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/sci/ - Science & Math


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7995787 No.7995787 [Reply] [Original]

If you really are an academic then you should be able to prove this.

Take your time, I'll wait.

>> No.7995793

>>7995787
Can't be proven, but
[math]\Delta 1 \Delta 1 \ge \hbar m[/math]
gives us a lower bound.

>> No.7995795

We start with the axiom that by definition x:=x

Hence, by substituting x = 1 into the equation we get 1 := 1 which degenerates into 1 = 1

>> No.7995802
File: 132 KB, 900x900, megamune.jpg [View same] [iqdb] [saucenao] [google]
7995802

>>7995793

>> No.7995809

>>7995787
Things equaling themselves is an axiom. You assume to be true you dont need to prove it

>> No.7995810

>>7995795
>proof by example
>>/trash/

>> No.7995814

>>7995809
Is there a place where the axiom of identity fails?

>> No.7995819

>>7995814
it's a primitive of logic. so you would need to go beyond logic.

realize what you're asking. when you need to deal with the axioms of math you go out of math and into logic. when you need to deal with the axioms of logic, where do you go? the answer is nowhere I think.

>> No.7995934

>>7995810
There weren't any examples in that post.

>>7995814
>Is there a place where the axiom of identity fails?
Axioms can't "succeed" or "fail". They're just part of the definition of a system.

>> No.7995941

>>7995787
1 = 1

proof: my 1 bbc fits into your 1 bbc sized asshole.

feel free to find a flaw.

>> No.7995942

>>7995795
this is the proof. this anon knows their stuff.

>>7995934
well worded

>>7995787
>OP gets shit on in the second post
>already butthurt

>> No.7995944

>>7995934
Yeah but I mean is there a proof or argument in some higher formalism that is capable of causing the axiom of identity to be the one logical fallacy of a proof-by-contradiction?

>> No.7995947

Suppose x = y, then x-y = 0
Let x = 1, then y = 1
1-1 = 0, therefore 1 = 1

>> No.7995950

OR

Suppose not, that is 1 != 1, then 1-1 != 0, but we have a contradiction because x-x = 0

>> No.7995951

>>7995947
>Suppose
>=
>Let
>=
You can't use equality to prove equality. >>7995795 correctly used:
>:=

>> No.7995958
File: 244 KB, 718x1024, 1426518206695.jpg [View same] [iqdb] [saucenao] [google]
7995958

I haven't constructed the numbers in a while. seems like a good opportunity to practice /sci/ latex

I'll work through it a few ways

Without Peano first:

Define the set of primitive natural numbers [math] \mathbb{N} [/math] as a set containing the empty set, and for each element [math] x [/math] in [math] \mathbb{N} [/math], the set [math] \{ x \} [/math] is also in [math] \mathbb{N} [/math]
so now [math] \mathbb{N} [/math] looks like [math] \{ \emptyset , \{ \emptyset \} , \{ \{ \emptyset \} \} \ldots \} [/math]
now define a function
[math] \mathbf{S} : \mathbb{N} \longrightarrow \mathbb{N} [/math]
by
[math] \mathbf{S}(x) \mapsto \{x\} [/math]
this is called the successor function, and in terms of numbers, it's a function that just adds 1 to x
you can define the natural numbers without the following notation, i'm just using it for convenience
the reason you would want to avoid this in a rigorous construction seems obvious: you're using numbers when they haven't been made yet
anyway
define [math] x^{n} \text{ for } x \in \mathbb{N} [/math] like this:
[math] x^{0} = x, ~x^{n} = \mathbf{S}(x^{n-1}) [/math]
this is just a convenient way to denote repeated succession
now define
[math] + : \mathbb{N}^{2} \longrightarrow \mathbb{N} [/math]
as such:
[math] (x+y) \overset{ \underset{ \mathrm{ def } } { } } { = } (+(x,y)) \mapsto x^{ y } [/math]
this reads: x+y is definitionally equal to +(x,y). x+y is the result of calling the successor function on x, y times

the natural numbers together with this addition operator form a semigroup, which isn't too important yet, but should be kept in mind
now if you're a pleb you'll probably have a hard time with the next part, where i construct the integers

the reason i'm constructing the numbers here is to provide a construction and show that it fits with your intuition properly, so i can prove that 1 = 1 intuitively

post is too long, should i continue or go to sleep?

