# Homework Help: Proving a function is bijective

1. Dec 13, 2012

### avg5

1. The problem statement, all variables and given/known data

The relation R, on the set ℝxℝ, is defined by (x,y)R(a,b) if and only if x+b=a+y for any real number x,y,a, and b.

Let G denote the set of equivalence classes of the relation R.

Question: Let f: ℝ -> G be defined by f(a)=[(0,a)]. Prove that f is a bijection.

3. The attempt at a solution

I think I have to come up with an equation, for f(a)=[(0,a)]. I also think I could prove this using induction.

I am very confused and need a leg up on how to start! Any help would be greatly appreciated.

2. Dec 13, 2012

### Dick

I think you need to understand what the question is actually asking first. [(0,a)] is the equivalence class consisting of all (x,y) such that (x,y)R(0,a). What does that set look like if you graph it in the xy plane?

3. Dec 13, 2012

### avg5

Do you mean the graph of the set (0,a)? The x-values are stationary at 0 and the y-values are varying.

I know that I need to prove that the function is both onto and 1:1. Ergh, am I really confused

4. Dec 13, 2012

### Dick

I still don't think you really understand what they are asking. No. (0,a) is just a point. I mean the graph of the equivalence class [(0,a)] with a fixed. The brackets mean 'equivalence class'. (0,a)R(x,y) means 0+y=x+a, right?

5. Dec 13, 2012

### avg5

Yes.. so the graph is y=x+a then?

6. Dec 13, 2012

### Dick

Right. So [(0,a)] is a straight line with y-intercept a and slope 1. So your function f takes a in R and maps it to that straight line. Now can you see why it's a bijection?

7. Dec 13, 2012

### avg5

Yes. Thank you so, so much.