1. ## Simple question about functions

Suppose that:
1) g={(0,0),(1,1/2),(2,1),(3,3/2),(4,2)}
2) f={(0,0),(1/2,0),(1,1),(3/2,1),(2,2)}

Can someone explain the logic behind figuring out whether a function is an inverse function?

I know that #1 is an inverse function and #2 is NOT an inverse function.

IS the correct way to figure this out to simply flip the x and y values and then see if it is still one-to-one and onto?

Also,
Can someone explain how I would find f-of-g?

THANK YOU.

2. Suppose that:
1) g={(0,0),(1,1/2),(2,1),(3,3/2),(4,2)}
2) f={(0,0),(1/2,0),(1,1),(3/2,1),(2,2)}

Can someone explain the logic behind figuring out whether a function is an inverse function?
Do you mean whether the inverse of a function is a function?

I know that #1 is an inverse function and #2 is NOT an inverse function.
Maybe you mean the inverse of #1 is a function and the inverse of #2 is not a function?

IS the correct way to figure this out to simply flip the x and y values and then see if it is still one-to-one and onto?
If all you're checking is to see whether the inverse of a function is a function, then just flip the x and y values to see whether you have a function (you don't have to check 1-1 nor onto).

To review:

inverse(f) = {<y x> | <x y> in f}

So, inverse(f) is a function if and only if {<y x> | <x y> in f} is a function.

{<y x> | <x y> in f} is a function
if and only if
for all x. y, z, if <y x> and <y z> in {<y x> | <x y> in f} then x = z.

Can someone explain how I would find f-of-g?
Do you mean f-comp-g?

f-comp-g = {<x y> | exists z such that <x z> in g & <z y> in f}.

So [f-comp-g](x) = f(g(x)). (I.e., determine g(x) = y, then determine f(y).)

3. Originally Posted by MoeBlee
Do you mean whether the inverse of a function is a function? YES

Maybe you mean the inverse of #1 is a function and the inverse of #2 is not a function? YES

If all you're checking is to see whether the inverse of a function is a function, then just flip the x and y values to see whether you have a function (you don't have to check for 1-1 adn onto).

To review:

inverse(f) = {<y x> | <x y> in f}

So, inverse(f) is a function if and only if {<y x> | <x y> in f} is a function.

{<y x> | <x y> in f} is a function
if and only if
for all x. y, z, if <y x> and <y z> in {<y x> | <x y> in f} then x = z.

I DONT UNDERSTAND THIS
Can you use the conditions and numbers that I have provided to create an example?

Do you mean f-comp-g?

f-comp-g = {<x y> | exists z such that <x z> in g & <z y> in f}.

So [f-comp-g](x) = f(g(x)). (I.e., compute g(x) = y, then compute f(y).)

I ALSO DON'T UNDERSTAND THIS please try to help explain given the information I know b/c the notation and the way you are explaining this is above my level of understanding, this is all really new to me, thank you for trying though

4. By the way, as to 1-1 and onto, probably what you have in mind are these facts:

If f is a 1-1 function then inverse(f) is a function.

If f is a function and inverse(f) is a function, then f is a 1-1 function.

If f is a 1-1 function onto S, then inverse(f) is a 1-1 function from S.

5. Let's review the definitions:

first coordinate in <x y> = x.

second coordinate in <x y> = y.

f is a relation iff every member of f is an ordered pair.

I.e., every member of f is of the form <x y> for some x and y.

f is a function
iff
f is a relation and for all x, y, z, if <x y> in f and <x z> in f, then y=z.

I.e., every member of f is of the form <x y> for some x and y, and there is no pair <x y> and <x z> in f unless y=z.

So, in your examples, both g and f are functions, and inverse(g) is a function. But inverse(f) is not a function because both <0 0> and <0 1/2> are in inverse(f), also because both <1 1> and <1 3/2> are in inverse(f).

/

I wrote "f-comp-g = {<x y> | exists z such that <x z> in g & <z y> in f}."

That means f-comp-g is the set of ordered pairs <x y> such that there exists a z such that <x z> in g and <z y> in f.

I.e., f-comp-g is formed by taking all the pairs <x y> where for some z there is a pair <x z> in g and <z y> in f.

In other words, look for any z that is the second coordinate in a pair in g and is the first coordinate in a pair in f.

In other words, find all the z's that link from the "end" in a pair in g to the "start" of a pair in f. Then the "comp" is the set of ordered pairs that have z as a "link" as just described.

For example,
suppose g = {<1 3> <4 2> <5 1> <6 2>}
and
f = {1 0> <3 2> <9 5>}.

Then f-comp-g = {<1 2> <5 0>}
because
3 is the second coordinate in <1 3> in g and the first coordinate in <3 2> in g, and 1 is the second coordinate in <5 1> in g and the first coordinate in <1 0> in f.