Equivalence Relations, Cardinality and Finite Sets.

Click For Summary

Homework Help Overview

The discussion revolves around three problems related to equivalence relations, cardinality, and finite sets, focusing on the properties of relations and functions in set theory.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the properties of an equivalence relation, specifically checking reflexivity, symmetry, and transitivity. There is a question about the implications of failing reflexivity on equivalence classes and partitions.
  • In the second question, there is a discussion about constructing a bijective function and the necessary steps to demonstrate its properties, with hints about using a diagonal argument.
  • The third question raises a need for clarification on notation and concepts, with a request for guidance on how to approach the proof.

Discussion Status

Some participants have provided insights into the properties of equivalence relations and the requirements for bijections. There is ongoing exploration of definitions and approaches, particularly regarding the construction of functions and the implications of notation.

Contextual Notes

Participants note the importance of clarity in notation and suggest that separate threads might be more effective for distinct questions to facilitate focused discussion.

3=MCsq
Messages
3
Reaction score
0
Hey everyone, I have three problems that I'm working on that are review questions for my Math Final.

Homework Statement



First Question: Determine if R is an equivalence relation: R = {(x,y) \in Z x Z | x - y =5}
and find the equivalence classes.
Is Z | R a partition?

Homework Equations



An equivalence class R is reflexive, symmetric and transitive.

The Attempt at a Solution



First Question: So I know I have to check reflexivity, symmetricity and transitivity. It seems this relation fails the reflexivity test. Since xRx = {(x,x) -> x-x=5} and x-x ≠ 5. Since it fails, if I'm correct, can I still find equivalence classes and partitions?

Homework Statement



Second Question: Show that |Z+ x Z +| → |Z+| (Where Z+ represents all positive integers)...
by showing that Z+ x Z+ → Z+ is a bijection.


Homework Equations



A function is bijective if it is one-to-one and onto.

The Attempt at a Solution



Second Question: I know that I have to show that if f(x)=f(y) then x=y but I don't know how to structure it. Also I have to show that f(x) = y \in Z+.

Homework Statement



Third Question: Prove that if n \in N , f:In → B and f is onto then B is finite and |B| < n.

Homework Equations





The Attempt at a Solution



Third Question: I don't even know where to begin. I'm not looking for an answer, just more of where to look, where to start..a break down of the concepts addressed in the question would be most helpful.


Thanks!
 
Physics news on Phys.org
3=MCsq said:
First Question: So I know I have to check reflexivity, symmetricity and transitivity. It seems this relation fails the reflexivity test. Since xRx = {(x,x) -> x-x=5} and x-x ≠ 5. Since it fails, if I'm correct, can I still find equivalence classes and partitions?
Your example shows that this is not an equivalence relation, because it is not reflexive. So there are no equivalence classes and there is no partition. End of story.
 
3=MCsq said:
Second Question: I know that I have to show that if f(x)=f(y) then x=y but I don't know how to structure it. Also I have to show that f(x) = y \in Z+.
Well, you can't show these properties of ##f## without first defining ##f##. Try to construct a function ##f : \mathbb{Z}^+ \times \mathbb{Z}^+ \rightarrow \mathbb{Z}^+## which is bijective, and then verify that it is bijective. Hint: consider a diagonal argument.
 
3=MCsq said:
Third Question: Prove that if n \in N , f:In → B and f is onto then B is finite and |B| < n.
Can you clarify the notation? What does "In" refer to?

P.S. In the future, it would be better to create separate threads for each question, unless they are closely related.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
Replies
6
Views
3K