>oh lord odin please let my latex work

>> No.7995962

>>7995951
Semantics

>> No.7995969

>>7995787
My teacher told me it is true. I believe my teacher, because he is an academic.

>> No.7995975

>>7995969
This is the only defensible reason to believe anything

>> No.7995980

>>7995962
This is a semantics thread. This is the one place you're not allowed to fail at semantics.

>> No.7995988

>>7995787
The = relation is reflexive.

>> No.7996005

>>7995787
That's not true though.
As we all know,
1=0.999...

>> No.7996069
File: 142 KB, 907x960, triple fields are the new triple integrals.jpg [View same] [iqdb] [saucenao] [google]
7996069

well i'm off to bed, but i'll just go ahead and say
if you've defined the numbers like i did here >>7995958
and assigned the symbol "1" to the element [math] \{ \emptyset \} [/math]
then 1 = 1 follows directly from the axiom of extensionality

[math] \forall{A}. \forall{B}. (\forall{C}.(C \in A \iff C \in B) \implies A=B) [/math]

which says "if you take any sets A and B, and if for every set C, C is in A only if C is in B, then A is equal to B"
which, put less autistically, says "if A and B have the same elements, then they are equal"

so there's my first proof that 1 = 1

i get bored in class and do a lot of messing around with foundations and low-level stuff so i've got a lot more of this if anybody's interested

i kinda want to post it all TЬH but i'd rather post it in a thread where there's interest
i have category theoretic and type theoretic proofs that 1=1 as well, but they're pretty much the same as that proof up there^

also can somebody tell me where pic related is from?
i've image searched it and stuff but i can't find anything
i just want to know why the dodecahedron group is off by itself as if it's as important or even as interesting as

>> No.7996360

>>7995947
You don't understand logic or implications.

>> No.7996376

= is reflexive, Q.E.D.

>> No.7996382

>>7996376
Thx for the copypasta, see
>>7995988

>> No.7996427

>>7995819
philosophy

>> No.7996437

>open 4chan
>this post is on the front page

>> No.7996438

>>7995944
>some higher formalism
Is there a higher formalism than logic? No

>> No.7997707

>>7996438
Do you even know what that word means? If logic is truly just formalism then there is inherently no logic until we define it as such

Anyways, all this (including logic itself and the axoims it relies on, and thus mathematics) comes from philosphy

>> No.7997867

>>7995787
But it's false.
1 = .999...

>> No.7997948

>>7995958

The successor of x is xu{x} though, not {x}.

>>7996069

I think that picture's from Max Tegmark's Our Mathematical Universe

>> No.7997966

>>7996069
> no arrow from groups to linear operators
> no arrow from something on the topological side to to manifolds

What are you doing senpai?

>> No.7997977
File: 124 KB, 600x900, 1458584114071.jpg [View same] [iqdb] [saucenao] [google]
7997977

>>7997867
wow!

>> No.7998123

>>7997948
why is successor defined like that though?

looking at the definition in >>7995958 , it seems like it would work just fine

>> No.7998181

>>7995787
I have an apple. My friend has an apple. We have the same number of apples. As proof, if he gives me his apple, I'll have two apples, or double the number of apples i had in the first place (x+x=2x). Could write more, but only if OP remains unconvinced.

>> No.7998317

X^0=1
X/X=1
0!=1

Look into the proofs behind these things to find out the significance of the number one.

Make one proof equal another and then you can prove 1=1 or something like that

Here I'll get you started down this rabbit hole

https://www.youtube.com/watch?v=Mfk_L4Nx2ZI

>> No.7998455

>>7995958
Kinda offtopic, but I have a question for mathfags here.
I was once asked by some stoner if I could prove numbers were real. I used this example (simplified and using reality too much=, that I now think is more or less Peano arithmetic.
I showed him a coin, then I took it away. I then said that objects had a property that we observe, and that is the property that changes from a coin being and not being. He agreed with that bullshit so I guess I didnt fuck up too much.
Then I said I was making a set of objects, containing no object being there, and calling it 0. I claimed this was part of a group which I called numbers (just defined, no other set on it I guess). I then also defined that, for every number, there is a next one (successor), and that 0 is next to no one.
How good is this layman's proof? Also, what's the difference between Peano and common arithmetic?

