To prove that the rational number $\frac{p}{q}$ is the $\frac{1}{2}(p + q - 1)(p + q - 2) + q$th term in the given sequence using mathematical induction, we will break the problem down into three steps:

1. Understanding the structure of the sequence: The sequence of rational numbers arranges all positive rational numbers in such a way that each pair $\frac{p}{q}$ (where $p$ and $q$ are positive integers) appears exactly once. The order follows a pattern that lists numbers with smaller values of $p + q$ first, and within each group of the same $p + q$, numbers are listed from largest numerator to smallest numerator.

   Here's how the first few terms appear:

   • $p + q = 2$: $\frac{1}{1}$
   • $p + q = 3$: $\frac{2}{1}, \frac{1}{2}$
   • $p + q = 4$: $\frac{3}{1}, \frac{2}{2}, \frac{1}{3}$
   • $p + q = 5$: $\frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}$

2. Inductive Hypothesis: We need to show by mathematical induction that the $\frac{p}{q}$ term corresponds to the expression:

   $n = \frac{1}{2}(p + q - 1)(p + q - 2) + q$

   where $n$ is the position of $\frac{p}{q}$ in the series.

3. Base Case: We start with the base case when $p = 1$ and $q = 1$, i.e., $\frac{p}{q} = \frac{1}{1}$. For $p = 1$ and $q = 1$, we have: $n = \frac{1}{2}(1 + 1 - 1)(1 + 1 - 2) + 1 = \frac{1}{2}(1)(0) + 1 = 1$ This matches the first term in the sequence, which is $\frac{1}{1}$. Thus, the base case holds.

4. Inductive Step: Assume that the formula holds for all terms up to $p + q = k$, i.e., for any rational number $\frac{p}{q}$ where $p + q = k$, the position is correctly given by:

   $n = \frac{1}{2}(p + q - 1)(p + q - 2) + q$

   We need to prove that it holds for $p + q = k + 1$.

   The number of terms for $p + q = k$ is $k - 1$, and the total number of terms up to $p + q = k$ is:

   $N_k = \frac{1}{2}(k)(k - 1)$

   Now, for $p + q = k + 1$, the first term will be $\frac{k + 1}{1}$ and the last term will be $\frac{1}{k + 1}$. The number of new terms added when $p + q = k + 1$ is exactly $k$, and the total number of terms up to this point is:

   $N_{k+1} = N_k + k = \frac{1}{2}(k)(k - 1) + k = \frac{1}{2}(k + 1)(k)$

   This confirms the formula holds for the next step in the induction. Thus, by mathematical induction, the formula is correct for all $p$ and $q$.

Conclusion:

By the principle of mathematical induction, we have proven that the rational number $\frac{p}{q}$ is the $\frac{1}{2}(p + q - 1)(p + q - 2) + q$th term in the sequence.

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Would you like more details or have any questions?

Here are some related questions to expand on the topic:

1. How are rational numbers ordered in the sequence based on their sum $p + q$?
2. Can we use a different method, such as a geometric argument, to visualize the sequence of rational numbers?
3. What is the significance of using mathematical induction in this context?
4. How can this sequence be applied to real-world problems involving rational numbers?
5. How would the formula change if we considered all rational numbers, including negative ones?

Tip: When using mathematical induction, always check