# Injective one to one proof?

1. May 13, 2012

### Tomp

The problem statement, all variables and given/known data

Let E ⊆ R and f : E → R a decreasing function for all x ∈ E. Prove that f is injective.

The attempt at a solution

I tried that f were not injective.
Then, there exist x < y such that f(x) = f(y)
-This contradicts f being a decreasing function.
I think this is right, but I am unsure what to do now

2. May 13, 2012

### Dick

In what way are you unsure that x<y and f(x)=f(y) contradicts f being a decreasing function? What's your definition of 'decreasing function'?

3. May 13, 2012

### Tomp

For a one to one function that is decreasing i have that x < y so therefore f(x) < f(y). So f(x) is not equal to f(y)).

or have i got this the wrong way around?

4. May 13, 2012

### Dick

So, if f(x)=f(y) for x<y then that would contradict the assumption that f is decreasing, yes? I'm still confused about where you are confused.

5. May 13, 2012

### Tomp

Yes it would. I thought it would be easier to prove though contradiction my question, but I am unsure whether this is the best way to prove this/right approach.

Sorry for the confusion

6. May 13, 2012

### Dick

If you are assuming f(x)=f(y) for x<y then you ARE doing a proof by contradiction. You've assumed f is not injective. And it is the right way to go. Maybe it would be better if you showed your whole proof step by step.