>> No.7999030

>>7998123
not him but the reason we use Sx=xu{x} is because it allows for sets which go beyond the natural numbers (further reading: Von Nuemann Ordinals).
so a set is ordinal if all its elements are ordinal.

clearly all natural numbers defined as xu{x} are ordinal.

but also the set of all finite ordinals is ordinal (let this be w).

what about the set w u { w } ?? also ordinal. ect.

if sx = {x} then you can only have finite ordinals because sets cannot have infinite layers of sets

>> No.7999048

I thought this board was for science, not philosophy.

>> No.7999060

>>7998317
>numberphile
lol

>> No.7999094

>>7998455
>I was once asked by some stoner if I could prove numbers were real.
Until you define what "real" means this is nonsensical, because odds are that any two people have different ideas of what it means for numbers to be real.

>> No.7999119

>>7995795
Assuming you're working over first-order logic and ":=" is the formal nonlogical symbol usually written as [math]\overdot{=}[/math], then yes, it follows from the equality axioms as you stated.

>> No.7999393

>>7999060
Okay, without watching that video, explain in your own words why 0!=1

>> No.7999400

1=1
1-1=0
0=0

>> No.7999466

>>7999393
The only real reason is that deciding so makes a lot of formulas much nicer and simpler to state.

If you want to hack together a sensible explanation, then you notice that if n is a positive integer, n! counts the number of ways to order n objects, and the only way to order 0 objects is to do absolutely nothing.
>inb4 this is what the video says and you bitch at me

>> No.7999484

>>7999466
if that's what the video says then I would commend you for showing me the video offers nothing unique and so it's unnecessary and probably sucks

Unless you're some godly mathematician in which case the video would be saying something only the likes of you could say...

However Idk what you're even saying. What do you mean do absolutely nothing? The first half sounds about the same as what the video is saying. If what you mean is the same, the video explains it much more eloquently and intuitively

>> No.7999549

>>7998123

It gives a nice structure to the natural numbers. Defining succession this way means that every natural number is the set of all natural numbers which "precede" it, meaning "precession" is really just membership. For example, 1={0}, 2={0,1} etc. It also means that every natural number has cardinality equal to the quantity it's supposed to represent.

>> No.7999565

>>7995787
I take it that you're using the natural number 1 in your definition. Then let's define the following system of natural numbers:
(N,{},s) where s is an injection on N and maps x to {x}. I won't bother listing all the properties.
To show that 1=1 it suffices to show that {{}} is a subset of {{}}. Let x be a member of {{}}. Then x is a member of {{}}. Hence, we have our result, and by the antisymmetry of the subset relation, {{}}={{}}. Equivalently, 1=1 since 1:=s(0):=s({})={{}}

>> No.7999601

>>7999048
I am not op,
but this is actually math. Although very close to philosophy .

>> No.7999759

You guys are fucking autistic nerds.
It's just one.

>> No.7999763

>>7999601
math is philosophy. math is not science.

anon didn't read past /sci/ to see that it stands for science AND math.

But yea, math is not science m8

>>7999759
you are just an ignorant pleb. nothing "just is" and everything comes from axioms

>> No.7999924

>>7999393
Γ(1)=1

>> No.8000023

>>7999393
There is one way to arrange zero objects.

>> No.8000058

>>7999400
Nygga you don't understand.

>> No.8000082

>>7999393
The empty product is necessarily 1, as [math] (\prod_{i=1}^0 1 ) \cdot 1 = (\prod_{i=1}^0 1 ) \cdot (\prod_{i=1}^1 1 ) = \prod_{i=1}^1 1 = 1 [/math], whence [math] \prod_{i=1}^0 1 [/math], the empty product, must be the multiplicative identity [math]1[/math].

