# Discrete Mathematics: Is relation $f$ a function?

1.) Determine whether each of the following relations is a function with domain $\{1,2,3,4\}$. For any relation that is not a function, explain why it isn't.

a.) $f=\{(1,1), (2,1), (3,1), (4,1), (3,3)\}$. - The answer in back of the book states the following: "Not a function; $f$ contains two different pairs of the form $(3,-)$." What does the dash mean?

b.) $f=\{(1,2), (2,3), (4,2)\}$ - ?
d.) $f=\{(1,1), (1,2), (1,3), (1,4)\}$ - ?
e.) $f=\{(1,4), (2,3), (3,2), (4,1)\}$ - ?

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A function is well defined, that means $x=y\implies f(x)=f(y)$ but you have once $f(3)=1$ and $f(3)=3$ try for the rst. for your control, b is no function, d is no function but e is a function.

In b) you have no function because $f(3)$ does not exist, but a function assigns to every object of your domain, an object of your codomain.

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The dash means "unbound" or "not fixed," so $(3,-)$ matches $(3,1)$ and $(3,3)$

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A function is defined in this context to be a relation (= a set of pairs) such that "each source has only one image", that is, since $(3,1)$ is interpreted as $f(3)=1$, it's impossible for $(3,3)$ to be in the same relation, as that would imply $f(3)=3$.

The dash in the book's answer attempts to represent a general pair $(3,-)$ should be read "3, something", meaning an ordered pair where the first entry is $3$. This is not standard notation, but is clear from the context.

As for the other relations, check them against the same criterion: $f$ is a function if there are no contradicting $(a,b), (a,c)$, because that would imply $f(a)=b$ and $f(a)=c$ at the same time.

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A function is a rule that assigns to every $x$ value in its domain exactly one $y$ value in its range. We write this as $f(x)=y$. By definition of a function, it is not possible to have $f(3)=1$ and $f(3)=3$, because then the value $3$ in the domain is assigned to two values $y$ in the range.

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It represents an open ended match or "value".

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