/sci/ - Science & Math » Thread #14012288

66KiB, 2126x626, Capture.png

View Same Google iqdb SauceNAO

REALLY? THIS IS THE FOUNDATIONS OF EVERYTHING THAT MATHEMATICIANS AGREED ON??

Anonymous Sat 25 Dec 20:49:47 2021 No.14012288 View Reply Original Report

Quoted By: >>14012295 >>14012406 >>14014311 >>14015415 >>14017293 >>14019007

Here's what I can prove with these so called foundations

4?7
2?15
5?23=5
1?(1,6) but 6?(1,6)

Incomprehensible garbage. Pathetic.

Anonymous

Anonymous Sat 25 Dec 2021 20:51:59 No.14012295 Report

Quoted By: >>14012314

>>14012288
>6?(1,6)
y not?

Anonymous

View Same Google iqdb SauceNAO Capture.png, 31KiB, 902x131

Anonymous Sat 25 Dec 2021 20:57:22 No.14012314 Report

Quoted By: >>14012386 >>14012406 >>14012463 >>14019007

>>14012295
My god man. Your "foundations" only has a single operation, membership, and you bungled it. No wonder mathematics is in such a mess.

Note, you will need to apply both the definition of integers and the definition of an ordered pair. Can you accomplish this? Doubtful.

Anonymous

View Same Google iqdb SauceNAO calvin-hobbes.jpg, 69KiB, 1200x390

Anonymous Sat 25 Dec 2021 21:13:53 No.14012369 Report

Quoted By: >>14012446 >>14012451

no no no
you add numbers to the signature as constants
then you add the binary operation symbols for the numbers like plus and minus
you also need a unary operation successor for the natural numbers
so you have
0 - constant
S(.) - successor
a + b a.k.a. plus(a,b) - plus
a - b a.k.a. minus(a,b) - minus
now you add an axiom for the natural numbers
there exists a set Na such that 0 <- Na and
S(n) <- Na for all n <- Na
you also need another axiom that says the natural numbers are equal to the intersection of all subsets of Na that have this property
next you have to construct the direct product of the natural numbers, Na x Na
then define the set for the function plus
plus = { (a,b,c) : a,b,c <- Na and a + b = c }
that's just for notation; if you write it out it's a circular definition, i.e. tautology
in order to get a math definition, use induction like above for the natural numbers, but do it for one base case and two inductive cases
(0,0,0) <- plus - assume this as an axiom
if (a,b,c) <- plus then (S(a),b,S(c)) <- plus
if (a,b,c) <- plus then (a,S(b),S(c)) <- plus
furthermore plus is the intersection of all subsets of plus that have this property
this is how you build up the operations on natural numbers

Anonymous

Anonymous Sat 25 Dec 2021 21:18:22 No.14012386 Report

Quoted By: >>14012459

>>14012314
no, this is a pedagogical bug that is acknowledged in Halmos - Naive Set Theory
what Halmos says is correct
the reason he says it is because it wasn't common at the time to teach the model theory proof style; the standard had not yet been created to make this really sturdy and iron-clad
now we can say that ordered pair is simply a binary function symbol
it's that simple
if you protest, introduce a trinary relation symbol
P(x,y,z) to stand for (x,y) = z
now introduce axioms that say ordered pairs exist and are unique
these are simply the admissibility conditions; it's only until a very early stage of development of model theory that you see any difference between these

Anonymous

Anonymous Sat 25 Dec 2021 21:22:29 No.14012406 Report

Quoted By: >>14012431

>>14012288
>>14012314
>1?(1,6)
Wrong

Anonymous

Anonymous Sat 25 Dec 2021 21:29:13 No.14012431 Report

Quoted By: >>14012450 >>14012459

>>14012406

1 = {O}

(1, b) = {{O}, {{O}, b}}

so

1?(1,b) but for b != 1, b?(1,b)

Just definitions. The state of this board....
wtf?

