# One to one function

1. Jun 9, 2010

### staw_jo

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

A function from the real numbers to the real numbers is one to one on an interval I if it is strictly increasing on that interval.

Any help please!!!

2. Relevant equations

3. The attempt at a solution

I am not quite sure how to prove it, I know that the use of strictly increasing is important as far as if x1 < x2, then f(x1) < f(x2). A hint I was told to use is contradiction.

2. Jun 9, 2010

### VeeEight

If you would like to prove it by contradiction, assume there is such a time that f(x) = f(y) where x is not y. Then either x < y or y < x. Now use the strictly increasing property of the function.

3. Jun 9, 2010

### JG89

To prove it by contradiction, negate the definition of one-to-one. Suppose that there are distinct points, a and b, in I, such that f(a) = f(b). You know that either a < b or b < a, right?

4. Jun 9, 2010

Great

5. Jun 9, 2010

### staw_jo

Okay so:

Assume that f(x1) = f(x2), but x1 does not equal x2, then either x1 < x2 or x2 < x1, since it is strictly increasing, this implies that f(x1) < f(x2) or f(x2) < f(x1), so f(x1) can never equal f(x2), therefore the function must be one to one.

Is this what you are saying?

6. Jun 9, 2010

### VeeEight

Yes, that is the right idea.

For clarity, instead of saying "f(x1) can never equal f(x2)", just state that "f(x1) < f(x2) or f(x2) < f(x1)" is a contradiction with the fact that f(x1) = f(x2) and thus the function is one-to-one.

7. Jun 9, 2010

### staw_jo

Alright, thank you SO much for your help!

8. Jun 9, 2010

### VeeEight

Great, glad I could help.

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