Proving that an equivalence relation is a bijection

In summary: Okay, this may sound really bad... but I do not know how to prove bijection in equivalence classes (I know how to do it for functions but my book never showed me examples about proving bijection in equivalence classes). I tried googling examples but I cannot find any. My finals is tomorrow! Can anyone help me?
  • #1
number0
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Homework Statement



Let (a, b), (c, d) be in R x R. We define (a, b) ~ (c, d) iff a^2 + b^2 = c^2 + d^2.

Let R* = all positive real numbers (including 0).

Prove that there is a bijection between R* and the set of all equivalence classes for this equivalence relationship.

Homework Equations


The Attempt at a Solution



I do not know how the formula for the relationship ~ looks like. I tried mapping from R* x R* -> R* where f: x^2 + y^2 but I got stuck. Any help please?
 
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  • #2
number0 said:

Homework Statement



Let (a, b), (c, d) be in R x R. We define (a, b) ~ (c, d) iff a^2 + b^2 = c^2 + d^2.

Let R* = all positive real numbers (including 0).

Prove that there is a bijection between R* and the set of all equivalence classes for this equivalence relationship.

Homework Equations




The Attempt at a Solution



I do not know how the formula for the relationship ~ looks like. I tried mapping from R* x R* -> R* where f: x^2 + y^2 but I got stuck. Any help please?

Doesn't (a, b) ~ (c, d) mean the two points are on the same circle centered at the origin? Is that what you mean by "what it looks like"?
 
  • #3
LCKurtz said:
Doesn't (a, b) ~ (c, d) mean the two points are on the same circle centered at the origin? Is that what you mean by "what it looks like"?

I knew what it looks like... but I do not know the equation is supposed to be set up. For example, I did f(x,y) = x^2 + y^2 but I could not find injection.
 
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  • #4
number0 said:
I knew what it looks like... but I do not know the equation is supposed to be set up. For example, I did f(x,y) = x^2 + y^2 but I could not find injection.

Well, it's true that the radius is common to all the elements in an equivalence class. So I would say you are on the right track. Try the map that maps the equivalence class for (a,b) into the common radius or radius squared. If you denote the equivalence class for (a,b) by
[(a,b)] your map could map that to a2+b2. Then you have to show your function is well defined and has all the necessary properties to be a bijection.
 
  • #5
LCKurtz said:
Well, it's true that the radius is common to all the elements in an equivalence class. So I would say you are on the right track. Try the map that maps the equivalence class for (a,b) into the common radius or radius squared. If you denote the equivalence class for (a,b) by
[(a,b)] your map could map that to a2+b2. Then you have to show your function is well defined and has all the necessary properties to be a bijection.

Okay, this may sound really bad... but I do not know how to prove bijection in equivalence classes (I know how to do it for functions but my book never showed me examples about proving bijection in equivalence classes). I tried googling examples but I cannot find any. My finals is tomorrow! Can anyone help me?
 

FAQ: Proving that an equivalence relation is a bijection

1. What is an equivalence relation?

An equivalence relation is a mathematical concept that states that a relation between elements of a set is reflexive, symmetric, and transitive. This means that for any element a in the set, a is related to itself, if a is related to b then b is related to a, and if a is related to b and b is related to c, then a is related to c.

2. What is a bijection?

A bijection is a mathematical function that has both an injective (one-to-one) and surjective (onto) mapping between two sets. This means that each element in the first set is paired with a unique element in the second set, and every element in the second set has a corresponding element in the first set.

3. How do you prove that an equivalence relation is a bijection?

To prove that an equivalence relation is a bijection, you must demonstrate that the relation is reflexive, symmetric, and transitive, as well as show that it has both an injective and surjective mapping between the two sets. This can be done using mathematical proofs and examples.

4. What are some real-life examples of an equivalence relation that is also a bijection?

One example would be the relationship between a person's name and their social security number. Each person has a unique name, and each social security number is assigned to only one person, making it a bijection. Another example is the relationship between a driver's license number and the driver's personal information, such as name and address.

5. Why is it important to prove that an equivalence relation is a bijection?

Proving that an equivalence relation is a bijection is important because it ensures that the relationship between the two sets is well-defined and that there are no contradictions or inconsistencies. It also allows for the use of mathematical techniques and formulas that are specifically designed for bijections, making problem-solving and analysis easier and more accurate.

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