Injective Function Proof for Decreasing Functions

In summary, we are given a set E ⊆ R and a decreasing function f : E → R. To prove that f is injective, we will first assume that f is not injective. This means that there exist x < y such that f(x) = f(y). However, this contradicts f being a decreasing function, as for a one-to-one function that is decreasing, we have x < y implying f(x) < f(y). Thus, f(x) cannot be equal to f(y). Therefore, our initial assumption that f is not injective is false, and thus, f must be injective.
  • #1
Tomp
27
0
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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  • #2
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
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
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.
 
  • #5
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
 
  • #6
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.
 

1. What is an injective (one-to-one) function?

An injective function is a type of mathematical function where each element in the function's range corresponds to exactly one element in its domain. In other words, no two elements in the domain can map to the same element in the range.

2. How do you prove that a function is injective?

To prove that a function is injective, you must show that for any two distinct elements in the domain, the corresponding elements in the range are also distinct. This can be done by assuming that the function is not injective and then arriving at a contradiction through logical steps.

3. What are some common techniques for proving injectivity?

Some common techniques for proving injectivity include using the definition of an injective function, using algebraic manipulation, and using the contrapositive or contradiction method. It is also important to clearly define the domain and range of the function before attempting to prove injectivity.

4. Can a function be both injective and surjective?

Yes, a function can be both injective (one-to-one) and surjective (onto). This type of function is called a bijective function, where each element in the domain corresponds to exactly one element in the range, and every element in the range has at least one corresponding element in the domain.

5. Are there any real-life examples of injective functions?

Yes, real-life examples of injective functions include a one-to-one correspondence between a person's social security number and their identity, or the relationship between a person's fingerprint and their identity. In both cases, each person has a unique identifier that cannot be shared with anyone else.

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