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Topic: A more stupid Italian mythmatician exists? Peano was a moron of
galactic proportions.

Replies: 10   Last Post: Oct 3, 2017 12:36 PM

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Registered: 1/23/16
Re: A more stupid Italian mythmatician exists? Peano was a moron of
galactic proportions.

Posted: Sep 30, 2017 9:26 AM
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On Saturday, September 30, 2017 at 2:28:13 PM UTC+2, John Gabriel wrote:

> [JvN] also thought of his [ingenious] set theoretic construction of natural
> numbers. [We] can't think of anything more [.]logical ...


He DEFINEs:

> 0 := {}
> 1 := { {} } = {0}
> 2 := { {}, { {} } } = {0, 1}
> 3 := { {}, { {} }, { {} , { {} } } } = {0, 1, 2}


etc.

> [to] add 0 and 1, there are two approaches:
>
> [A] 0 + 1 = {} + { {} } = { {}, { {} } }


Huh?! Nonsense. 0 + 1 = s(0) (due to Peano) = 0 U {0} (due to von Neumann)
= {} U { {} } = { {} } = 1.

Got it, idiot?

Now which is the second appoach, idiot?

I can show you one:

Let n, m e IN. Then

n + m := card({(0, k) : k e n} U {(1, k) : k e m} .

in this context we may replace /card/ with /#/ and define /#/ for any finite set the following way:

#M := the number n e IN such that n ~ M .

Hence:

n + m := #({(0, k) : k e n} U {(1, k) : k e m})

Now let's try this for say, n = 0 and m = 1. Then {(0, k) : k e n} = {} and {(1, k) : k e m} = { (1,0) }. Hence {(0, k) : k e n} U {(0, k) : k e n} = { (1,0) } and hence #({ (1,0) }) = { {} } = 1, since { {} } ~ { (1,0) } (both sets just contains one element). Hence 0 + 1 = 1.



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