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What does it mean for a function to be well-defined? I encountered with this term in an excersice asking to check if a linear transformation is well-defined.

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It usually means that you have to check that the definition provides a unique output value for each input value. –  Stephen Montgomery-Smith Dec 14 '13 at 19:51
    
It means the function doesn't depend on the way an element of the domain set is expressed. For your particular example one should check what's the map and the vector spaces involved. –  DonAntonio Dec 14 '13 at 19:51
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A function $f:A \to B$ is well defined if for all $a \in A$, $f(a) \in B$ –  Amihai Zivan Dec 14 '13 at 19:51
    
please write it as an answer, so I'll accept it. thanks –  AndrePoole Dec 14 '13 at 19:52
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@AmihaiZivan No, that just means it is defined. –  Thomas Andrews Dec 14 '13 at 19:53
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up vote 6 down vote accepted

All functions are well-defined; but when we define a function, we don't always know (without doing some work) that our definition really does give us a function. We say the function (or, more precisely, the specification of the function) is 'well-defined' if it does.

That is, $f : A \to B$ is well-defined if for each $a \in A$ there is a unique $b \in B$ with $f(a)=b$.

This often comes up when defining functions in terms of representatives of equivalence classes, or in terms of how an element of the domain is written. For example, the 'function' $f : \mathbb{Z} \to \mathbb{Z}$ defined by $$f(n) = \text{the first digit of the decimal expansion of}\ n\ \text{after the decimal point}$$ is not a well-defined function: we get $f(1)=0$ and $f(0.999\dots)=9$, even though $0.999\dots = 1$. We could turn it into a well-defined function by saying that the chosen decimal expansion must not have recurring $9$s.

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@AndrePoole: It depends how you define it... it is well-defined as long as you declare something like $$\mathtt{abs}(x) = \text{the greatest integer}\ \le x$$ This only depends on the value of $x$, not its representation as a number, so it's well-defined. –  Clive Newstead Dec 14 '13 at 20:02
    
cheers, you made it very clear –  AndrePoole Dec 14 '13 at 20:05
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