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Guys is this TRUE?!!!!

Anonymous Sat 12 Dec 01:50:20 2020 No.12455304 View Reply Original Report

Quoted By: >>12455580 >>12456296 >>12456345

under Q
(1) q_1 + q_2 = q_3 \in Q
(2) q_1/q_2 = q_3 \in Q
I mean do you guys really think these rules can be true about rational numbers?
>inb4 people criticize q_n/0, we're not talking about hitomi's number in this thread, we don't need to talk about that right now.

Anonymous

Anonymous Sat 12 Dec 2020 03:14:03 No.12455530 Report

Quoted By: >>12455914

One possible answer:
q1 = 1/2
q2 = -1
q3 = -1/2

Anonymous

Anonymous Sat 12 Dec 2020 03:34:21 No.12455574 Report

Quoted By:

Hello guys this is Youbuntu my sub-Saharan friend, say something Youbuntu.
EQULIBRUM
Ugh, what did you say Youbuntu I didn't quite get that?
I say A BIG EQULIBRUM
An equilibrium? but how is that...
ME SMARTER THAN YOU
Haha Youbuntu is soo eloquent, right guys???

Anonymous

Anonymous Sat 12 Dec 2020 03:36:46 No.12455580 Report

Quoted By: >>12455914 >>12456092

>>12455304
>guys are closure axioms true?
yes, they're axioms

Anonymous

Anonymous Sat 12 Dec 2020 06:15:36 No.12455914 Report

Quoted By: >>12456092 >>12456443

>>12455530
they aren't related they are separate axioms.
>>12455580
THEY ARE TRUE?
so then would any combination of them be true??
like an algorithm of 1,2,1,1?

Anonymous

Anonymous Sat 12 Dec 2020 07:52:06 No.12456092 Report

Quoted By:

>>12455914
>>12455580
WELLL!???
C'MON GUY I JUST WANT TO UNDERSTAND

Anonymous

Anonymous Sat 12 Dec 2020 08:27:36 No.12456143 Report

Quoted By: >>12456186

You can prove it if you accept the definition of numbers and how to compute them.
Do you know the definition of rational number?
It's the key of understanding this problem.
If you don't know you may look it up.

Anonymous

Anonymous Sat 12 Dec 2020 08:58:15 No.12456186 Report

Quoted By: >>12456212

>>12456143
I thought it was the set of number that were consistent with addition and multiplication and their inverses.

Anonymous

Anonymous Sat 12 Dec 2020 09:14:22 No.12456212 Report

Quoted By: >>12456226

>>12456186
I think that is the explanation of the field.
Yes, Q is a field. But before stating that, You should prove the axioms of fields in Q and one of them is Closure axioms OP refers to.

Anonymous

Anonymous Sat 12 Dec 2020 09:18:31 No.12456226 Report

Quoted By: >>12456231

>>12456212
you can't prove axioms.

Anonymous

Anonymous Sat 12 Dec 2020 09:21:49 No.12456231 Report

Quoted By: >>12456236

>>12456226
I mean, you should check that any q in Q satisfy the axioms.

Anonymous

Anonymous Sat 12 Dec 2020 09:23:51 No.12456236 Report

Quoted By: >>12456246

>>12456231
I'm pretty sure that's not possible.

Anonymous

Anonymous Sat 12 Dec 2020 09:29:59 No.12456246 Report

Quoted By: >>12456291

>>12456236
Well, check it together.
The definition of rational number is
Q = { a / b | a in Z, b in Z\{0}}
Is this OK?

Anonymous

Anonymous Sat 12 Dec 2020 09:56:40 No.12456291 Report

Quoted By: >>12456321

>>12456246
I believe it
so if a, b in Z where a/b defines an element of Q say q
then q1 + q2 = a/b + c/d = a/b * d/d + c/d * b/b = ad + cd / bd
where ad + cd in Z and bd in Z.
thus z1 / z2 holds true
does this prove it for all rationals?

Anonymous

Anonymous Sat 12 Dec 2020 09:58:19 No.12456296 Report

Quoted By:

>>12455304
Is that senator armstrong and the heckin valid and cute enby girl?

Anonymous

Anonymous Sat 12 Dec 2020 10:16:02 No.12456321 Report

Quoted By: >>12456345

>>12456291
>does this prove it for all rationals?
Yes, if you accept my the definition of Q.
you may accept ANY q in Q is a form of a / b (a in Z, b in Z \ {0}).
As you show me, the sum of any two rational numbers q1,q2 is a rational number.
This implies that if you take two random rational numbers from Q and add them, it always becomes rational number.

Anonymous

Anonymous Sat 12 Dec 2020 10:27:18 No.12456345 Report

Quoted By: >>12456409 >>12456443

>>12456321
so is an algorithm based on there rules >>12455304
a rational number too?

Anonymous

Anonymous Sat 12 Dec 2020 10:52:08 No.12456409 Report

Quoted By: >>12457213

>>12456345
Yeah, you are keen.
I should've mentioned binary operation + and * is defined on Q.
A binary operation is a kind of map(or function)like this.
+:S x S -> T
The Same goes with *.
In this case, I assumed + and * is an operator that we use ordinary.

