Proving a function is bijective

[avg5]

Homework Statement

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.

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.

 

[Dick]

Homework Statement

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.

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.

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?

 

[avg5]

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?

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

 

[Dick]

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

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?

 

[avg5]

Slow down you are jumping to conclusions. 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?

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

 

[Dick]

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

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?

 

[avg5]

Yes. Thank you so, so much.