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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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David Bernier

Posts: 3,317
Registered: 12/13/04
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 6, 2013 5:51 PM
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On 10/06/2013 04:42 PM, Peter Percival wrote:
> Arturo Magidin wrote:
>

>>
>> Technically, two functions are equal if and only if they have the
>> same domain, and they take the same value at each point on the
>> domain.

>
> I wonder why you don't require the codomains to be the same as well.
> Two functions might be equal according to that definition but one could
> be invertible and the other not.
>


In the way I think of a function 'f', no co-domain
attaches to 'f' in a natural way. For example,
for the function f with domain R (= the set of real numbers)
and such that, for any x in R,

f(x) = sin(x),

the domain of f is R, but I would write:
"f with co-domain R", "f with co-domain [-1, 1]",
or by abuse of language,
"f(x) with co-domain R", "f(x) with co-domain [-1,1]".

Nevertheless, the default, primary, definition of "function"
at PlanetMath provides for a co-domain being attached to
a function. They say that "in set theory",
a function is a special type of relation, so along the
line of thought in Arturo Magidin's way of thinking
or my way of thinking.

The "no co-domain included" definition is convenient,
because for example we can say

f: C -> C is the polynomial function defined
by f(z):= z^137 + 202 z^23 - 45z^4 + 32 z^3 - 1001 z^2 + 7z ,
all this without bothering to specify a co-domain.

And in any case, if we say "f maps C onto C", it's equivalent
to saying f(C) = C.

cf.:

http://planetmath.org/node/30360 (def. of function)

David Bernier
--
Let us all be paranoid. More so than no such agence, Bolon Yokte K'uh
willing.


Date Subject Author
9/28/13
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Hetware
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10/16/13
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10/19/13
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Virgil
10/18/13
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10/19/13
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Peter Percival
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Peter Percival
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10/19/13
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10/19/13
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10/20/13
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