Question about a proof that the Cartesian product $ \mathbb{Z}^{+} \times \mathbb{Z}^{+} $ is countably infinite

Question

I encountered this question along with its proof in which I have to prove that the Cartesian product $ \mathbb{Z}^{+} \times \mathbb{Z}^{+} $ is countably infinite. (I don't know any advanced set theory and so my only understanding is that I have to find a bijection between $ \mathbb{Z}^{+} \times \mathbb{Z}^{+} $ and $ \mathbb{Z}^{+}. $)

Here's the proof from my textbook (Munkres Topology), quoted verbatim: First we define a bijection $ f : \mathbb{Z}^{+} \times \mathbb{Z}^{+} \to A $ where $ A $ is a subset of $ \mathbb{Z}^{+} \times \mathbb{Z}^{+} $ consisting of pairs $ (x,y) $ for which $ y \leq x, $ by the equation $ f(x, y) = (x + y - 1, y). $ Then we define a bijection $ g : A \to \mathbb{Z}^{+} $ by the rule $ \displaystyle g(x, y) = \frac{1}{2}(x - 1)x + y. $

I have successfully showed that $ f $ is a bijection and that $ g $ is one-to-one. To show that $ g $ is onto, suppose that $ a \in \mathbb{Z}^{+}, $ I need to find a pair $ (x, y) $ such that $ y \leq x $ and that $ \displaystyle a = \frac{1}{2}(x - 1)x + y. $ Solving this in terms of $ x $ yields $ \displaystyle x = \frac{1 \pm \sqrt{1 - 8(y - a)}}{2}, $ but there's no guarantee that you can find such integers $ x $ and $ y $ that satisfies these requirements. Am I doing anything wrong? If I redefine $ g(x, y) = (y - 1)y + x, $ then for every $ a \in \mathbb{Z}^{+}, $ I can simply choose $ x = a $ and $ y = 1 \; (x \geq y $ in this case) and it follows that $ g(a, 1) = a. $

Also, if anyone could help me understand the motivation behind this proof, i.e. why the author defines $ f $ and $ g $ in the first place. I seem to get the proof but not the process of coming up with such a proof.

Answer 1

Consider the sequence $$ 0, 1, 3, 6,\ldots$$ of triangular numbers, its $n$-th term given by $$ \dfrac{n(n-1)}{2}.$$

Note that the difference between the $n$-th term and the $(n+1)$-th term is $n$. That means that if $a$ is a positive integer, and $\dfrac{x(x-1)}{2}$ is the largest triangular number that doesn't exceed $a$, then

$$a - \dfrac{x(x-1)}{2} \leq x$$

Now let $y = a - \dfrac{x(x-1)}{2}$ and you have that $g$ is surjective.

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On the subject of the motivation, it's very common in the literature to prove that $\mathbb{Z}^+$ is isomorphic to $\mathbb{Z}^+\times \mathbb{Z}^+$ by first creating a table

$$\begin{array}{ccccccccc} 1,1 && 1,2 && 1,3 && 1,4 && \cdots \\ 2,1 && 2,2 && 2,3 && 2,4 && \cdots \\ 3,1 && 3,2 && 3,3 && 3,4 && \cdots \\ 4,1 && 4,2 && 4,3 && 4,4 && \cdots \\ \vdots && \vdots && \vdots && \vdots && \ddots \end{array}$$

And then creating a path that goes through every element:

$$\begin{array}{ccccccccc} 1,1 & \rightarrow& 1,2 && 1,3 & \rightarrow & 1,4 && \cdots \\ & \swarrow & & \nearrow & & \swarrow && \nearrow\\ 2,1 && 2,2 && 2,3 && 2,4 && \cdots \\ \downarrow & \nearrow && \swarrow && \nearrow && \swarrow\\ 3,1 && 3,2 && 3,3 && 3,4 && \cdots \\ & \swarrow & & \nearrow & & \swarrow && \nearrow\\ 4,1 && 4,2 && 4,3 && 4,4 && \cdots \\ \downarrow & \nearrow && \swarrow && \nearrow && \swarrow\\ \vdots && \vdots && \vdots && \vdots && \ddots \end{array}$$

However, coming up with a general formula for this bijection is messy at best. Munkres aproach might be less visual, but more formal than a picture and simpler than whatever mess comes from its formula.