How to prove that this function is a surjection?

In summary, the conversation discusses the process of finding the inverse of a function with one variable and how it becomes more complicated with two variables. The goal is to prove that for any given positive integer p, there exist two positive integers m and n that satisfy the equation 1/2(m+n-2)(m+n-1)+n=p. It is suggested to draw a sketch to better understand the function and find explicit solutions for m and n. The definition of a bijection for a function of two variables is also mentioned.
  • #1
Anezka
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Member advised that the homework template must be used
If f were a function of 1 variable only, then this would be straight forward as I can try to find its inverse by reversing the operations defined in f. I know I need to show that for any given positive integer,p, there exists two positive integers, m and n such that 1/2(m+n−2)(m+n−1)+n=p. However, I am not so sure what to do in the case of two variables. Do I have to show that if I assume that if m or n are some fixed positive integer, then there is a solution for p?

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  • #2
Did you draw a sketch? That should make clear how the function "works". With that you can also find explicit n,m for every p if you want.
 
  • #3
Anezka said:
If f were a function of 1 variable only, then this would be straight forward as I can try to find its inverse by reversing the operations defined in f. I know I need to show that for any given positive integer,p, there exists two positive integers, m and n such that 1/2(m+n−2)(m+n−1)+n=p. However, I am not so sure what to do in the case of two variables. Do I have to show that if I assume that if m or n are some fixed positive integer, then there is a solution for p?

View attachment 224576

Did anyone ever mention that your ##n## looks very similar to your ##m##?

Anyway, the definition of a bijection for a function, ##f##, of two variables is:

1) ##f## is a surjection: ##\forall \ p, \ \exists m, n \ ## such that ##f(m, n) = p##
2) ##f## is an injection: ##f(n_1, m_1) = f(n_2, m_2) \ \Rightarrow \ n_1 = n_2 \ ## and ##m_1 = m_2##
 

Related to How to prove that this function is a surjection?

1. How can I prove that a function is a surjection?

To prove that a function is a surjection, you need to show that every element in the range of the function has a corresponding element in the domain.

2. What is the definition of a surjective function?

A surjective function, also known as a "onto" function, is a function where every element in the range is mapped to by at least one element in the domain.

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

A surjective function ensures that every element in the range is mapped to, while an injective function ensures that no two elements in the domain are mapped to the same element in the range.

4. How do I prove that a function is not a surjection?

If a function is not a surjection, it means that there is at least one element in the range that does not have a corresponding element in the domain. To prove this, you can provide a counterexample or use a proof by contradiction.

5. 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 and it means that every element in the range is mapped to by exactly one element in the domain, and no two elements in the domain are mapped to the same element in the range.

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