Unit 1 - 3 practice Problems
exc. 2.5.2. Find (1) Xz(0) (2) Xz^-1 (0) (3) Xz([3, 5]) (4)Xz^-1 ([3, 5])
(1) 1, (2) R Z, (3) {0, 1}, (4) ∅
exc. 1.9.1. Let A = {a, b, c, d, e} and let B = {1, 2}. Find (1) B × A. (2) |B × A| (3) Is (a, 2) ∈ B × A? (4) Is (2, a) ∈ B × A? (5) Is 2a ∈ B × A?
(1) B⇥A = {(1, a),(1, b),(1, c),(1, d),(1, e),(2, a),(2, b),(2, c),(2, d),(2, e)} (2) |B ⇥ A| = 10, (3) No, (4) Yes, (5) No
Chp.2. exc. 1.10.1. p is the statement "I will prove this by cases", q is the statement "There are more than 500 cases," and r is the statement "I can find another way." (1) State (¬r ∨ ¬q) → p in simple English. (2) State the converse of the statement in part 1 in simple English. (3) State the inverse of the statement in part 1 in simple English. (4) State the contrapositive of the statement in part 1 in simple English.
(1) If I cannot find another way or there are not more than 500 cases, then I will prove this by cases. (2) If I prove this by cases, then I could not find another way or there are not more than 500 cases. (3) If I can find another way and there are more than 500 cases, then I will not prove this by cases. (4) If I cannot prove this by cases, then I can find another way and there are more than 500 cases.
2.p.2.10.1. Find the conjunctive normal form of the proposition (p∧¬q)∨r.
(1) Negate: ¬[(p ∧ ¬q) ∨ r] ⇔ (¬p ∨ q) ∧ ¬r. (2) Find the disjunctive normal form of (¬p ∨ q) ∧ ¬r: Truth Table The disjunctive normal form for (¬p ∨ q) ∧ ¬r is (p ∧ q ∧ ¬r) ∨ (¬p ∧ q ∧ ¬r) ∨ (¬p ∧ ¬q ∧ ¬r). (3) The conjunctive normal form for (p∧ ¬q)∨r is then the negation of this last expression, which, by De Morgan's Laws, is (¬p ∨ ¬q ∨ r) ∧ (p ∨ ¬q ∨ r) ∧ (p ∨ q ∨ r).
exc. 2.4.3. Let g : R → [0,∞) be defined by g(x) = |-x^2-|. Let A = {x ∈ [0,∞)|3.2 < x < 8.9}. (1) domain (2) codomain (3) range (4) Find g(A). (5) Find g^ −1 (A).
(1) R (2) [0, 1) (3) N (4) g(A) = {11, 12, 13,..., 80} (5) g^-1(A)=[ --spr8, --spr3) ∪ ( sqr3, sqr8]
2.c.2.3.2. Use the method of Solution 2 in Example 2.3.2 to show that the propositions p ↔ q and ¬(p ⊕ q) are equivalent.
(1) Suppose p $ q is true while ¬(p q) is false. For p $ q to be true, p and q have to both be true or both be false. (a) If both are true, then (p q) will be false, and ¬(p q) will be true, contradicting our assumption that ¬(p q) is false. Thus this case is not possible. (b) If both are false, then (p q) will be false, and ¬(p q) will be true, again contradicting our assumption that ¬(p q) is false. Thus this case is not possible. (2) Suppose p $ q is false while ¬(p q) is true. Then for p $ q to be false, p and q have di↵erent truth values. (a) Assume p is true and q is false. Then (pq) is true so ¬(pq) is false, contradicting our assumption that ¬(p q) is true. Thus this case is not possible. (b) If p is false and q is true. Then we still have (p q) is true so ¬(p q) is false, again contradicting our assumption that ¬(p q) is true. Thus this case is not possible. We have shown that it is not possible for p $ q and ¬(p q) to have di↵erent truth values and so they are equivalent.
exp. 2.4.2. Let h : [0,∞) → R be defined by h(x) = |_3x − 1c_|. Let A = {x ∈ R|4 < x < 10}. (1) The domain is (2) The codomain is (3) The range is (4) g(A) = (5) g −1 (A) =
(1) The domain is [0,∞) (2) The codomain is R (3) The range is {−1, 0, 1, 2, . . . } (4) g(A) = {11, 12, 13, 14, . . . , 28}. (5) g −1 (A) = [2, 11/3).
