Injective Function Proof for Decreasing Functions

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    Injective Proof
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Homework Statement

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
 
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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'?
 
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?
 
Tomp said:
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)).

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.
 
Dick said:
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.

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
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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