91Ó°ÊÓ

It's given that the dot diagram has \(k\) columns and \(m\) rows. Therefore, there are clearly \(m\) dots in the first column (\(p_1\)). By definition, the first row has \(k\) dots in it (\(r_1\)). So, we have: 1. \(m=p_1\) 2. \(k=r_1\)

We will prove \(p_{j}=\max\left\{i: r_{i} \geq j\right\}\) for \(1\leq j\leq k\) and \(r_{i}=\max\left\{j: p_{j} \geq i\right\}\) for \(1\leq i\leq m\) using mathematical induction on \(m\). Base Case: For \(m = 1\), we have \(k = r_1\) and \(p_{j} = 1\) for all \(1\le j\le k\). Therefore, the given formulas hold true. Inductive Step: Assume that the statement is true for some \(m = n\), now we will prove it true for \(m = n+1\). By the induction hypothesis, we know that for the diagram with \(n\) rows and \(k\) columns, the following is true: \(p_{j}=\max\left\{i: r_{i} \geq j\right\}\) for \(1\leq j\leq k\) \(r_{i}=\max\left\{j: p_{j} \geq i\right\}\) for \(1\leq i\leq n\) Now, let's add a new row (the \((n+1)^{st}\) row) to the dot diagram. There are two possibilities: 1. The new row is the same as one of the earlier rows, in which case nothing changes in terms of the position of the last non-zero dot in each column, and the induction hypothesis remains true. 2. The new row is different from the earlier rows. In this case, the last non-zero dots of some columns will shift to the new row. But since the row numbers increase by only one (\(n+1\)), \(\max\left\{i: r_{i} \geq j\right\}\) and \(\max\left\{j: p_{j} \geq i\right\}\) still hold true for the \(n+1\) rows. Thus, by mathematical induction, the statement is true for all \(m\).

We need to prove that \(r_1\ge r_2\ge \cdots \ge r_m\). Look at the dot diagram and observe the following: 1. \(r_1\) is the number of dots in the first row, which includes all columns. 2. \(r_2\) is the number of dots in the second row, which can have dots only from columns \(2\) to \(k\). 3. \(r_3\) is the number of dots in the third row, which can have dots only from columns \(3\) to \(k\). And so on... Each row from the second to the last row can have dots in fewer columns than the row above it. This implies that the number of dots in successive rows decreases. Thus, we have \(r_1\ge r_2\ge \cdots \ge r_m\).

Now we have to show that the number of dots in each column of a dot diagram is completely determined by the number of dots in the rows. From Part (b), we know that: \(p_{j}=\max\left\{i: r_{i} \geq j\right\}\) for \(1\leq j\leq k\) \(r_{i}=\max\left\{j: p_{j} \geq i\right\}\) for \(1\leq i\leq m\) These two formulas show that given any dot diagram, we can find the number of dots in each column by using the number of dots in the rows. Therefore, the numbers \(p_j\) are completely determined by the numbers \(r_i\), and the number of dots in each column is completely determined by the number of dots in the rows.