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Math 215 Worksheet 2: Logical operations
- Paula has baseball practice every Friday and Saturday, so her statement “I play baseball ev- ery Friday” is true. Rephrase this as a conditional statement. State the converse and the contrapositive, and find their truth values.
Solution: “If today is Friday, then I play baseball.” The converse is “If I play baseball, then today is Friday,” which is false since Paula also plays on Saturday. The contrapositive is “If I don’t play baseball today, then today isn’t Friday,” which is true; the contrapositive is logically equivalent to the original statement so it has the same truth value.
- Let A represent “6 is an even integer” and B represents “4 divides 6.” Express each of the following compound propositions in an ordinary English sentence, and then determine its truth value.
(a) A ∧ B (b) A ∨ B (c) ¬A (d) ¬B (e) ¬(A ∧ B) (f) ¬(A ∨ B) (g) A ⇒ B
Solution: Notice that A is true and B is false. Translating each statement to English gives
(a) “6 is an even integer and 4 divides 6,” which is false, (b) “6 is an even integer or 4 divides 6,” which is true, (c) “6 is not an even integer” or “6 is an odd integer” which is false, (d) “4 does not divide 6” which is true, (e) By de Morgan’s laws, this is equivalent to (¬A) ∨ (¬B), which in English is “6 is an odd integer or 4 doesn’t divide 6” (true), (f) Again by de Morgan this is equivalent to (¬A) ∧ (¬B), which means “6 is an odd integer and 4 doesn’t divide 6” (false),
(g) “If 6 is an even integer, then 4 divides 6,” a conditional statement with a true hy- pothesis but a false conclusion, which is false.
- Find an example of a true conditional proposition whose converse is false.
Solution: “If n and m are even numbers, then n + m is even.” In the first worksheet, we conjectured that this is true, after trying several cases. It is in fact true (we’ll prove it next week sometime), so for now let’s accept that it is. The converse is “If n + m is even, then n and m are even.” This is demonstrably false; 1 + 1 = 2 is even, but 1 is not even.
- Prove de Morgan’s first law ¬(A ∧ B) ≡ (¬A) ∨ (¬B) using a truth table.
Solution: To avoid parsing a giant truth table, let’s use two: one for ¬(A ∧ B) and the other for (¬A) ∨ (¬B). For the first, we have
A B A ∧ B ¬(A ∧ B) T T T F T F F T F T F T F F F T
and for the second, A B ¬A ¬B (¬A) ∨ (¬B) T T F F F T F F T T F T T F T F F T T T Since the column for ¬(A∧B) and (¬A)∨(¬B) are the same, these statements are logically equivalent. Notice that when splitting into two truth tables, it’s essential that the first two columns (the possible T/F combinations of A and B) are the same in both.
- Consider the statement “every even number is divisible by 4.”
(a) Write this as quantified statement, and find its truth value. (b) Formally negate the quantified statement, translate it back to English, and find its truth value.
Since the statement is false for each T/F combination of A and B, it is a contradiction.
