Definition of a unique function


by jonsploder
Tags: definition, function, unique
jonsploder
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#1
Jan27-14, 04:12 AM
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Hi all, I'm wondering whether an expression which is used to describe a function in a certain domain is a different function for the same expression with a differing domain.

For example: expression; x^2.
f(x) = x^2 for domain {1 < x < 10}
f(x) = x^2 for domain {10 < x < 11}

Are these two f(x)'s the same function, or different functions, by definition. I couldn't be sure by Wikipedia, and it's a difficult question to type into a search engine.
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tiny-tim
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Jan27-14, 05:23 AM
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hi jonsploder! welcome to pf!
Quote Quote by jonsploder View Post
… Are these two f(x)'s the same function, or different functions …
they're different

they're both restrictions of the same function defined on the whole of R
jonsploder
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Jan27-14, 05:33 AM
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Thanks for the welcome, and the reply.
I know that they are different, however I was wondering, by the most formal definition of a function, whether they are different functions, or if indeed the domain of a function constitutes its identity as a function.

tiny-tim
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Jan27-14, 05:42 AM
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Definition of a unique function


Quote Quote by jonsploder View Post
I know that they are different, however I was wondering, by the most formal definition of a function, whether they are different functions, or if indeed the domain of a function constitutes its identity as a function.
they're functions, and they're different

so they're different functions

the definition of a function includes its range and domain: different range and/or domain, different functions
pasmith
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Jan27-14, 06:31 AM
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Quote Quote by jonsploder View Post
Thanks for the welcome, and the reply.
I know that they are different, however I was wondering, by the most formal definition of a function, whether they are different functions, or if indeed the domain of a function constitutes its identity as a function.
The domain and codomain are part of the definition of a function.

Two functions [itex]f : A \to B[/itex] and [itex]g : C \to D[/itex] are equal if and only if [itex]A = C[/itex] and [itex]B = D[/itex] and for all [itex]a \in A[/itex], [itex]f(a) = g(a)[/itex].
economicsnerd
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Jan27-14, 06:02 PM
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Quote Quote by pasmith View Post
Two functions [itex]f : A \to B[/itex] and [itex]g : C \to D[/itex] are equal if and only if [itex]A = C[/itex] and [itex]B = D[/itex] and for all [itex]a \in A[/itex], [itex]f(a) = g(a)[/itex].
This definition is exactly correct. That should be your definition.

....

It's worth noting, however, that sometimes people get lazy about codomains and say [itex]f : A \to B[/itex] and [itex]g : C \to D[/itex] are equal when [itex]A = C[/itex] and for all [itex]a \in A[/itex], [itex]f(a) = g(a) \in B\cap D[/itex].
pwsnafu
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Jan27-14, 06:33 PM
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Quote Quote by economicsnerd View Post
It's worth noting, however, that sometimes people get lazy about codomains and say [itex]f : A \to B[/itex] and [itex]g : C \to D[/itex] are equal when [itex]A = C[/itex] and for all [itex]a \in A[/itex], [itex]f(a) = g(a) \in B\cap D[/itex].
To explain why this definition is bad, consider
##f : \mathbb{R} \to \mathbb{R}##, ##f(x) = 0##
##g : \mathbb{R} \to \{0\}##, ##g(x) = 0##.
Note that under the definition economicnerd gave these would be considered equal. However, g is a surjection while f is not.


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