Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

5 Answers 5

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.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

It represents an open ended match or "value".

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.