/sci/ - Science & Math » Thread #11859893

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Anonymous Thu 02 Jul 12:28:38 2020 No.11859893 View Reply Original Report

Quoted By: >>11859903 >>11859906 >>11859966 >>11859999

if A=C and C=B then A=B

Anonymous

Anonymous Thu 02 Jul 2020 12:34:07 No.11859903 Report

Quoted By:

>>11859893
Transitivity level 1, good job ball sack head

Anonymous

Anonymous Thu 02 Jul 2020 12:35:51 No.11859906 Report

Quoted By: >>11859913

>>11859893
Set theory is the most important topic in mathematics.

Anonymous

Anonymous Thu 02 Jul 2020 12:37:22 No.11859913 Report

Quoted By:

>>11859906
Category Theory obstructs your route to a destination

Anonymous

Anonymous Thu 02 Jul 2020 13:03:20 No.11859966 Report

Quoted By: >>11860307 >>11861073

>>11859893
Let x be the function f(x) = f^2. Then f'(x) = 2f

Anonymous

Anonymous Thu 02 Jul 2020 13:13:35 No.11859999 Report

Quoted By: >>11860019 >>11860135 >>11860206 >>11860315 >>11860565 >>11861065

>>11859893
Prove it

Anonymous

Anonymous Thu 02 Jul 2020 13:20:24 No.11860019 Report

Quoted By: >>11860060

>>11859999
let A = 1, B = 1, C =1
A=C, 1=1
C=B, 1=1
A=B, 1=1
QED.

Anonymous

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Anonymous Thu 02 Jul 2020 13:31:19 No.11860060 Report

Quoted By: >>11860092

>>11860019
That's only proof for A,B,C = 1

Is there a generalized consistent proof? Godel thinks not.

Anonymous

Anonymous Thu 02 Jul 2020 13:48:13 No.11860092 Report

Quoted By:

>>11860060
Let A=x and B=y and C=z
If A=C then x=z
If C=B then z=y
So then x=y meaning that A=B
Someone check me because I'm not good at proofs

Anonymous

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Anonymous Thu 02 Jul 2020 13:50:10 No.11860096 Report

Quoted By:

>if A=C and C=B then A=B

Anonymous

Anonymous Thu 02 Jul 2020 13:51:45 No.11860102 Report

Quoted By: >>11860300

disproof by counterexample
Euler is the goat.
The goat is a subspecies of Capra aegagrus.
Euler is a subspecies of Capra aegagrus.
conjecture rekt

Anonymous

Anonymous Thu 02 Jul 2020 14:06:27 No.11860135 Report

Quoted By:

>>11859999
take A=C, and because C=B you can substitute C for B and get A=B

checkmate nerd

Anonymous

Anonymous Thu 02 Jul 2020 14:33:00 No.11860206 Report

Quoted By:

>>11859999
A=C & C=B
A=C=B
A(=C)=B
A=B(=C)
A=B

Anonymous

Anonymous Thu 02 Jul 2020 15:14:30 No.11860300 Report

Quoted By:

>>11860102
massive if true

Anonymous

Anonymous Thu 02 Jul 2020 15:16:45 No.11860307 Report

Quoted By:

>>11859966
Actually, f^2 is a constant.

Anonymous

Anonymous Thu 02 Jul 2020 15:21:01 No.11860315 Report

Quoted By:

>>11859999
A=B=C

Anonymous

Anonymous Thu 02 Jul 2020 15:43:05 No.11860360 Report

Quoted By:

Wrong premise. A and B are clearly different letters.

Anonymous

Anonymous Thu 02 Jul 2020 16:44:42 No.11860565 Report

Quoted By:

>>11859999
let A=C and C=B but A=/=B
Because C=B, we can write A=/=C, which is wrong according to (A=C), so we get:
A=B

Anonymous

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Anonymous Thu 02 Jul 2020 19:35:58 No.11861065 Report

Quoted By:

>>11859999
= is an equivalence relation, and equivalence relations are transitive

El Arcón

El Arcón Thu 02 Jul 2020 19:37:54 No.11861067 Report

Quoted By:

>if A=C and C=B then A=B
statement of the transitivity of the equivalence relation (which is defined to be transitive)

Anonymous

Anonymous Thu 02 Jul 2020 19:39:31 No.11861073 Report

Quoted By:

>>11859966
hmmmmm

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