Injective one to one proof?

  • Thread starter Tomp
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  • #1
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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
 

Answers and Replies

  • #2
Dick
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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'?
 
  • #3
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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
Dick
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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.
 
  • #5
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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
 
  • #6
Dick
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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.
 

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