Answer
See explanation
Work Step by Step
A classic proof that \(\mathbb{Q}\) (the set of all rational numbers) is countable goes as follows:
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## 1. Expressing Rationals in Lowest Terms
Every rational number \(r\) can be uniquely written (up to a sign convention) as
\[
r \;=\; \frac{p}{q},
\quad
\text{where }p \in \mathbb{Z},\, q \in \mathbb{Z}^+,
\text{ and } \gcd(p,q)=1.
\]
So it suffices to count all such pairs \((p,q)\).
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## 2. Counting Integer Pairs
We know \(\mathbb{Z}\) is countable and \(\mathbb{Z}^+\) is countable. Hence their Cartesian product \(\mathbb{Z} \times \mathbb{Z}^+\) is countable. Indeed, you can **list** all pairs \((p,q)\) in a two‐dimensional grid (with \(p\) on one axis and \(q\) on the other) and then enumerate them by “diagonals” or another systematic path.
However, not all pairs \((p,q)\) in \(\mathbb{Z} \times \mathbb{Z}^+\) correspond to different rationals: you must ensure \(\gcd(p,q)=1\) and account for sign. But these conditions only **remove** some pairs (those that are not in lowest terms or have negative denominators), which means you are taking a **subset** of a countable set.
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## 3. Subsets of Countable Sets
A subset of a countable set is at most countable. More precisely, if \(A\) is countable and \(B \subseteq A\), then \(B\) is either finite or countably infinite (hence “countable”).
Here, the set of all \((p,q)\) with \(\gcd(p,q)=1\) and \(q>0\) is a subset of the countable set \(\mathbb{Z} \times \mathbb{Z}^+\). Therefore, it is **countable**.
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## 4. Conclusion
Because each rational number corresponds to exactly one such pair \((p,q)\) in lowest terms (ignoring a finite set of sign conventions), we have a **bijection** between \(\mathbb{Q}\) and a **countable subset** of \(\mathbb{Z}\times \mathbb{Z}^+\). Hence \(\mathbb{Q}\) is **countable**.
Therefore, the set of all rational numbers is countable.
