
your 2nd fields medal is in the mail

Anonymous

Anonymous Sat 09 May 2020 13:17:04 No.11653165 Report

Quoted By: >>11653171

>>11653094
>implying 1 is not a real number

Anonymous

Anonymous Sat 09 May 2020 13:19:13 No.11653167 Report

Quoted By: >>11653171

>>11653094
2 is clearly a natural number
2 is clearly a real number
therefore 2 =|= 2

See a problem?

Anonymous

Anonymous Sat 09 May 2020 13:20:10 No.11653171 Report

Quoted By: >>11653182

>>11653165
>>11653167
to be more accurate
1 is a real and natural number
0.999... is exclusively real

Anonymous

Anonymous Sat 09 May 2020 13:30:52 No.11653182 Report

Quoted By: >>11653219

>>11653171
>0.999... is exclusively real
can you prove it ?

Anonymous

Anonymous Sat 09 May 2020 13:37:10 No.11653194 Report

Quoted By:

>>11653094
Simpler and to avoid the meme of 0.999...

I ) 1 is clearly a natural number
II )2/2... is clearly a rational number
therefore 2/2... =|= 1

${a \over a } \in \mathbb{N}$ ?

How is writing a number should change what it is?

In the case of negative integers, how you write changes what they are, but that is inconsistent with the notation and use of ^.
$-2^2 "=" -4$
$(-2)^2 = 4$
$-2 \ne (-2)$
$lim_{\space n \to \infty} \sum ^n _{i = 1} {9 \over 10^i}$ = 3/3

Anonymous

Anonymous Sat 09 May 2020 13:51:10 No.11653219 Report

Quoted By: >>11653221

>>11653182
we know 0.999... > 0, 0.999... < 1 and that there is no natural number between 0 and 1
therefore 0.999... cant be a natural number

Anonymous

Anonymous Sat 09 May 2020 13:53:00 No.11653221 Report

Quoted By: >>11653230 >>11653281

>>11653219
>0.999... > 0
>0.999... < 1
can you prove it ?

Anonymous

Anonymous Sat 09 May 2020 13:56:03 No.11653230 Report

Quoted By: >>11653238

>>11653221
that is the definition of "<" and ">"

Anonymous

Anonymous Sat 09 May 2020 13:58:32 No.11653238 Report

Quoted By: >>11653241

>>11653230
so you can't, okay

Anonymous

Anonymous Sat 09 May 2020 13:59:36 No.11653241 Report

Quoted By: >>11653245 >>11653247 >>11653272

>>11653238
it is not necessary

Anonymous

Anonymous Sat 09 May 2020 14:01:36 No.11653245 Report

Quoted By: >>11653248

>>11653241
well I disagree with 0.999... < 1, so it kind of is necessary

Anonymous

Anonymous Sat 09 May 2020 14:02:07 No.11653247 Report

Quoted By: >>11653250

>>11653241
neither is your life
lmao rekt

Anonymous

Anonymous Sat 09 May 2020 14:02:13 No.11653248 Report

Quoted By: >>11653254 >>11653263

>>11653245
well I disagree with your disagreement so why dont you prove it then

Anonymous

Anonymous Sat 09 May 2020 14:03:15 No.11653250 Report

Quoted By:

>>11653247
rudeposters are not welcome in this thread

Anonymous

Anonymous Sat 09 May 2020 14:06:44 No.11653254 Report

Quoted By: >>11653261

>>11653248
0.999... = $\sum_{n=1}^{\infty}\frac{9}{10^n}$ = $\lim_{n\to\infty}\sum_{k=1}^n \frac{9}{10^k}$ = $\lim_{n\to\infty}(1-\frac{1}{10^n})$ = 1 - 0 = 1
therefore 0.999... = 1 $\nless$ 1

Anonymous

Anonymous Sat 09 May 2020 14:11:30 No.11653261 Report

Quoted By: >>11653267

>>11653254
Wrong
0.999... does converge against 1 but it is not 1

Anonymous

Anonymous Sat 09 May 2020 14:11:39 No.11653262 Report

Quoted By: >>11653269

>>11653094

For once, a schizo is right. If you pick the typical encodings of ZFC, and interpret 1 ?
? and 0.9999 ? R, then they are definitely going to be different.

