Math Is Fun Forum

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Functions.. the logical way

heya, I was just wondering if someone could explain this to me because I don't get it.

My logic prof said that there is a function f : {T,F} x {T,F} -> {T,F} (where A -> B means A is independent and B is dependent).
He says that f = {((F,F),F),((F,T),T),((T,T),T)} and that this is not a function because there is no pair whos 1st element is (T,F). Did my prof just chose 1 example out of many possible ones from A to B ? And if he did, does this mean other examples you can pull out would be functions? Also, why is it not a function if there is no pair whos 1st element is (T,F) ?

Thx if you understand lol

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Re: Functions.. the logical way

wow... I just got it lol. Sorry for wasting space.

I realize now that the prof chose only 3/4 possible A's and so since all A's have to be used up to be a function, and (T,F) is not used in his example, then it is not a function. Is this reasoning correct? Thx (P.S. I think a discontinuous function is not a function in logic)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Functions.. the logical way

That sounds about right. So if f had had ((T,F),T) added to the end then it would be a function (commonly known as AND), but it didn't so it isn't.

By the way, I think you're using discontinuous incorrectly. Discontinuous functions are still defined everywhere, it's just that the points don't necessarily 'join up'.
The notion of continuity wouldn't make sense with the domain {T,F} x {T,F} anyway.

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Why did the vector cross the road?
It wanted to be normal.