Randy MacDonald wrote: > Do you have a more detailed version of that statement? I can > see how ?(arg,seed) is a function but I can't, in general, > see how foo(arg,range) would be a function, short of making > the range argument a function of the argument: > > foo(arg,bar(arg)) > > which presupposes bar, which I don't see as a given.
Consider the relationship x,y where x is 0 or 1 and y is 0 or 1. Then, y is not a function of x, but y is a function of x and y. Consider the relationship X,Y where x is any real number and y is any real number. Then, Y is a function of X,Y, even though Y is not a function of X. Etc. Note that there is no need to have a function bar(x) for this condition to hold. If this still does not make sense to you, I challenge you to come up with an example relationship which cannot be made into a function using this approach -- I'll describe the trivial function for you. -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