Anonymous

Anonymous Sat 09 May 2020 14:12:11 No.11653263 Report

Quoted By: >>11653375

>>11653248
There is no natural number n such that 0.999.... + 1/n < 1. This means that 1-0.999.... < e for all real e > 0 (Archimedean property), meaning 0.9999.... >= 1.

Anonymous

Anonymous Sat 09 May 2020 14:14:39 No.11653267 Report

Quoted By:

>>11653261
didn't you say that 0.999... was a real number ?

Anonymous

Anonymous Sat 09 May 2020 14:16:38 No.11653269 Report

Quoted By: >>11653275 >>11653277

>>11653262
"limits"? yeah right

Please prove via Peano axioms right here right now 0.999... = 1 otherwise nobody's gonna believe you

Anonymous

Anonymous Sat 09 May 2020 14:18:43 No.11653272 Report

Quoted By:

>>11653241
> can you prove it?
> it is not necessary

gargantuan kek

Anonymous

Anonymous Sat 09 May 2020 14:18:52 No.11653275 Report

Quoted By:

>>11653269
You can't even answer to the right person, how do you hope understanding Peano's axioms?

Anonymous

Anonymous Sat 09 May 2020 14:19:22 No.11653277 Report

Quoted By: >>11653284

>>11653269
define 0.999... using peano axioms first

Anonymous

Anonymous Sat 09 May 2020 14:22:12 No.11653281 Report

Quoted By:

>>11653221

The proof is left as an exercise to the reader

Anonymous

Anonymous Sat 09 May 2020 14:23:05 No.11653284 Report

Quoted By: >>11653287 >>11653291 >>11653300 >>11653322

>>11653277
>define 0.999... using peano axioms first
Since the logical strength of PA is ?, you're going to have a hard time defining reals. You need at least PA2 for Dedekind cuts IIRC.

If your point is that anything stronger than arithmetic doesn't exist, then truth is PA + AC is already pretty fucked up.

Anonymous

Anonymous Sat 09 May 2020 14:25:15 No.11653287 Report

Quoted By: >>11653301 >>11653322

>>11653284
my point is that it's impossible to prove 0.999... = 1 using peano axioms, because 0.999... is a symbol which cannot be defined using peano axioms

Anonymous

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Anonymous Sat 09 May 2020 14:25:25 No.11653288 Report

Quoted By: >>11653298

>>11653119

im currently homeless can you send it to my postal box instead?

Anonymous

Anonymous Sat 09 May 2020 14:27:43 No.11653291 Report

Quoted By:

>>11653284
maybe real numbers don't exist then

Anonymous

Anonymous Sat 09 May 2020 14:30:40 No.11653298 Report

Quoted By: >>11653303 >>11653307

>>11653288
I keep seeing this loli all over 4chan. What's her name?

Anonymous

Anonymous Sat 09 May 2020 14:31:41 No.11653300 Report

Quoted By:

>>11653284
all natural numbers can be defined with the peano axioms
since 0.999... cant be defined with them its obviously not a natural number

Anonymous

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Anonymous Sat 09 May 2020 14:31:49 No.11653301 Report

Quoted By:

>>11653287
>my point is that it's impossible to prove 0.999... = 1 using peano axioms, because 0.999... is a symbol which cannot be defined using peano axioms

Anonymous

Anonymous Sat 09 May 2020 14:32:44 No.11653303 Report

Quoted By:

>>11653298
I will tell you if you admit that 0.999... =|= 1

Anonymous

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Anonymous Sat 09 May 2020 14:34:05 No.11653307 Report

Quoted By: >>11653316

>>11653298

Anonymous

Anonymous Sat 09 May 2020 14:38:25 No.11653316 Report

Quoted By: >>11653318

>>11653307
Quit being so smug, your shirt is too big for you

Anonymous

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Anonymous Sat 09 May 2020 14:40:17 No.11653318 Report

Quoted By:

>>11653316

Anonymous

Anonymous Sat 09 May 2020 14:44:01 No.11653322 Report

Quoted By: >>11653333

>>11653287
>>>11653284 (You)
>my point is that it's impossible to prove 0.999... = 1 using peano axioms, because 0.999... is a symbol which cannot be defined using peano axioms

Actually, this is not even true. You can define computable real numbers in PA as some Turing machine number, plus additional properties. Then you define the quotient relation, and show that any computable decimal expansion is a real. In particular, 0.999... is trivially computable, and it's easy to show that it is equivalent to 1.

