1.
1
True or False. Justify your answer.
1.1
A. The Contrapositive of A → B is ¬B → ¬A.
True.
A B
T T
T F
F T
F F
1.2
¬A
F
F
T
T
¬B
F
T
F
T
A→B
T
F
T
T
¬B → ¬A.
T
F
T
T
B. If you want to prove a statement is true, it is enough to ﬁnd
867 examples where it is true.
False
The case may not be true for the 868th example.You must show it is true for all cases.
However, to disprove it, you only need to show one counter example.
1.3
C. If P ∧ Q is true, then P ∨ Q is true.
True
P ∧ Q is true means that P is true and Q is true. In order for P ∨ Q to be true, we
only need only one of P, Q to be true. Since we have both, then P ∨ Q is true.
1.4
D. P → Q and Q → P are logically equivalent.
False
P Q P → Q Q → P.
T T
T
T
T F
F
T
F T
T
F
F F
T
T
Consider the bolded cases that have diﬀerent truth values.
1
2.
1.5
E. If p → q is false, then the truth value of (¬p ∨ ¬q) → (p ↔ q) is
also false.
True.
Realize that the only case where p → q is false is when p is true and q is false. So ¬p
is false and ¬q is true. So (¬p ∨ ¬q) is true since ¬q is true.
Also, (p ↔ q) means (p → q) ∧ (q → p). However, since p → q is false, then (p ↔ q) is
false.
Therefore, since the ﬁrst part of the implication(¬p∨¬q) is true and the second (p ↔ q)
is false, we have that (¬p ∨ ¬q) → (p ↔ q) is false when p → q is false.
2
2.1
Prove that if a and b are odd integers, then a ∗ b is an odd integer.
Proof. Assume that a, b are odd integers. Then a = 2k + 1 for some integer k. Also,
b = 2m + 1 for some integer m.
We have that
a ∗ b = (2k + 1)(2m + 1).
= 4km + 2k + 2m + 1.
= 2(2km + m + k) + 1
.
Since k, m are integers, so is 2km + k + m.
So a ∗ b is odd.
3
3.1
For any mathematical statement, say C, fn (C) denotes the mathematical statement: ¬¬ . . . ¬C, where there are n¬ symbols in front
of C.
Prove that if A is a True mathematical statement and B is a False mathematical
statement, then ¬(f3 (A) ∨ f2 (B)) is a True mathematical statement.
2
3.
Proof. Assume that A is a True mathematical statement and B is a False mathematical
statement.
We have that f3 (A) = ¬¬¬A = ¬(¬¬)A = ¬A since every pair of ¬ symbols cancels
out by the Double Negative Law.
We also have f2 (B) = ¬¬B = (¬¬)B = B.
So we have ¬(¬A ∨ B), which is (¬¬A ∧ ¬B).
So we have A ∧ ¬B.
Since A is true and B is false (that is, ¬B is true), we have two true statements.
The joining of two true statements with ”and” is true, so ¬(f3 (A) ∨ f2 (B)) is a True
mathematical statement
4
4.1
4.1.1
p
T
T
T
T
F
F
F
F
Find the Disjunctive Normal Form of ((p → q) ∧ (q → r)) → (p → r)
Method: Truth Table
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
¬p
F
F
F
F
T
T
T
T
¬q
F
F
T
T
F
F
T
T
¬r
F
T
F
T
F
T
F
T
p→q
T
T
F
F
T
T
T
T
q→r
T
F
T
T
T
F
T
T
p→r
T
F
T
F
T
T
T
T
((p → q) ∧ (q → r))
T
F
F
F
T
F
T
T
((p → q) ∧ (q → r)) → (p → r)
T
T
T
T
T
T
T
T
So we have (p ∧ q ∧ r) ∨ (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ r) ∨ (p ∧ ¬q ∧ ¬r) ∨ (¬p ∧ q ∧ r) ∨ (¬p ∧
q¬r) ∨ (¬p ∧ ¬q ∧ r) ∨ (¬p ∧ ¬q ∧ ¬r).
Note that this the original statement is true no matter what the truth values of p, q, r.
So Disjunctive Normal Form addresses each of the possible cases represented in the truth
table.
3
4.
5
5.1
Let a, b, c be real numbers. Prove that if a + b ≥ c, then a ≥
b ≥ c2.
c
2
or
Note: This can be proven directly but then you must address each of the following cases:
c
c
(i). a = 2 , b = 2 .
c
c
(ii) a = 2 , b > 2 .
c
c
(iii) a > 2 , b = 2 (similar to ii)
c
c
(iiii) a > 2 , b > 2 .
This is why contradiction proofs are easier to use in situations where you have OR
statements to deal with. In this situation, a contradiction gives us an AND statement,
which is a single case. When writing proofs it is often helpful to ask yourself ”When can
this be false?” Then show that that case cannot happen.
So,
c
c
Proof. Suppose for contradiction that a + b = c and a < 2 and b < 2 .
c
c
Then a + b < 2 + 2 .
So a + b < c, which contradicts our original assumption that a + b = c.
4
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