Let E = {4n | n ∈ N} and consider the characteristic function Xe : Z -> Z. What is the . . . (1) domain (2) codomain (3) range (4) Xe ({2n | n ∈ N}) (5) Xe^-1 ({2n | n ∈ N})
(1) Z (2) Z (3) {0, 1} (4) {0, 1} (5) Z - E
exc. 2.4.2. Suppose f : R → R is defined by f(x) = |_x_|. Find (1) the range of f. (2) the image of Z, the set of integers. (3) f −1 (π). (4) f −1 (−1.5). (5) f −1 (N), where N is the set of natural numbers. (6) f −1 ([2.5, 5.5]). (7) f([2.5, 5.5]). (8) f(f −1 ([2.5, 5.5])). (9) f −1 (f([2.5, 5.5])).
(1) Z, (2) Z, (3) ∅ (4)∅ (5) [0, 1), ( 6) [3, 6), 7) {2, 3, 4, 5}, (8) {3, 4, 5}, (9) [2, 6)
2.p.2.8.1Construct a proposition in disjunctive normal form that is true precisely when (1) p is true and q is false (2) p is true and q is false or when p is true and q is true. (3) either p is true or q is true, and r is false
(1) p is true and q is false Solution. p ∧ ¬q (2) p is true and q is false or when p is true and q is true. Solution. (p ∧ ¬q) ∨ (p ∧ q) (3) either p is true or q is true, and r is false Solution. (p ∨ q) ∧ ¬r ⇔ (p ∧ ¬r) ∨ (q ∧ ¬r) (Distributive Law)
exc. 2.3.1 Suppose f : R → R is defined by f(x) = |x|. Find (1) the range of f. (2) the image of Z, the set of integers. (3) f −1 (π). (4) f −1 (−1). (5) f −1 (Q), where Q is the set of rational numbers.
(1)[0, 1), (2) N, (3), {⇡, ⇡}, (4) ;, (5) Q
Chp. 2. exc.1.8.1. Which of the following statements are equivalent to "If x is even, then y is odd"? There may be more than one or none. (1) y is odd only if x is even. (2) x is even is sufficient for y to be odd. (3) x is even is necessary for y to be odd. (4) If x is odd, then y is even. (5) x is even and y is even. (6) x is odd or y is odd.
(2) and (6) are the only equivalent statements.
2.exp.1.14.1. (a) -11010- = (b) 11010 ∨ 10001 = (c) 11010 ∧ 10001 = (d) 11010 ⊕ 10001 =
(a) -11010- = 00101 (b) 11010 ∨ 10001 = 11011 (c) 11010 ∧ 10001 = 10000 (d) 11010 ⊕ 10001 = 01011
Chp. 2. 1.13.2. Suppose p is the proposition "the apple is delicious" and q is the proposition "I ate the apple." Notice the difference between the two statements below. (a) ¬p ∧ q = (b) ¬(p ∧ q)
(a) ¬p ∧ q = The apple is not delicious, and I ate the apple. (b) ¬(p ∧ q) = It is not the case that: the apple is delicious and I ate the apple.
Chp.2. exp.1.4.1. p: This book is interesting. ¬p can be read as: Truth table:
(i.) This book is not interesting. (ii.) This book is uninteresting. (iii.) It is not the case that this book is interesting. Truth Table: p | ¬ p T | F F | T
exp. 1.6.2. The cardinality of ∅ is _______, but the cardinality of {∅, {∅}} is ____________.
0, 2
Chp.2. Ex. 1.2.1-13 What is a proposition? 1. "Drilling for oil caused dinosaurs to become extinct." 2. "Look out!" 3. "How far is it to the next town?" 4. "x + 2 = 2x" 5. "x + 2 = 2x when x = −2" 6. All cows are brown 7. The Earth is further from the sun that venus 8. There is life on Mars 9. 2 x 2 = 5 10. "Do you want to go to the movies?" 11. "Clean up your room." 12. "2x = 2 + x." 13. "This sentence is false."
