Sorry, I am not sure what you are asking. Could you please clarify?

[XodoX]

Trying to prepare for an exam...

4)
Let f : A -> B be any function from the set A to the set B. How is the equivalence relation ~f
on A defined?

5. Let f : R -> R, x -> x^2, (Couldn't find the R symbol - real numbers) be the parabola function. What does the partition for the equivalence relation of this function look like?

Does anybody know how I do this? I'm not understanding this at all!
Thank you in advance

 

[pwsnafu]

The only one I can think of is x \sim y \iff f(x) = f(y) which is an equivalence on the domain of f.

 

[SW VandeCarr]

Trying to prepare for an exam...

4)
Let f : A -> B be any function from the set A to the set B. How is the equivalence relation ~f
on A defined?

5. Let f : R -> R, x -> x^2, (Couldn't find the R symbol - real numbers) be the parabola function. What does the partition for the equivalence relation of this function look like?

Does anybody know how I do this? I'm not understanding this at all!
Thank you in advance

If you're talking about sets, two sets (A,B) are equal if they have the same members. For any two functions; F(A),F(B) are equivalent if the equality A=B still holds. So two functions which impose different orderings on A and B are still equivalent if the two sets still have the same members.

 

[XodoX]

The only one I can think of is x \sim y \iff f(x) = f(y) which is an equivalence on the domain of f.

Thank.s No clue. It's a start. I'll try to do it with this one.

If you're talking about sets, two sets (A,B) are equal if they have the same members. For any two functions; F(A),F(B) are equivalent if the equality A=B still holds. So two functions which impose different orderings on A and B are still equivalent if the two sets still have the same members.

Do you mean the first question?

 

[SW VandeCarr]

Do you mean the first question?

Yes. If two sets are equal, any partition function(s) on one or both set(s) will leave both sets with the same elements. If you're defining new sets, that's a different question. Does A=B? This would satisfy question one but not question two. If A is {2,3,4} and B is {4,9,16} they are not the same. However f(2)=4, f(3)=9, f(4)=16. f(2,3)=(4,9) etc. That is, the function holds for any partition of A onto B. Perhaps that's what they mean. In this case, both sets must have same cardinality so equivalence means a one to one mapping is possible and that it is bijective (an inverse function exists).