1. ## Set Problems

Using ordered-pair notation, let g be the function defined by
g = {(5n, 3n) | n element of Z}
a) State the domain and a reasonable codomain of g.
b) Evaluate g(10), g(g(50)), and g^-1(-6).

Isn't the domain all real numbers?

How do I go about evaluating and finding the domain and codomain

2. ## Re: Set Problems

Originally Posted by xmathlover
Using ordered-pair notation, let g be the function dened by
g = f(5n, 3n) | n element of Z
a) State the domain and a reasonable codomain of g.
b) Evaluate g(10), g(g(50)), and g^-1(-6).
Isn't the domain all real numbers?
How do I go about evaluating and finding the domain and codomain

What you have posted leaves a great many questions.

What in the world is the definition of $f(5n,3n)~?$

Do you mean $g(n)=f(5n,3n)~?$.

Where do ordered pairs come in to it at all?

3. ## Re: Set Problems

I fixed it.

{5n, 3n | n elements of Z}

4. ## Re: Set Problems

Originally Posted by xmathlover
I fixed it.

{5n, 3n | n elements of Z}
If $g=\{(5n,3n):n\in\mathbb{Z}\}$ then the domain of $g$ is the set of multiples of five.

The image set (final set or codomain) is the set of multiples of three.

EX: $g(10)=g(5\cdot 2)=3\cdot 2)=6$

5. ## Re: Set Problems

How did you come to that conclusion? do you look at the ordered pairs as x input, y output?

How would I evaluate g(g(50)) and g^-1(-6)?

g(50) = g(5*10) = 3 * 10 = 30, now how do I run it through the outer g?

6. ## Re: Set Problems

Originally Posted by xmathlover
How did you come to that conclusion? do you look at the ordered pairs as x input, y output?

How would I evaluate g(g(50)) and g^-1(-6)?

g(50) = g(5*10) = 3 * 10 = 30, now how do I run it through the outer g?

$g(30)=g(5\cdot 6)=3\cdot 6=18$ and $-6=3\cdot ?$ so ?

7. ## Re: Set Problems

what do you do with the g^-1?

8. ## Re: Set Problems

Originally Posted by xmathlover
what do you do with the g^-1?

From your questions, it seems that you need a good review of function notation.

$g(g^{-1}(t))=t$. That is if $g^{-1}(t)=s$ then $g(s)=t$

Thus $g^{-1}(-21)=-35$, now can you explain why?

9. ## Re: Set Problems

you lost me! lol...

10. ## Re: Set Problems

Originally Posted by xmathlover
you lost me! lol...

I really don't think that you are ready to do these questions.
Why not give your notes and/or textbook a good review?

Asking for $g^{-1}(-21)$ is the same as asking $g(?)=-21$, fill in the blank?

11. ## Re: Set Problems

for this set {5n, 3n}?

g(7) = -21

wouldn't g^-1(-6) = 2?

12. ## Re: Set Problems

Originally Posted by xmathlover
for this set {5n, 3n}?

g(7) = -21
No. $g(7)$ does not exist, because seven is not in the domain of $g$. WHY?

13. ## Re: Set Problems

because it isn't a multiple or 5 or 3?

But by that assumption nothing will work.

14. ## Re: Set Problems

Originally Posted by xmathlover
because it isn't a multiple or 5 or 3?

But by that assumption nothing will work.

See, as I told you: You do not understand the notation.

$g^{-1}(-21)=-35$. It does work: $g(-35)=-21$.

15. ## Re: Set Problems

oh so then, g^1(-6) = -10

Is the domain of g express liked this {x | x +- 5} or should it be expressed different? My notes suck...they don't explain anything about domain and codomain in orderpair notation

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