1. proposition 2. not a proposition 3. not a proposition 4. not a proposition 5. proposition 6. proposition 7. proposition 8. proposition 9. proposition 10. not a proposition. Since a question is not a declarative sentence, it fails to be a proposition. 11. not a proposition. Likewise, an imperative is not a declarative sentence; hence, fails to be a proposition. 12. not a proposition. This is a declarative sentence, but unless x is assigned a value or is otherwise prescribed, the sentence neither true nor false, hence, not a proposition. 13. not a proposition. What happens if you assume this statement is true? false? This example is called a paradox and is not a proposition, because it is neither true nor false.
exp. 1.9.1. Let A = {a, b, c, d, e} and let B = {1, 2}. Then (1) A × B = (2) |A × B| = (3) is {a, 2} ∈ A × B? (4) is (a, 2) ∈ A ∪ B?
1. {(a, 1),(b, 1),(c, 1),(d, 1),(e, 1),(a, 2),(b, 2),(c, 2),(d, 2),(e, 2)}. 2. 10 3. {a, 2} /∈ A × B 4. (a, 2) /∈ A ∪ B
exp. 1.6.1. Suppose A = {a, b}. Then |A| =
2
exc. 1.3.1. Which of the following sets are equal to the set of all integers that are multiples of 5. There may be more than one or none. (1) {5n | n ∈ R} (2) {5n | n ∈ Z} (3) {n ∈ Z | n = 5k and k ∈ Z} (4) {n ∈ Z | n = 5k and n ∈ Z} (5) {−5, 0, 5, 10}
2 and 3 only
Chp.2. exc.1.13.1. How many rows should a truth table have for a statement involving n different propositions?
2n It is not always so clear cut how many columns one needs. If we have only three propositions p, q, and r, you would, in theory, only need four columns: one for each of p, q, and r, and one for the compound proposition under discussion, which is (p ∨ q) → ¬r in this example. In practice, however, you will probably want to have a column for each of the successive intermediate propositions used to build the final one. In this example it is convenient to have a column for p ∨ q and a column for ¬r, so that the truth value in each row in the column for (p ∨ q) → ¬r is easily supplied from the truth values for p ∨ q and ¬r in that row.
2.c.2.5.1. Which of the following are equivalent to ¬(p → r) → ¬q? There may be more than one or none. (1) ¬(p → r) ∨ q (2) (p ∧ ¬r) ∨ q (3) (¬p → ¬r) ∨ q (4) q → (p → r) (5) ¬q → (¬p → ¬r) (6) ¬q → (¬p ∨ r) (7) ¬q → ¬(p → r)
4 only. The statement is equivalent to (p ^ ¬r) → ¬q and (p → r) ∨ ¬q as well.
2.c.2.5.2.Which of the following is the negation of the statement "If you go to the beach this weekend, then you should bring your books and study"? (1) If you do not go to the beach this weekend, then you should not bring your books and you should not study. (2) If you do not go to the beach this weekend, then you should not bring your books or you should not study. (3) If you do not go to the beach this weekend, then you should bring your books and study. (4) You will not go to the beach this weekend, and you should not bring your books and you should not study. (5) You will not go to the beach this weekend, and you should not bring your books or you should not study. (6) You will go to the beach this weekend, and you should not bring your books and you should not study. (7) You will go to the beach this weekend, and you should not bring your books or you should not study.
7 You will go to the beach this weekend, and you should not bring your books or you should not study
p.3.2.3 Suppose Q(y) is the predicate "y holds a world record," and the universe of discourse for y is the set of all competitive swimmers. Notice that the universe of discourse must be defined for predicates. This would be a different predicate if the universe of discourse is changed to the set of all competitive runners.
?
exp. 1.5.1. Let A = {∅, {∅}}. What are A's elements and subsets
A has two elements ∅ and {∅} and the four subsets ∅, {∅}, {{∅}}, {∅, {∅}}.
2.c.2.6.1. Prove (p → q) ∧ ¬q ⇒ ¬p.
Answers 6
2.c.2.6.2. Prove p ∧ (p → q) → ¬q is a contingency using a truth table.
Answers 6
2.p.2.9.1. Find the disjunctive normal form for the proposition p → q.
Construct a truth table for p → q: p | q | p → q | T | T | T ← T | F | F F | T | T ← F | F | T ← p → q is true when either p is true and q is true, or p is false and q is true, or p is false and q is false. The disjunctive normal form is then (p ∧ q) ∨ (¬p ∧ q) ∨ (¬p ∧ ¬q)
2.c.2.5.3. p is the statement "I will prove this by cases", q is the statement "There are more than 500 cases," and r is the statement "I can find another way." State the negation of (¬r ∨ ¬q) → p. in simple English. Do not use the expression "It is not the case."
I cannot find another way or there are not more than 500 cases, but I will not prove this by cases.
p.3.2.2. Suppose P(x) is the sentence "x has fur" and the universe of discourse for x is the set of all animals. In this example is P(x) statement true if - x is a cat - x is an alligator.
In this example P(x) is a true statement if x is a cat. It is false, though, if x is an alligator.
2.c.2.4.2. Use truth tables to verify De Morgan's Laws.
Look at answers pg 5.
2.c.2.6.5. Prove [(p → q) ∧ (q → r)] ⇒ (p → r) using a truth table.
Proof. To prove this statement we must show [(p → q) ∧ (q → r)] ⇒ (p → r) is a tautology. See answers pg. 7
2.c.2.6.3. Prove p → (p ∨ q) is a tautology using a verbal argument.
Proof. To prove this we will show that the statement cannot be false, so suppose p → (p ∨ q) is false. Then p has to be true while p ∨ q is false. But If p is true, then p ∨ q <=> T ∨ q and we know T ∨ q <=> T by the domination law. This contradict the statement that p ∨ q is false and so we must have p → (p ∨ q) is always true.
2.c.2.6.6. Prove [(p ∨ q) ∧ ¬p] ⇒ q using a verbal argument.
Proof. We will show this by proving that when the hypothesis, (p ∨ q) ∧ ¬p, is true the conclusion, q, must also be true. Suppose (p ∨ q) ∧ ¬p is true. Then both p ∨ q and ¬p are true by the definition of ^. Since ¬p is true, p must be false. Since p is false and p ∨ q is true, q must be true by the definition of _. This shows [(p ∨ q) ∧ ¬p] ⇒ q.
2.c.2.4.1. Use the propositional equivalences in the list of important logical equivalences above to prove [(p → q) ∧ ¬q] → ¬p is a tautology.
Proof. [(p → q) ∧ ¬q] → ¬p ⇔¬[(¬p ∨ q) ∧ ¬q] _ ¬p Implication Law (2 times) ⇔[¬(¬p _ q) _ ¬¬q] _ ¬p DeMorgan's Law ⇔[(¬¬p ^ ¬q) _ ¬¬q] _ ¬p DeMorgan's Law ⇔[(p ^ ¬q) _ q] _ ¬p Double Negation Law (2 times) ⇔[q _ (p ^ ¬q)] _ ¬p Commutative Law ⇔[(q _ p) ^ (q _ ¬q)] _ ¬p Distributive Law ⇔[(q _ p) ^ T] _ ¬p Tautology ⇔(q _ p) _ ¬p Identity ⇔q _ (p _ ¬p) Associative Law ⇔q _ T Tautology ⇔T Domination
Chp.2. exp. 1.13.3. Express the proposition "If you work hard and do not get distracted, then you can finish the job" symbolically as a compound proposition in terms of simple propositions and logical operators.
Set • p = you work hard • q = you get distracted • r = you can finish the job In terms of p, q, and r, the given proposition can be written (p ∧ ¬q) → r.