>> No.8000152

Threadly reminder that the longer a thread is on /sci/ the less /sci/ knows about the topic of the thread

>> No.8000979

>>7995947
Into the trash it goes.

>> No.8000988

>>7996069
That's an autistic way to prove it.
Just use the 2nd Peano axiom.

>> No.8001003

>>7995787

Can you define the " = " operator?

>> No.8001277

>>8000152
Still waiting for someone to prove that 0.999... = 1

>> No.8002071

>>8001277
1/3=0.3333....
3(1/3)=1
3(0.33333....)=1
0.9999...=1

>> No.8002074

>>8002071
Define "1/3" without using ... please.

>> No.8002090

>>8002074
tell that to
>>8001277

BTW 0.999... = 1 is not actually true lol. i was trolling

>> No.8002150

>>8001277
[math]0.999... = 0.9 + 0.09 + 0.009 + ... = 9(0.1 + 0.01 + 0.001 + ...) = 9\sum_{n=1}^{\infty}0.1^n = 9(-1 + \sum_{n=0}^{\infty}0.1^n)[/math]

Since 0.1 < 1,

[math]9(-1 + \sum_{n=0}^{\infty}0.1^n) = 9(-1 + \frac{1}{1-0.1}) = 9(\frac{0.1}{0.9}) = 9(\frac{1}{9}) = 1[/math]

>> No.8002160

>>8001277
If [math]0.999... \ne 1[/math], then we can find a number [math]0.999... < x < 1 [/math], for instance [math]x = \frac {1 + 0.999...} {2}[/math]. But each digit to the right of the decimal in the expansion of [math]x=0.a_1 a_2 a_3...[/math] must be 9. Otherwise, let [math]a_k[/math] be the first non 9. Then [math]0.a_1 a_2 ... a_k - 0.99...9 = -0.00...a'_k[/math], where [math]a'_k = 9-a_k[/math]. Then [math]x - 0.999... = (0.a_1 a_2 ... a_k - 0.99...9 ) + (x - 0.a_1a_2 ... a_k) + (0.999... - 0.99...9) < 0[/math].

>> No.8002162

>>8001277
You really can't, you have to assume .999 rounds up to something which is absurd as it's infinite.

>> No.8002166

>>8002150
>>8002160
lol what. do you know what asymptotes are?

>> No.8002167

Test

>> No.8002218

e^(i pi) - e^(i pi) - e^(i pi) = e^(i pi) - e^(i pi) - e^(i pi)

>> No.8002282

>>7995787
We all agree that '1' represents a specific value, which is to be exact, one, we all agree on to read '1' as one. We also agree on that the '=' symbol is used for representing an equation. Then 1=1 is true, since both '1' symbols have the same exact specific value, which is equal to eachother.

>> No.8002288

>>8002282
that's no better than >>7995795

>> No.8002294

>>8002288
both are correct, and as correct as you can get
identity is a primitive in logic. you can't say anything else about it. equality of two symbols which are the same is always true.

you can't reason about logic from within logic, and there's nothing behind. so what the fuck are you looking for?

>> No.8002299
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8002299

>>8002294

>> No.8002308

>>8002294
So basically, it can't be proved and it's just an axiom

>> No.8002318

>>8002090
Nice meta-trolling.

>> No.8002327

>>8000000

>> No.8002346

>>8002308
Of course

>> No.8002400
File: 120 KB, 350x270, Ok.png [View same] [iqdb] [saucenao] [google]
8002400

>>8002166

>> No.8003965

>>8002162
Τhere are infinite numbers that equal to 1. 1 is the equivalence class of those numbers.

>> No.8003987
File: 19 KB, 443x364, main-qimg-4f6a9133a19b7456071a7636c4795b6f.gif [View same] [iqdb] [saucenao] [google]
8003987

This was actually a question on the British TV Series QI ["Quite Interesting"] a few years ago.
It was "proven" symbolically.

>> No.8004013

>>7995787
According to Barnett's Identity,
[math] 1 = 0.99999... = e^{i \pi} = 1 [/math]

>> No.8004014

>>8004013
[math] - e^{i \pi} [/math] *