Set Theory and Logic

a) Set operations

Set operations are the operations performed on two or more sets to form a new set. The main ones are union, intersection, difference and complement.

Let $U=\{1,2,3,4,5,6,7,8\}$, $A=\{1,2,3,4\}$ and $B=\{3,4,5,6\}$. Two overlapping circles inside the rectangle $U$ divide it into four regions:

 +------------------------------------+
 |  U                                 |
 |      ___________   ___________     |
 |     /     A     \ /     B     \    |
 |    /   (I)       X       (III)  \   |
 |    |  1  2   |  3  4  |  5  6  |   |
 |    \          / \  (II)         /   |
 |     \________/___\______________/   |
 |                                    |
 |   (IV)  7  8  (outside both)       |
 +------------------------------------+

Region I $=A-B$, region II $=A\cap B$, region III $=B-A$, region IV $=(A\cup B)'$. The Venn diagram of each operation is obtained by shading these regions:

b) Difference between mean and median

Example: for the data 2, 4, 6, 8, 100, the mean is $\dfrac{120}{5}=24$ but the median (the $3^{rd}$ item) is $6$. The extreme value 100 pulls the mean up but does not change the median, so the median is the better representative here.

How I teach the median of discrete data

1. Recall: five students of different heights stand in a line from shortest to tallest, and I ask who stands in the middle.
2. Rule: I state that the data must first be arranged in ascending order. If $N$ is odd, median $=\left(\dfrac{N+1}{2}\right)^{th}$ item; if $N$ is even, median $=$ average of the $\left(\dfrac{N}{2}\right)^{th}$ and $\left(\dfrac N2+1\right)^{th}$ items.
3. Demonstration: 7, 3, 9, 5, 11 arranged as 3, 5, 7, 9, 11 gives median $=3^{rd}$ item $=7$; 12, 8, 15, 10 arranged as 8, 10, 12, 15 gives median $=\dfrac{10+12}{2}=11$.
4. Practice: in pairs, students find the median of marks from a recent class test.
5. Evaluation: a short exercise of three items (one odd, one even, one unarranged) with correction of the common mistake of forgetting to arrange the data.

a) For the statement “All scientists are honest”:

Let (P): a person is a scientist, and (Q): a person is honest.
The statement is (P \Rightarrow Q).

• Converse: (Q \Rightarrow P)
  “All honest persons are scientists.”

• Inverse: (\neg P \Rightarrow \neg Q)
  “All non-scientists are not honest” / “No non-scientist is honest.”

• Contrapositive: (\neg Q \Rightarrow \neg P)
  “All non-honest persons are non-scientists” / “No dishonest person is a scientist.”

• Negation: (P \land \neg Q)
  “Some scientists are not honest” / “At least one scientist is dishonest.”

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b) Prove that for any two non-empty sets (A) and (B), [ (A \cup B)^c = A^c \cap B^c ]

Let (U) be the universal set and (A,B \subseteq U).

Take any element (x \in U).

[ \begin{aligned} x \in (A \cup B)^c &\iff x \notin A \cup B \ &\iff x \notin A \text{ and } x \notin B \ &\iff x \in A^c \text{ and } x \in B^c \ &\iff x \in A^c \cap B^c \end{aligned} ]

Since (x) was arbitrary, every element of ((A \cup B)^c) belongs to (A^c \cap B^c), and every element of (A^c \cap B^c) belongs to ((A \cup B)^c).

Hence, [ (A \cup B)^c = A^c \cap B^c ]

This is De Morgan’s law for sets. It is not a contradiction; it is a valid identity for any sets (A) and (B).