2.p.2.3.1. Show that (p → q) ∧ (q → p) is logically equivalent to p ↔ q. Solution 1. Show the truth values of both propositions are identical. Truth Table: Solution 2. Examine every possible case in which the statement (p → q) ∧ (q → p) may not have the same truth value as p ↔ q
Solution 1 p | q | p → q | q → p | T | T | T | T T | F | F |. T F | T | T |. F F | F | T |. T (p → q) ∧ (q → p). | p ↔ q T | T F | F F | F T | T Solution 2 Case 1. Suppose (p → q) ∧ (q → p) is false and p ↔ q is true. There are two possible cases where (p → q) ∧ (q → p) is false. Namely, p → q is false or q → p is false (note that this covers the possibility both are false since we use the inclusive "or" on logic). (a) Assume p → q is false. Then p is true and q is false. But if this is the case, the p ↔ q is false. (b) Assume q → p is false. Then q is true and p is false. But if this is the case, the p ↔ q is false. Case 2. Suppose (p → q) ∧ (q → p) is true and p ↔ q is false. If the latter is false, the p and q do not have the same truth value and the there are two possible ways this may occur that we address below. (a) Assume p is true and q is false. Then p → q is false, so the conjunction also must be false. (b) Assume p is false and q is true. Then q → p is false, so the conjunction is also false.
2.p.2.3.2. Show ¬(p → q) is equivalent to p ∧ ¬q. Solution 1: Build a truth table containing each of the statements. Solution 2: We consider how the two propositions could fail to be equivalent. This can happen only if the first is true and the second is false or vice versa.
Solution 1 p | q | ¬q | p → q | T | T | F | T T | F | T |. F F | T | F |. T F | F | T |. T ¬(p → q) | p ∧ ¬q F | F T | T F | F T | F Solution 2 Case 1. Suppose ¬(p → q) is true and p ∧ ¬q is false. ¬(p → q) would be true if p → q is false. Now this only occurs if p is true and q is false. However, if p is true and q is false, then p ∧ ¬q will be true. Hence this case is not possible. Case 2. Suppose ¬(p → q) is false and p ∧ ¬q is true. p ∧ ¬q is true only if p is true and q is false. But in this case, ¬(p → q) will be true. So this case is not possible either.
Chp.2. 1.13.1. Write the following statement symbolically, and then make a truth table for the statement. "If I go to the mall or go to the movies, then I will not go to the gym." Truth Table
Suppose we set • p = I go to the mall • q = I go to the movies • r = I will go to the gym The proposition can then be expressed as "If p or q, then not r," or (p ∨ q) → ¬r.
Chp2. exp.1.13.1. Write the following statement symbolically, and then make a truth table for the statement. "If I go to the mall or go to the movies, then I will not go to the gym." Truth Table:
Suppose we set • p = I go to the mall • q = I go to the movies • r = I will go to the gym The proposition can then be expressed as "If p or q, then not r," or (p ∨ q) → ¬r. Truth Table: p | q | r | (p ∨ q) | ¬r | (p ∨ q) → ¬r T | T | T | T | F | F T | T | F | T | T | T T | F | T | T | F | F T | F | F | T | T | T F | T | T | T | F | F F | T | F | T | T | T F | F | T | F | F | T F | F | F | F | T | T
Chp. 2. exc.1.13.2. . Find another way to express Example 1.13.2 Part b without using the phrase "It is not the case."
The apple is not delicious or I did not eat the apple.
exc.. 2.5.2.Graph the function Xz given in the previous example in the plane.
The graph should have points at (n, 1) for each integer n and the rest of the graph is a horizontal line on the x-axis with holes at the integers.
exp. 2.5.1. Consider the set of integers as a subset of the real numbers. Then χZ(y) will be __ when y is an integer and will be _______ otherwise.
Then χZ(y) will be 1 when y is an integer and will be zero otherwise.
2.p.2.5.2. Find a simple form for the negation of the proposition "If the sun is shining, then I am going to the ball game."
This proposition is of the form p → q. As we showed in Example 2.3.2 its negation, ¬(p → q), is equivalent to p ∧ ¬q. This is the proposition "The sun is shining, and I am not going to the ball game."
exc 1.8.1. Let A = {1, 2, {1}, {1, 2}} and B = {1, {2}}. True or false: (a) 2 ∈ A ∩ B (b) 2 ∈ A ∪ B (c) 2 ∈ A − B (d) {2} ∈ A ∩ B (e) {2} ∈ A ∪ B (f ) {2} ∈ A − B
a - F, b-T, c-T, d-F, e-T, f-F
Chp.2. exc. 1.13.3 Insert a comma into the sentence "If the fish is cooked then dinner is ready and I am hungry." to make the sentence mean (a) f → (r ∧ h) (b) (f → r) ∧ h
a) If the fish is cooked, then dinner is ready and I am hungry. b) If the fish is cooked then dinner is ready, and I am hungry
Exp. 2.4.1 (a) |_3.5_| = (b) |-3.5-| = (c) |_−3.5_| = (d) |-−3.5-| = (e) notice that the floor function is the ________ as truncation for positive numbers.
