/sci/ - Science & Math » Thread #11466150

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Anonymous Fri 13 Mar 15:29:16 2020 No.11466150 View Reply Original Report

Quoted By: >>11466426 >>11466441 >>11466448 >>11467852 >>11467858 >>11470547 >>11470557

Proof elementary arithmetic

Anonymous

Anonymous Fri 13 Mar 2020 17:34:16 No.11466426 Report

Quoted By: >>11466438

>>11466150
Axioms are there to be axioms, not to be proven.

Anonymous

Anonymous Fri 13 Mar 2020 17:39:35 No.11466438 Report

Quoted By:

>>11466426
sounds like cope to me

Anonymous

Anonymous Fri 13 Mar 2020 17:42:13 No.11466441 Report

Quoted By:

>>11466150
Gödel pls

Anonymous

Anonymous Fri 13 Mar 2020 17:45:00 No.11466448 Report

Quoted By:

>>11466150
The set of natural numbers forms a model of the axioms, hence the axioms are consistent and true for the natural numbers.

Anonymous

Anonymous Sat 14 Mar 2020 04:21:34 No.11467852 Report

Quoted By:

>>11466150
Following Apostol's Calculus, take real numbers as primitive terms and construct the natural numbers as the generalized intersection of all real algebraic inductive sets. Peano axioms follow as theorems.

Anonymous

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Anonymous Sat 14 Mar 2020 04:24:25 No.11467858 Report

Quoted By: >>11469166 >>11469185

>>11466150
Principia Mathematica you stupid nigger

Anonymous

Anonymous Sat 14 Mar 2020 17:44:55 No.11469166 Report

Quoted By:

>>11467858
lmao fucking retard

Anonymous

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Anonymous Sat 14 Mar 2020 18:07:03 No.11469185 Report

Quoted By:

>>11467858
go back

Anonymous

Anonymous Sun 15 Mar 2020 01:50:36 No.11470547 Report

Quoted By:

>>11466150
Let:
$0= \emptyset$
$n^{+}=n \cup \{ n \}$
Then:
$1=0^{+} = 0 \cup \{ 0 \} = {0}$
$2=1^{+} = 1 \cup \{ 1 \} = {0,1}$
$3=2^{+} = 2 \cup \{ 2 \} = {0,1, 2}$
Et cetera.

Let the sum:
1. $a+0=a$
2. $a+b^{+}=(a+b)^{+}$

Theorem. $1+1=2$

Proof:
$1+1=1+0^{+}=(1+0)^{+}=1^{+}=2 \square$

Anonymous

Anonymous Sun 15 Mar 2020 01:53:33 No.11470557 Report

Quoted By:

>>11466150
Let:
$0= \emptyset$
$n^{+}=n \cup \{ n \}$
Then:
$1=0^{+} = 0 \cup \{ 0 \} = \{ 0 \}$
$2=1^{+} = 1 \cup \{ 1 \} = \{ 0,1 \}$
$3=2^{+} = 2 \cup \{ 2 \} = \{ 0,1, 2 \}$
Et cetera.

Let the sum:
1. $a+0=a$
2. $a+b^{+}=(a+b)^{+}$

Theorem. $1+1=2$

Proof:
$1+1=1+0^{+}=(1+0)^{+}=1^{+}=2 \square$

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