Anonymous

Anonymous Sat 25 Dec 2021 21:33:37 No.14012446 Report

Quoted By:

>>14012369
>b a.k.a
SUSSY BAKA

Anonymous

Anonymous Sat 25 Dec 2021 21:35:10 No.14012450 Report

Quoted By: >>14012457

>>14012431
>(1, b) = {{O}, {{O}, b}}
Wrong

Anonymous

Anonymous Sat 25 Dec 2021 21:35:13 No.14012451 Report

Quoted By: >>14012466

>>14012369
How does this invalidate any of my original "theorems"? You just added additional operations, no constraints. I can still haphazardly apply the membership operator whenever I want to.

Now we have the power to say (1 + 1)?7!

You have fixed nothing.

Anonymous

Anonymous Sat 25 Dec 2021 21:37:53 No.14012457 Report

Quoted By: >>14012462

>>14012450
{O} ? {{O}, {{O}, b}} same as 1 ? {1, b}

Anonymous

Anonymous Sat 25 Dec 2021 21:38:26 No.14012459 Report

Quoted By:

>>14012431
see
>>14012386

Anonymous

Anonymous Sat 25 Dec 2021 21:38:46 No.14012462 Report

Quoted By:

>>14012457
(1, b) = {{1}, {1, b}}, not {1, {1, b}}

Anonymous

Anonymous Sat 25 Dec 2021 21:39:20 No.14012463 Report

Quoted By: >>14012473

>>14012314
My definition of ordered pair is (a,b) = {{{a},{b},{a,b}},{{a},{a,b},{a,b,{a,b}}}, 6} so I'm confused since 6 is a member of (1,6)

Anonymous

Anonymous Sat 25 Dec 2021 21:39:42 No.14012466 Report

Quoted By:

>>14012451
you used the wrong definitions
the theorems aren't wrong
you simply used the symbols to refer to the wrong thing

Anonymous

Anonymous Sat 25 Dec 2021 21:40:10 No.14012468 Report

Quoted By: >>14012479 >>14017311

Why can't abstract algebra be a foundation of mathematics? Constructing the field of divisors of an integral domain was pretty simple compared to the set-theoretic construction of Q.

Anonymous

Anonymous Sat 25 Dec 2021 21:41:30 No.14012473 Report

Quoted By:

>>14012463
ordered pairs are undefined constants
we can prove using grammatical arguments that there are no theorems of the form a <- (b,c)
likewise the constant 0, the least natural number, was introduced as an undefined constant

Anonymous

Anonymous Sat 25 Dec 2021 21:42:51 No.14012479 Report

Quoted By:

>>14012468
it is, if you consider formal systems to be a subfield of monoid theory

Anonymous

Anonymous Sun 26 Dec 2021 10:23:01 No.14014311 Report

Quoted By:

>>14012288
>jew von Neumann basically defined all modern mathematics
oh no...

Anonymous

Anonymous Sun 26 Dec 2021 16:59:59 No.14015415 Report

Quoted By:

>>14012288
>there is only one foundation considered in modern math
You gotta lot to learn, buddyboyo

Anonymous

Anonymous Mon 27 Dec 2021 04:48:55 No.14017293 Report

Quoted By:

>>14012288
ZFC isnt even universally adopted as our set theoretic axioms, they agree on even less

Anonymous

Anonymous Mon 27 Dec 2021 04:57:01 No.14017311 Report

Quoted By:

>>14012468
usually someone deriving a foundation works by starting with a couple set theoretic axioms, then derives either the real or complex numbers and then starts deriving the concepts of fields, groups, and isomorphisms so AA typically comes second or third depending on what taxonomy of sub-disciplines youre using

Anonymous

Anonymous Mon 27 Dec 2021 15:39:44 No.14019007 Report

Quoted By:

>>14012288
>>14012314
It is merely a construction, nobody uses the set theoretic definition unless declaring it beforehand, you can also define: 0 := ø and 1 = {0}, 2 = {1}, 3 = {2} and so on, but the von Neumann definition is useful for defining ordinals.
If you wanted to be consistent you can also criticize the formal definition of quotient set (under equivalence relations) because, under that definition, you have that +1 = { (1, 0), (2, 1), (3, 2), ... } therefore
>omg guys you are so stupid, if the definition of the integers are such then (4, 3) ? +1 lawlz xd

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches