Show that a multivariate function is bijective

In summary, the function ##f: \mathbb{Z}^{+} \times \mathbb{Z}^{+} \longrightarrow \mathbb{Z}^{+}## defined by ##\displaystyle f(m,n) = \frac{(m+n-2)(m+n-1)}{2}+m## is bijective. This is shown by first proving that if ##f(a,b) = f(c,d)##, then ##a=c## and ##b=d##. Then, it is shown that for every positive integer a, there exists an m and n such that ##f(m,n) = a##. However, there is a potential issue with the second part as ##2-m
  • #1
Mr Davis 97
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Homework Statement


Show that ##f: \mathbb{Z}^{+} \times \mathbb{Z}^{+} \longrightarrow \mathbb{Z}^{+} ##where ##\displaystyle f(m,n) = \frac{(m+n-2)(m+n-1)}{2}+m## is bijective

Homework Equations

The Attempt at a Solution


First, we show that ##f(a,b) = f(c,d) \implies a=c \land b=d##.

##f(a,b) = f(c,d)##

##\displaystyle \frac{(a+b-2)(a+b-1)}{2}+a = \frac{(c+d-2)(c+d-1)}{2}+c##

After simplification, we find that

##(a+b)^2 + a - b = (c+d)^2 + c - d##

Comparing sides, then ##a + b = c+d, ~a-b = c-d##
Then ##a=c,~b=d##

Second, we show that for all positive integers a, there exist an m and an n such that f(m,n) = a.
We see that ##f(m,2-m) = m##, so the function maps to every element in the codomain

Does this show that the function is bijective?
 
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  • #2
For the second part 2-m might not necessarily be a positive integer which seems to be a problem since the function is defined for pairs of positive integers.

For first part this conclusion
Mr Davis 97 said:
Comparing sides, then a+b=c+d, a−b=c−da+b=c+d, a−b=c−da + b = c+d, ~a-b = c-d
doesn't necessarily follow from the previous line. It might be correct but needs some additional justification me thinks.
 

1. What does it mean for a multivariate function to be bijective?

A multivariate function is said to be bijective if it is both injective (one-to-one) and surjective (onto). This means that each element in the function's range has a unique preimage in its domain, and every element in the function's domain has at least one corresponding element in its range.

2. How can you prove that a multivariate function is bijective?

To prove that a multivariate function is bijective, you must show that it is both injective and surjective. This can be done by showing that the function is one-to-one, meaning that no two distinct elements in its domain map to the same element in its range, and onto, meaning that every element in its range is mapped to by at least one element in its domain.

3. What are some techniques for showing that a multivariate function is injective?

Some common techniques for showing that a multivariate function is injective include using the horizontal line test, proving that the function's derivative is always positive or always negative, or using algebraic techniques such as substitution or solving for the inverse function.

4. How can you demonstrate that a multivariate function is surjective?

To demonstrate that a multivariate function is surjective, you can use techniques such as finding the range of the function and showing that it is equal to the entire codomain, or using algebraic techniques such as solving for the inverse function and showing that it maps to the entire domain.

5. Are there any special cases where a multivariate function may not be bijective?

Yes, there are a few special cases where a multivariate function may not be bijective. One example is if the function is not defined for certain values in its domain, such as when dividing by zero. Another example is if the function has multiple outputs for a single input, making it not one-to-one. Additionally, if the function's domain and codomain are of different sizes, it cannot be bijective.

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