Question

which sets of ordered pairs represent functions from a to b , A={u,v,w} and B={-2,-1,0,1,2}

Answer 1

@Nnesha

Answer 2

@Loser66 @Luigi0210

Answer 3

@mathmate

Answer 4

can you please help @mathmate

Answer 5

There are many possible answers.
If you have answer choices, choose the choices where there is no repetition in the set A, i.e. u, v, w appear at most once in the set of ordered pairs.
This is because a function can only have one single value of the image given a value of the preimage.

Answer 6

the choices are

(a) {(v,-1) , (u,2) , (w,0) , (u,-2)}
(b) (u,-2) , (v,2) , (w,1)
(c) (u,2) , (v,0) , (w,2)

Answer 7

@mathmate

Answer 8

Please try to understand my previous post, and reread if necessary. You will have no problem making your choices. :)

Answer 9

thank you i got it

Answer 10

i had missed out one of the answer choices and it happened to be the right one

Answer 11

If we examine each of the answers choices,
(a) {(v,-1) , (u,2) , (w,0) , (u,-2)}
(b) (u,-2) , (v,2) , (w,1)
(c) (u,2) , (v,0) , (w,2)

For the relation
{(v,-1) , (u,2) , (w,0) , (u,-2)}
Ask yourself, what would be the value of f(v)? It is clear that f(v)=-1 because the ordered pair (v,-1) tells us that.
However, what is the value of f(u)? There are two ordered pairs, (u,2), and (u,-2).
We do not know which one is true. A relation that contains ambiguities like this is NOT considered a function. So this relation is not a function.

For both of the following relations,
(u,-2) , (v,2) , (w,1)
(u,2) , (v,0) , (w,2)
the ambiguity does not exist, given any value of u, v or w, we can find a UNIQUE value of the image, so both are functions.