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Prove that 2 + 2 = 4 brainlets need not apply.

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Thread replies: 23
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Prove that
2 + 2 = 4
brainlets need not apply.
>>
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>>9076147
$I I = 2$
$I I I I = 4$
$2+2 = I I + I I = I I I I = 4$
>>
Two men have gay sex. Each one has 2 balls but when they fuck there's 4 balls total. Atheists BTFO.
>>
>>9076147
Set theory already did my guy.
>>
I count two fingers on my one hand. I count two fingers on my other hand. I count all fingers erected on both hands and count the humber 4.

Thus two fingers on one hand, counted with two fingers on another hand equals four fingers in total.

PRESCHOOL KNOWLEDGE WHATUKNOBOUTDATHAHA
>>
>>9076147
First we need to define addition:
In general, for each [math]m\in\mathbb{N}[/math] there exists a unique function [math]A_m[/math] such that, for all [math]n\in\mathbb{N},~A_m(0)=m~\text{and}~A_m(S(n))=S(A_m(n)),~S[/math] denoting the successor function. Then [math]+:=\{((n,\,m),\,k):k=A_n(m)=k,~n,m\in\mathbb{N}\}[/math]. Addition now being defined, we can move on the the problem at hand: 2+2=[math]A_2(2)=A_2(S(S(0)))=S(S(A_2(0)))=S(S(2))=4[/math].
>>
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>>9076147
It's trivial
>>
>>9076154

engineer detected lol
>>
>>9076147
4 = 4 • 1= 1 + 1 + 1 + 1
2 = 2 • 1= 1 + 1
4 = (1 + 1) + (1 + 1)
4 = (2) + (2)
4 = 2 + 2
>>
>>9076147
2+2 = s(2) + 1 = s(s(2)) + 0 = s(s(2)) = s(3) = 4
>>
>>9076153
I want to make an addition to that ass, if you can contemplate what I'm implying
>>
>>9076147
2+2=1+1+2=1+1+1+1=4
>>
>>9076147
what axioms are you willing to accept?
>>
>>9077548
No axioms are allowed.
>>
>>9076147
f(x)=x+x=2x
Let x = 2
f(2)=2(2)
2*2=4=2+2
>>
>>9076182
>/math]denotingthesuccessorfunction
...rather than set cardinality?
>>
>>9076165
>I count two fingers
I count one finger extended upward to you.
>>
>>9076147
because of how we define = + 2 and 4
>>
>>9076147
Definition 1
[math]\forall n \in \mathbb{N} \exists ! s(n) \in \mathbb{N} | s(n) = n+1 [/math]
>For all natural number n exists only one natural number called successor of n s(n) such as [math]s(n) = n + 1[/math].
Definition 2
[math]s(1)= 2 = 1+1[/math] , [math]s(2)= 3 = 2+1[/math] , [math] s(3)= 4 = 3+1[/math]
Theorem [math]2+2= 4.[/math]
Proof
[math]4 = 3+1 = 2+1+1 = 2+2 \to 2+2=4 \blacksquare[/math]
>>
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>>9076182
[s4s] visiting
>>
>>9077609
>■
>>
>>9076153
wuts cumming out of her butt?
>>
>>9077602
I don't know what construction of [math]\mathbb{N}[/math] you're using, but with the von Neumann construction, the cardinality of [math]n[/math] is [math]n[/math].
Thread posts: 23
Thread images: 4


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