For example
f(4) = 3, 2
f(5) = 4
f(6) = 10Please forgive me if my example is atrocious, hopefully it gets my point across.
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$\begingroup$This seems like a good time to learn about what it means for an expression to be well-defined. No doubt you are comfortable with integer arithmetic; we may define addition of fractions by $$ \frac{x}{y}+\frac{p}{q}=\frac{xq+py}{yq}.\tag{1} $$ This is familiar. To work correctly though, the definition in $(1)$ should not depend on which fractions we insert. For example, $\frac{2}{3}+\frac{10}{2}$ should be equivalent to, say, $\frac{-4}{-6}+\frac{5}{1}$. The definition in $(1)$ is well-defined because it always works and there is no ambiguity concerning the output. However, consider a different operation that may be ambiguous. Let's denote this operation by $\boxplus$ and define it as follows: $$ \frac{x}{y}\boxplus\frac{p}{q}=\frac{xy}{pq}.\tag{2} $$ Using $(2)$, we would have $\frac{2}{3}\boxplus\frac{10}{2}=\frac{6}{20}$, but we would also have $\frac{-4}{-6}\boxplus\frac{5}{1}=\frac{24}{5}$. Uh oh. This produces a different output than before because $\frac{6}{20}\neq\frac{24}{5}$. Thus, the operation $\boxplus$ is said to not be well-defined because its outputs are ambiguous based on its inputs.
How does this relate to your question specifically? Well, $f(5)=4$, and we are certain of this. There is no ambiguity. However, if you wrote $f(4)$, then I wouldn't know if you meant $f(4)=3$ or $f(4)=2$. That is, $f(4)$ is ambiguous and hence we can say that, in your case, $f(x)$ is not well-defined.
Does that help?
$\endgroup$ 2 $\begingroup$A function $f: A \to B$ associates to each element $a \in A$ an element $b \in B$. So for each input in $A$, there is exactly one output in $B$.
To elaborate: writing $f(4)=3,2$ does not make sense, yet writing $f(4)=(3,2)$ could make sense in a different context: it could be that $B$ is a set containing ordered pairs (for instance, $B$ could be the set $\mathbb{R}^2 = \{(r,s): r,s\in \mathbb{R}\}$ of ordered pairs of real numbers). In that case, $a$ is send to an ordered pair of two numbers, but again to exactly one such pair. Furthermore, each $a \in A$ would have to be sent to such a pair.
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