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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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    $\begingroup$ It usually means that you have to check that the definition provides a unique output value for each input value. $\endgroup$ Dec 14 '13 at 19:51
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    $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Dec 14 '13 at 19:51
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    $\begingroup$ A function $f:A \to B$ is well defined if for all $a \in A$, $f(a) \in B$ $\endgroup$ Dec 14 '13 at 19:51
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    $\begingroup$ @AmihaiZivan No, that just means it is defined. $\endgroup$ Dec 14 '13 at 19:53
  • $\begingroup$ @ThomasAndrews Adding "If $a=b$ then $f(a)=f(b)$" will complete the definition? $\endgroup$ Dec 14 '13 at 19:56
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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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    $\begingroup$ @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. $\endgroup$ Dec 14 '13 at 20:02
  • $\begingroup$ cheers, you made it very clear $\endgroup$
    – AndrePoole
    Dec 14 '13 at 20:05
  • $\begingroup$ Very good example as well as the presentation of what makes a function/map well-defined - this is the way it was taught to me when I also came across this same predicament. $\endgroup$
    – Procore
    Sep 10 '16 at 20:02
  • $\begingroup$ @AndrePoole Note, that the inverse of statement 1 is not true: while for every a there is just one b (or: you only get one result when throwing a into the formula), this b could be the result of several (or all) a. b=a² is well defined, but b=1 is the answer to both a=1 and a=-1. $\endgroup$
    – Trish
    Nov 13 '17 at 9:07

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