a) |_3.5_| = 3 (b) |-3.5-| = 4 (c) |_−3.5_| = −4 (d) |-−3.5-| = −3 (e) notice that the floor function is the same as truncation for positive numbers.
exc. 1.5.1. Let A = {1, 2, {1}, {1, 2}}. True or false? (a) {1} ∈ A (b) {1} ⊆ A (c) {{1}} ∈ A (d) {{1}} ⊆ A (e) 2 ∈ A (f ) 2 ⊆ A (g) {2} ∈ A (h) {2} ⊆ A
a-T, b-T, c- T, d- F, e- F, f-T
exp. 1.8.1. Suppose A = {1, 3, 5, 7, 9, 11}, B = {3, 4, 5, 6, 7} and C = {2, 4, 6, 8, 10}. Then (a) A ∪ B = (b) A ∩ B = (c) A ∪ C = (d) A ∩ C = (e) A − B = (f ) |A ∪ B| = (g) |A ∩ C| =
a. {1, 3, 4, 5, 6, 7, 9, 11} b. {3, 5, 7} c. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} d. ∅ e. {1, 9, 11} f. 8 g. 0
Chp. 2. exp.1.8.1. p: This book is interesting. q: I am staying at home. p → q: Truth Table:
if...then... p → q: If this book is interesting, then I am staying at home. Truth Table: p | q | p → q T | T | T T | F | F F | T | T F | F | T
p.3.2.1. The propositional function P(x) is given by "x > 0" and the universe of discourse for x is the set of integers. To create a proposition from P, we
may assign a value for x. For example, • setting x = −3, we get P(−3): "−3 > 0", which is false. • setting x = 2, we get P(2): "2 > 0", which is true.
2.c.2.1.1. Build a truth table to verify that the proposition (p ↔ q)∧(¬p∧q) is a contradiction.
p | q | (p ↔ q) | ¬p | ¬p∧q T | T | T | F | F T | F | F | F | F F | T | F | T | T F | F | T | T | F (p ↔ q)∧(¬p∧q) F F F F
2.c.2.3.1. Use a truth table to show that the propositions p ↔ q and ¬(p⊕q) are equivalent.
p | q | p ↔ q | p ⊕q | ¬(p q) T | T | T | T | T T | F | F |. F | F F | T | F |. T | F F | F | T |. T | T
2.exp.1.13.4. build a truth table for p → (q → r) and (p ∧ q) → r.
p | q | r | q → r | p ∧ q | T | T | T | T | T | T | T | F | F | T | T | F | T | T | F | T | F | F | T | F | F | T | T | T | F | F | T | F | F | F | F | F | T | T | F | F | F | F | T | F | p → (q → r) | (p ∧ q) → r T | T F | F T | T T | T T | T T | T T | T T | T
2.c.1.13.4. Build one truth table for f → (r ∧ h) and (f → r) ∧ h.
