Hello Raul;
I don't see how you say Y is a function of X,Y. Is it: once you know x in
X, and y in Y, you know y?
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|\/| Randy A MacDonald | APL: If you can say it, it's done.. (ram)
|/\| [EMAIL PROTECTED] |
|\ | |If you cannot describe what you are doing
BSc(Math) UNBF'83 þas a process, you don't know what you're doing.
Sapere Aude | - W. E. Deming
Natural Born APL'er | Demo website: http://156.34.71.170/
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----- Original Message -----
From: "Miller, Raul D" <[EMAIL PROTECTED]>
To: "General forum" <[email protected]>
Sent: Saturday, July 01, 2006 2:06 PM
Subject: RE: [Jgeneral] significant digits
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
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