Question

Is {(-6,2),(7,1),(0,-3)} a function?

Answer 1

Yes, when thinking of functions as sets of ordered pairs. Each input `-6, 7, 0` has exactly one associated output.

Explanation:

Any function, technically-speaking, is really a special kind of set of ordered pairs (strange to realize if you've never thought about it that way before).

For example, the function `f` defined by the formula `f(x)=x^2` over the entire real number line `RR`, technically-speaking, is really the set `{(x,y):y=x^2\mbox{ for } x\in RR}`.

The key thing that makes a set of ordered pairs a function is that each input (first coordinate of each ordered pair) occurs in at most one spot in the set (or, if you wrote it down twice, it would have the same output (second coordinate), though technically-speaking elements of sets can't occur twice ...if you wrote an element twice it doesn't change what the set is).

Hence, the set of ordered pairs `{(-6,2), (7,1), (0,-3)}` represents a function with 3 inputs (and we'd write `f(-6)=2`, `f(7)=1`, and `f(0)=-3`) whereas the set of ordered paris `{(-6,2), (7,1), (0,-3), (-6,5)}` does not represent a function (there would be ambiguity in the value `f(-6)`).

Such a set still has a name, however. It's called a relation .

Things get stranger in the realm of complex analysis, however, where mathematicians study the oxymoronically-named multi-valued functions , which, technically speaking are relations that are not functions in the sense above. If that's too confusing to consider late at night when you want to go to sleep, you can just ignore it.

Answer 2

Yes, this set of points in the xy-plane describes a function

`f:{-6, 0, 7}->{-3, 1, 2}`

that maps `-6` to `2`, `7` to `1` and `0` to `-3`.

Explanation:

Since the first elements of each of the ordered pairs are distinct, this does describe a function.

Since the second elements of each ordered pair are also distinct, this function is also one-one with inverse described by the set:

`{(2, -6), (1, 7), (-3, 0)}`