If I have a function f(a,b) -> k and another function g(c,d) -> k
and k is equal to N and a,b,c,d are elements of N (so that for each tuple (a,b) and (c,d) there is a k in N)
Does always a function p(x) and u(x) exist so that p(a)=c and u(b)=d ?
>k is equal to N
>there is a k in N
I don't know what you're trying to say but [math]k = N [/math] and [math]k \in N [/math] can't both be true. There is a theorem that says that a set never contains itself as an element.
>>8768918
You mean is there a function from N to N? Yes...
the two functions have the same output for every k in N, so they must be the same function. hence, a = c and b = d, and the functions p(a) = x and u(b) = y give p(a) = a = c and u(b) = b = d.
QED
>>8768927
ok then pls help me to express it correctly
I want to say that k is every number in N and every number exactly only once. F.e. f(1,1) = 3 and f(5,2)=3 can not happen because f would have two tuples ((1,1) and (5,2)) pointing to the same element in N (3).
>>8768983
no they are not
example: f(x,y) = (2x+1)2^y and g(x,y) =1/6 (-3 + (1/2 (6 *y - (-1)^y + 3)) * 3^(2*x+1))
They both generate N but with different x's and y's
>>8768985
You mean that [math]f[/math] is a bijection from [math]\mathbb{N}^2 [/math] to [math]\mathbb{N} [/math]?
>>8768996
Yes, thanks
>>8768996
So you're asking for f(a,b) = g(c,d) if there is p(a)=c and u(b)=d ?
>>8769018
Yes
>>8769038
Of course not. Consider f(1,1) = 1 = g(1,1) and f(1,2) = 2 = f(2,1). Therefore p(1) = 1,2 and cannot be a function.
>>8769075
Sorry, that should be
f(1,2) = 2 = g(2,1)
>>8769116
Yes just take
(p(a,b),u(a,b)) = g^(-1)(f(a,b))
and
(t(c,d),h(c,d)) = f^(-1)(g(c,d))
Is the quote true, /sci/?
>>8769194
>formula a sentence
>>8769807
No one reads papers from the 19th century anymore