Anonymous

Anonymous Sat 09 May 2020 14:47:37 No.11653333 Report

Quoted By:

>>11653322
>>11653322
>Actually, this is not even true
it's possible, I'm a total brainlet when it comes to logic.
I'm pretty sure it doesn't change that 0.999... is 1 though.

Anonymous

Anonymous Sat 09 May 2020 15:00:21 No.11653360 Report

Quoted By: >>11653365 >>11653366 >>11653367 >>11653374 >>11653379 >>11653391 >>11653395

Here's my a proof that 0.999999... is not 1.

Suppose we want to prove that two numbers, a and b, are the same number. Then clearly for any function f, we should have that f(a) = f(b).

Let's define f to be the function that takes in a real number and outputs the tens digits. f(1) = 0, f(0.99...) = 9.

Therefore, 0.99... != 1.

Anonymous

Anonymous Sat 09 May 2020 15:02:24 No.11653365 Report

Quoted By:

>>11653360
wat

Anonymous

Anonymous Sat 09 May 2020 15:02:35 No.11653366 Report

Quoted By:

>>11653360
HOLY SHIT
B A S E D

Anonymous

Anonymous Sat 09 May 2020 15:02:41 No.11653367 Report

Quoted By: >>11653370

>>11653360
this assumes that every number has a unique decimal expansion

which isn't true because for example 0.999...=1

Anonymous

Anonymous Sat 09 May 2020 15:03:38 No.11653370 Report

Quoted By: >>11653375 >>11653379 >>11653389

>>11653367
circular argument

Anonymous

Anonymous Sat 09 May 2020 15:04:10 No.11653374 Report

Quoted By:

>>11653360
can you prove the function is well-defined ?

Anonymous

Anonymous Sat 09 May 2020 15:05:09 No.11653375 Report

Quoted By:

>>11653370
>>11653263

Anonymous

Anonymous Sat 09 May 2020 15:07:33 No.11653379 Report

Quoted By:

>>11653370
>circular argument
yes, but on >>11653360's side

Anonymous

Anonymous Sat 09 May 2020 15:08:12 No.11653381 Report

Quoted By:

>>11653094
Does 1=1.0?

Anonymous

Anonymous Sat 09 May 2020 15:11:01 No.11653389 Report

Quoted By: >>11653391

>>11653370
Not a circular argument, a false proof of a statement (>>11653350) isn't true if that statement is false (the statement being 0.999?1).

Anonymous

Anonymous Sat 09 May 2020 15:13:00 No.11653391 Report

Quoted By:

>>11653389
ouch that link wrong tho

correct link: >>11653360

Anonymous

Anonymous Sat 09 May 2020 15:15:01 No.11653395 Report

Quoted By: >>11653399 >>11653402 >>11653406

>>11653360
>>11653360
LMAO

> f(0.99...) = 9.
No.

$0.999... = lim_{\space n \to \infty} \sum ^n _{i = 1} {9 \over 10^i}$

->

$f(lim_{\space n \to \infty} \sum ^n _{i = 1} {9 \over 10^i}) [\math] ?$

Anonymous

Anonymous Sat 09 May 2020 15:16:15 No.11653399 Report

Quoted By:

>>11653395
circular argument retard

Anonymous

Anonymous Sat 09 May 2020 15:16:45 No.11653402 Report

Quoted By:

>>11653395
$f(lim_{\space n \to \infty} \sum ^n_{i = 1} {9 \over 10^i})$ ?

Anonymous

Anonymous Sat 09 May 2020 15:17:22 No.11653406 Report

Quoted By:

>>11653395
Even if f(0.999....)=9, the proof doesn't hold because it assumes that every number has a unique decimal expansion.

i.e. it assumes that 0.999...?1 to prove 0.999?1

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