Proving functions are surjective

In summary, the first function g: \mathbb{R} \rightarrow \mathbb{R} is not injective because it does not imply that x = x' when g(x') = g(x). It is surjective because for any y in the real numbers, there exists an x in the real numbers such that g(x) = y. For the second function g: \mathbb{Z} \rightarrow \mathbb{Z}, it is also not injective and its surjectivity can be verified using the intermediate value theorem.
  • #1
synkk
216
0
prove

whether or not the following functions are surjective or injective:
1) [tex] g: \mathbb{R} \rightarrow \mathbb{R}[/tex] [tex] g(x) = 3x^3 - 2x [/tex]

2)[tex] g: \mathbb{Z} \rightarrow \mathbb{Z}[/tex] [tex] g(x) = 3x^3 - 2x [/tex]my working for 1):

injective: suppose [tex] g(x') = g(x) [/tex] : [tex] 3x'^3 - 2x' = 3x^3 - 2x [/tex] this does not imply [tex] x = x' [/tex] hence not injective
surjective: need to show [tex] \forall y \in \mathbb{R} \exists x \in \mathbb{R} s.t. g(x) = y [/tex], [tex] y = 3x^3 - 2x [/tex] as cubics always have one real root then as ## y \in \mathbb{R} ## ## \exists x \in g(x) \in \mathbb{R} ## s.t. ## g(x) = y ## therefore it's surjective

2):
injective: same as 1:
surjective: I'm not sure how to phrase this for the integers,

overall I'm not happy with my proofs, for the injectivity I haven't really shown that x is not equal to x', how would I do it? And for surjectivity I have mainly written it in words, how would I write it out formally for both questions?
 
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  • #2
synkk said:
injective: suppose [tex] g(x') = g(x) [/tex] : [tex] 3x'^3 - 2x' = 3x^3 - 2x [/tex] this does not imply [tex] x = x' [/tex] hence not injective

Why doesn't it imply that it's not injective? You seem to have skipped a step here.
 
  • #3
FeDeX_LaTeX said:
Why doesn't it imply that it's not injective? You seem to have skipped a step here.

I don't really know I just saw it as that :\
 
  • #4
synkk said:
I don't really know I just saw it as that :\

Can you find an example which demonstrates that g is not injective?
 
  • #5
FeDeX_LaTeX said:
Can you find an example which demonstrates that g is not injective?

0 and rt(2/3)
 
  • #6
Yes, that works. The function's surjectivity can be verified via the intermediate value theorem.
 

1. What is a surjective function?

A surjective function, also known as an onto function, is a type of function in mathematics where every element in the range of the function corresponds to at least one element in the domain. This means that for every output of the function, there is at least one input that produces that output.

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

To prove that a function is surjective, you must show that every element in the range of the function has at least one corresponding element in the domain. This can be done by either using the definition of surjective functions or by using a proof by contradiction.

3. What is the difference between a surjective function and an injective function?

A surjective function ensures that every element in the range has at least one corresponding element in the domain, while an injective function ensures that every element in the domain has at most one corresponding element in the range. In other words, a surjective function covers all possible outputs, while an injective function covers all possible inputs.

4. Can a function be both surjective and injective?

Yes, a function can be both surjective and injective. This type of function is called a bijective function. It means that every element in the range has exactly one corresponding element in the domain, and every element in the domain has exactly one corresponding element in the range.

5. Why is proving a function is surjective important?

Proving that a function is surjective is important because it ensures that every output of the function has at least one corresponding input. This is essential in many mathematical applications, such as solving equations and finding inverse functions. It also helps in understanding the relationship between the domain and range of a function.

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