Question

Find the domain and range of the relation, and state whether or not the relation is a function.
{(1, 3), (2, 3), (3, 3), (4, 3)}

Answer 1

The domain is the set of the first values in each ordered pair
The range is the set of the second values in each ordered pair

It would be a function if for each value in the domain, there is only one not necessarily unique corresponding range value.

Answer 2

It is a function if every value of the domain has a unique value of the range.
If x=y, then f(x) =f(y)
for example if f(x) =2 and f(x) =5, then f is not a function.

The relation in that question is a function.

@AccessDenied the statement below is not true
It would be a function if for each value in the domain, there is only one not necessarily unique corresponding range value.

Answer 3

What I meant was, for each x in the domain, there is a y-value that may or may not be the same as another x-value's, but you may not have two y-values for one x-value.
I.e. f(x) = x^2, x=2 and x=-2 have the same y-value.

Maybe my words are mixed up there, or I'm missing something...

Answer 4

As far as I can tell though, that definition is correct. o.O

If each value \(x\) in the domain maps to \(\mathbf{one}\) (not necessarily unique) value \(y\) in the range, then the relation is a function.
\( \begin{array}{|c|c|}
\hline
\mathbf x & \mathbf y \\
\hline
1 & 3\\
2 & 3 \\
3 & 3 \\
4 & 3 \\
\hline
\end{array}
\qquad 1 \to 3,\ 2 \to 3,\ 3 \to 3,\ 4 \to 3\)
The relation is a function because for each \(x\), there is one value of \(y\), which is not unique in this case between the other \(x\)'s.

Answer 5

If a relation f is such
f(1) =2 and f(1) =3 then f is not a function.

Answer 6

Yep, definitely true. Two y-values for one x-value doesn't fit that definition. I guess it may be because I say 'not necessarily unique" somewhat ambiguously to clarify that its the x-value that cannot have two y-values and not the y-value having two x-values that matters..