Question 1184157
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.