Anonymous

Anonymous Sat 12 Dec 2020 11:07:34 No.12456443 Report

Quoted By: >>12457213

>>12456345
Or you mean
>>12455914
>so then would any combination of them be true??
>like an algorithm of 1,2,1,1?
Yes.
I assume 1 is addition and 2 is multiplication.
q1 + q2 = q1' q1' is in Q.
q'1*q'2 = q1'' q1'' is in Q.
q1'' + q2'' = q1''' q1''' is in Q.
q1''' + q2''' = q q is in Q.

In other words, the sum of two rational numbers is a rational number. So if you add or multiplicate it with a rational number, it becomes a rational number.

Anonymous

Anonymous Sat 12 Dec 2020 17:25:57 No.12457213 Report

Quoted By: >>12458316 >>12458384

>>12456409
>>12456443
hmmmmmmmmmmmmmmmmmm *doge hmmmmm face*
SOOOOOOO
if we were to agree 1/1 = 1 is a rational number,
then we would agree that the series of 1/2^n is a rational number algorithm, because it's comprised of q1 + q2 + ...
thus it's an algorithm such that A_1 := 1.
because we know it ends at hitomi's count limit.
SOOOOOOOOOOOOOO
now that we have that out of the way, we can use algorithms to express many ideas.
Such as an algorithm that uses bisections to approach sqrt(2), though there is infinite amount of algorithms for this in different ways and orders, it's generally true that A_sqrt(2) := sqrt(2)
under A_Q:1,2 this is trivially true.
And thus it is defined under rationals.
and we can agree that A_pi and A_e exists too.
reals once again BTFO

Anonymous

Anonymous Sat 12 Dec 2020 23:55:05 No.12458316 Report

Quoted By: >>12458351

>>12457213
>if we were to agree 1/1 = 1 is a rational number,
Yeah

>then we would agree that the series of 1/2^n is a rational number algorithm, >because it's comprised of q1 + q2 + ...
Yeah

>because we know it ends at hitomi's count limit.
What's hitomi's count?
And if you assume infinite series, the series never ends because it is infinite!

>Such as an algorithm that uses bisections to approach sqrt(2), though there is infinite amount of algorithms for this in different ways and orders, it's generally true that A_sqrt(2) := sqrt(2)
Some series approaches to sqrt(2) but is not equal to sqrt(2).
You should study analysis or even only limit concept.
And you should use an natural number for an index

>under A_Q:1,2 this is trivially true.
>A_Q:1,2
What does this notation mean?

>And thus it is defined under rationals.
>and we can agree that A_pi and A_e exists too.
If you mean A_pi is a series that converges to pi and A_e is a series that converges to e.
I agree though I'm not into numerical analysis.

>reals once again BTFO
NO. There's only a minunderstanding.

Anonymous

Anonymous Sun 13 Dec 2020 00:09:01 No.12458351 Report

Quoted By: >>12458362

>>12458316
>What's hitomi's count?
last number in N.
h is also a part of rationals because that's where I discovered it.
because it exists under Q 1/2^n at h will equal 0.
the count limit says an algorithm will end with a finite value.
real indefinitely BTFO'd

Anonymous

Anonymous Sun 13 Dec 2020 00:14:56 No.12458362 Report

Quoted By: >>12458372

>>12458351
>last number in N.
There's no last number(or the largest number) in N.
Look it up for the definition of natural number.

>h is also a part of rationals because that's where I discovered it.
>because it exists under Q 1/2^n at h will equal 0.
>the count limit says an algorithm will end with a finite value.
>real indefinitely BTFO'd
And please read the rest of my post.

Anonymous

Anonymous Sun 13 Dec 2020 00:20:28 No.12458372 Report

Quoted By: >>12458443

>>12458362
It's the limiting number.
If it doesn't exist then there is no infinite density of numbers.
and then n/n+1 doesn't equal one in the neighborhood of infinity.
the rest of your post didn't matter because it was all based on a misunderstanding.

Anonymous

Anonymous Sun 13 Dec 2020 00:26:14 No.12458384 Report

Quoted By: >>12458401 >>12458419

>>12457213
ive honestly never heard of hitomis count before and cant seem to find anything on the net about it.
can you link to some source?

Anonymous

Anonymous Sun 13 Dec 2020 00:33:02 No.12458401 Report

Quoted By:

>>12458384
Check the math talk archives on /sci/
hitomi's number was discovered on /sci/

Anonymous

Anonymous Sun 13 Dec 2020 00:41:13 No.12458419 Report

Quoted By:

>>12458384
https://warosu.org/sci/?task=search&ghost=&search_text=hitomi%27s+number

Anonymous

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Anonymous Sun 13 Dec 2020 00:51:08 No.12458443 Report

Quoted By: >>12458451

>>12458372
Ah, I was dealing with a schizo.
I thought he is some naive kid or at worst some troll.
I'm done. bye.

Anonymous

Anonymous Sun 13 Dec 2020 00:54:34 No.12458451 Report

Quoted By:

>>12458443
I am 18. And it's true you know.

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