- #1

- 2

- 0

## Main Question or Discussion Point

Good Morning,

I am trying to prove that any 2 open intervals (a,b) and (c,d) are equivalent.

Show that f(x) = ((d-c)/(b-a))*(x-a)+c is one-to-one and onto (c,d).

a,b,c,d belong to the set of Real numbers with a<b and c<d.

Let f: (a,b)->(c,d) be a linear function which i graphed to help me visualize, restricting the domain to (a,b). I know I have to prove that f is a bijection but I am not sure how to do this.

I know that I am supposed to show that any two open intervals are equivalent, even if the intervals are different lengths, so the open interval (3,4) is equivalent to the open interval (1,8), (3,4) is a subset of (1,9).

I would appreciate any help with this if you can get me on the right track...thanks so much!

I am trying to prove that any 2 open intervals (a,b) and (c,d) are equivalent.

Show that f(x) = ((d-c)/(b-a))*(x-a)+c is one-to-one and onto (c,d).

a,b,c,d belong to the set of Real numbers with a<b and c<d.

Let f: (a,b)->(c,d) be a linear function which i graphed to help me visualize, restricting the domain to (a,b). I know I have to prove that f is a bijection but I am not sure how to do this.

I know that I am supposed to show that any two open intervals are equivalent, even if the intervals are different lengths, so the open interval (3,4) is equivalent to the open interval (1,8), (3,4) is a subset of (1,9).

I would appreciate any help with this if you can get me on the right track...thanks so much!