SOLUTION: Construct two examples of bijective function from ZxZ to Q where Z is set of all integers and Q is set of rational numbers?

Algebra.Com
Question 1184157: Construct two examples of bijective function from ZxZ to Q where Z is set of all integers and Q is set of rational numbers?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
It's impossible to construct a bijective function from Z×Z to Q. Here's why:
* **Cardinality:** The set Z×Z (the set of all ordered pairs of integers) is countably infinite. This means its cardinality is the same as the set of integers (denoted by ℵ₀). The set of rational numbers Q is also countably infinite (cardinality ℵ₀).
* **Bijection Requirement:** A bijection requires a one-to-one and onto mapping between two sets. While both Z×Z and Q are countably infinite, that *alone* doesn't guarantee a bijection is possible. The *way* the elements are ordered or structured matters.
The crucial issue is that while both sets are countably infinite, any attempt to create an ordered list of elements in Z×Z and Q so that you can pair them off one-to-one will fail. You can list all the elements of Z×Z, and you can list all the elements of Q. But the fundamental difference is that between any two elements in Q, there are infinitely many other rational numbers. This "denseness" of Q is something Z×Z does not have. You can't create a mapping that preserves the "gaps" between integers when mapping to a set with no gaps.
**Why the "interleaving" approach doesn't work:**
You might think you could "interleave" the elements of Z×Z and Q. For example, you could list the elements of Z×Z in a spiral pattern and try to pair them with the rational numbers in some order. However, no matter how clever your interleaving method, you'll always run into the problem of the density of Q. You'll use up infinitely many pairs from ZxZ to cover a single "gap" between two elements in Q.
**In summary:** There is no bijective function from Z×Z to Q. The difference in their "denseness" properties prevents such a mapping from being constructed.

Answer by ikleyn(52779)   (Show Source): You can put this solution on YOUR website!
.
Construct two examples of bijective function from ZxZ to Q where Z is set
of all integers and Q is set of rational numbers?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        In his post,  @CPhill states and insists that there is no bijective function from  Z×Z  to  Q.
        "The difference in their  " denseness "  properties prevents such a mapping from being constructed."

        It is not correct,  and below I explain  .


The set ZxZ is countably infinite.  It means that there is a bijection 

    f: ZxZ  <---> Z


The set Q is also countably infinite. It means that there is a bijection

    g: Z <---> Q


It    that the composition  gof  is the bijection ZxZ and Q

            f             g
    ZxZ <--------> Z <--------> Q.



It disproves the statement by @CPhill.


Another question is IF it is possible to construct an explicit bijective function .

I asked Google. Google AI created this response under this link

https://www.google.com/search?q=Construct+two+examples+of+bijective+function+from+ZxZ+to+Q+where+Z+is+set+of+all+integers+and+Q+is+set+of+rational+numbers%3F&rlz=1C1CHBF_enUS1071US1071&oq=Construct+two+examples+of+bijective+function+from+ZxZ+to+Q+where+Z+is+set+of+all+integers+and+Q+is+set+of+rational+numbers%3F&gs_lcrp=EgZjaHJvbWUyBggAEEUYOTIGCAEQRRhA0gEJMTg1NWowajE1qAIIsAIB8QXIg8hlL4FHy_EFyIPIZS-BR8s&sourceid=chrome&ie=UTF-8


             - - - START of the Google response - - - (of 2/21/2025)


Two examples of bijective functions from ℤxℤ to ℚ (where ℤ is the set of all integers and ℚ is the set 
of all rational numbers) are: 

Example 1: Function using a numerator-denominator pairing 

• Definition: f(m, n) = 
	• If n ≠ 0, then f(m, n) = m/n 
	• If n = 0, then f(m, n) = m 



Explanation: 

• This function maps each ordered pair (m, n) from ℤxℤ to a unique rational number. 
• When n is not zero, it represents the rational number "m/n". 
• When n is zero, the function effectively maps the pair to infinity (or any other fixed value depending on your interpretation of division by zero). 

Example 2: Function using a "diagonal" mapping with a sign convention 

• Definition: f(m, n) = (2m + n) / (2*(abs(n) + 1)) 

Explanation: 

• This function creates a unique mapping by leveraging the "diagonal" pattern of adding 2m and n, while ensuring that the denominator always has a positive even value. 
• The absolute value of n in the denominator is used to handle cases where n is negative, preventing duplicate values. 

Key points about bijective functions: [1, 2]  

• A bijective function needs to be both injective (one-to-one) and surjective (onto), meaning each element in the domain maps to a unique element in the codomain, and every element in the codomain is reached by some element in the domain. [1, 2]  


Generative AI is experimental.


             - - - END of the Google response - - - 


                Below is my comment about the Google AI response


Example 1 above is  INCORRECT.  The function defined in  Example 1 is  NOT  bijective.
It is not bijective,  since the pairs  (m,1)  and  (m,0) map into the same rational/(integer)  number  m.


                Surely,  I reported to  Google about it.


Example 2  looks like to be correct.  At least,  I do not see the way to disprove it.



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