/sci/ - Science & Math

[Anonymous No.9306632]

If I have 2 functions f1(x,y) and f2(a,b) where x,y,a,b eof N does there always 2 different functions f3(k) and f4(i) exist where i,k eof N such that f1( f3(k) ,f4(i) ) = f2(a,b) ?

I'd say yes.

[Anonymous No.9306636]

>>9306632
No.

[Anonymous No.9306641]

>>9306632
I Think I got it wrong, it must be:

If I have 2 functions f1(x,y) and f2(a,b) where x,y,a,b eof N does there always 2 different functions f3(a) and f4(b) exist such that f1( f3(a) ,f4(b) ) = f2(a,b) ?

[Anonymous No.9306681]

>>9306641
>If I have 2 functions f1(x,y) and f2(a,b) where x,y,a,b eof N does there always 2 different functions f3(a) and f4(b) exist such that f1( f3(a) ,f4(b) ) = f2(a,b) ?
No.

[Anonymous No.9306697]

>>9306641
no consider f1 and f2 both constant functions, but different constants

[Anonymous No.9306733]

>>9306697
why tho

[Anonymous No.9306741]

>>9306733
if f1(x,y) = 1 for all x,y and f2(a,b)=2 for all a,b then f1(f3(a),f(b))=1 for all a,b, which can't equal f2(a,b)=2

[Anonymous No.9307163]

>>9306741
oh I forgot to say f1(x,y) and f2(a,b) are both bijective to N

[Anonymous No.9307976]

>>9307163
Then the answer is still no.

[Anonymous No.9307979]

>>9307163
the functions do not necessarily overlap in which case you can pick whatever i,k you like it won't change anything

[Anonymous No.9308556]

>>9307976
>>9307979
Its bijective to n. That means each point in Q(=N^2) hits each one in N exactly once. Ofc you'll be able to rearramge the input through some function f3 and f4 sich that they overlap at least in one point. Hell you should even be able to construct f3 and f4 such that they overlap everywhere

[Anonymous No.9308597]

>>9308556
>Hell you should even be able to construct f3 and f4 such that they overlap everywhere
No you shouldn't. Consider two functions with
f1(u,v)=f2(a,b)=1
f1(r,s)=f2(a,c)=2
with r != u.

Then if f1(f3(a),f4(b))=f2(a,b)=1, we must have f3(a)=u and f4(b)=v since f1 is injective.

Then if f1(f3(a),f4(c))=f2(a,c)=2 we must have f3(a)=r and f4(c)=s since f1 is injective

But now we have f3(a)=r and f3(a)=u, a contradiction.

[Anonymous No.9309229]

>>9308597
But f1 and f2 are bijective, therefore your example doesnt work. I mean f1(x,y)=2 is definetly not surjective.
I'm not 100% sure since I didnt construct f3 and f4 myself, but my intuition tells me it'll work.

[Anonymous No.9309246]

>>9309229
>But f1 and f2 are bijective, therefore your example doesnt work.
The f1 and f2 I posted are just any two bijections satisfying the conditions I listed, I didn't mean to imply they are constant. You can certainly construct two bijections from N^2 to N where

f1(1,1)=f2(1,1)=1
f1(2,1)=f2(1,3)=2

But now if you want f1(f3(1),f4(1))=f2(1,1)=1 then you must have f3(1)=1 and f4(1)=1 since f1 is injective.
And then if you want f1(f3(1),f4(3))=f2(1,3)=2 then you must have f3(1)=2 and f4(3)=3 since f1 is injective. But now f3(1)=1 and f3(2) is impossible, so there's no such f3 and f4.

>but my intuition tells me it'll work.
It doesn't work.

[Anonymous No.9309251]

>>9309246
>But now f3(1)=1 and f3(2) is impossible
But now f3(1)=1 and f3(1)=2 is impossible*

[Anonymous No.9309258]

>>9309246
>And then if you want f1(f3(1),f4(3))=f2(1,3)=2 then you must have f3(1)=2 and f4(3)=3 since f1 is injective.
f4(3)=1*

[Anonymous No.9309373]

>>9308556
this guy gets it

>>9308597
this prove is awesome, thanks


If a function doesn't always exist what kind of functions exist then such that >>9306641 and >>9307163

[Anonymous No.9309617]

>>9309373
>If a function doesn't always exist what kind of functions exist then such that >>9306641 and >>9307163
Given a bijection f2(x,y), there exists infinitely many f1(x,y) such that there exists f3 and f4 with f1(f3(x),f4(y))=f2(x,y). Letting f3 and f4 be any bijections from N to N, you can define f1(x,y)=f2(f3^(-1)(x), f4^(-1)(y)) so that f1(f3(x),f4(y))=f2(x,y). f1 picks up injectivity and surjectivity from f2, e.g.
f1(x,y)=f1(w,z) =>
f2(f3^(-1)(x), f4^(-1)(y))=f2(f3^(-1)(w), f4^(-1)(z)) =>
(f3^(-1)(x), f4^(-1)(y))=(f3^(-1)(w), f4^(-1)(z)) (by injectivity of f2) =>
x=w, y=z (by injectivity of f3 and f4).

This is likely the converse, as in if f1(f3(x),f4(y))=f2(x,y) then f3 and f4 are bijections of N.