Equivalence Relations, Cardinality and Finite Sets.

3=MCsq
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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!
 
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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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