Me
Posts:
1,716
Registered:
1/23/16


Re: A more stupid Italian mythmatician exists? Peano was a moron of galactic proportions.
Posted:
Sep 30, 2017 9:36 AM


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 {(1, k) : k e m} = { (1,0) } and hence n + m = #({(0, k) : k e n} U {(1, k) : k e m}) = #({ (1,0) }) = { {} } = 1, since { {} } ~ { (1,0) } (both sets just contain one element). Hence 0 + 1 = 1.