p | q | r | r ∧ h | f → r | T | T | T | T | T | T | T | F | F | T | T | F | T | F | F | T | F | F | F | F | F | T | T | T | T | F | T | F | F | T | F | F | T | F | T | F | F | F | F | T | f → (r ∧ h) | (f → r) ∧ h T | T F | F F | F F | F T | T T | F T | T T | F
Chp2. exp.1.11.1. p: This book is interesting. q: I am staying at home. p ↔ q: Truth Table:
p ↔ q: This book is interesting if and only if I am staying at home. Truth Table: p | q | p ↔ q T | T | T T | F | F F | T | T F | F | T
Chp.2. exp.1.5.1. p: This book is interesting. q: I am staying at home. p ∧ q: Truth Table:
p ∧ q: This book is interesting, and I am staying at home. Truth Table: p | q | p ∧ q T | T | T T | F | F F | T | F F | F | F
Chp.2. exp. 1.6.1. p: This book is interesting. q: I am staying at home. p ∨ q: Truth Table:
p ∨ q: This book is interesting, or I am staying at home. Truth Table: p | q | p ∧ q T | T | T T | F | T F | T | T F | F | F
Chp. 2. exp. 1.7.1. p: This book is interesting. q: I am staying at home. p ⊕ q: Truth Table:
p ⊕ q: Either this book is interesting, or I am staying at home, but not both. Truth Table: p | q | p ∧ q T | T | F T | F | T F | T | T F | F | F
2.p.2.9.2. Construct the disjunctive normal form of the proposition (p → q) ∧ ¬r
write out truth table The disjunctive normal form will be a disjunction of three conjunctions, one for each row in the truth table that gives the truth value T for (p → q) ∧ ¬r. These rows have been boxed. In each conjunction we will use p if the truth value of p in that row is T and ¬p if the truth value of p is F, q if the truth value of q in that row is T and ¬q if the truth value of q is F, etc. The disjunctive normal form for (p → q) ∧ ¬r is then (p ∧ q ∧ ¬r) ∨ (¬p ∧ q ∧ ¬r) ∨ (¬p ∧ ¬q ∧ ¬r), because each of these conjunctions is true only for the combination of truth values of p, q, and r found in the corresponding row. That is, (p ∧ q ∧ ¬r) has truth value T only for the combination of truth values in row 2, (¬p∧q∧¬r) has truth value T only for the combination of truth values in row 6, etc. Their disjunction will be true for precisely the three combinations of truth values of p, q, and r for which (p → q) ∧ ¬r is also true.
exc. 1.6.1. Let A = {1, 2, {1}, {1, 2}}, B = {1, {2}}, C = {1, 2, 2, 2}, D = {5n|n ∈ R} and E = {5n|n ∈ Z}. Find the cardinality of each set.
|A| = 4, |B| = 2, |C| = 2
exc. 2.4.1 Suppose x is a real number. Do you see any relationships among the values |_−x_|, −|_x_|, |-−x-|, and −|-x-|?
|_x_| = -|-x-| and -|_x_| = |-x-|
2.c.2.6.4. Prove (p ∧ q) → p is a tautology using the table of propositional equivalences.
¬(p ^ q) _∨ p by the implication law (¬p ∨ ¬q) ∨ p by DeMorgan's law (¬q ∨ ¬p) ∨ p by the commutative law ¬q ∨ (¬p ∨ p) by the associative law ¬q ∨ T by the tautology T by the domination law
2.p.2.4.1. Use the logical equivalences above and substitution to establish the equivalence of the statements in Example 2.3.2. Show ¬(p → q) is equivalent to p ∧ ¬q.
¬(p → q) ⇔ ¬(¬p ∨ q) Implication Law ⇔ ¬¬p ∧ ¬q De Morgan's Law ⇔ p ∧ ¬q Double Negation Law This method is very similar to simplifying an algebraic expression. You are using the basic equivalences in somewhat the same way you use algebraic rules like 2x−3x = −x or ((x + 1)(x − 3))/ (x − 3) = x + 1.
2.c.2.6.7. Prove (p ∧ q) → (p ∨ q) is a tautology using the table of propositional equivalences.
¬(p ∧ q) → (p ∨ q) by the implication law , T by the tautology
2.p.2.5.1. Use the logical equivalences above to show that ¬(p ∨ ¬(p ∧ q)) is a contradiction.
¬(p ∨ ¬(p ∧ q)) ⇔ ¬p ∧ ¬(¬(p ∧ q)) De Morgan's Law ⇔ ¬p ∧ (p ∧ q) Double Negation Law ⇔ (¬p ∧ p) ∧ q Associative Law ⇔ F ∧ q Contradiction ⇔ F Domination Law and Commutative Law
exp. 1.5.2. If S = {2, 3, {2}, {4}}, then what are or are not element or subset. 1 {1} {1} {{1}}
• 2 ∈ S • 3 ∈ S • 4 6∈ S • {2} ∈ S • {3} 6∈ S • {4} ∈ S • {2} ⊂ S • {3} ⊂ S • {4} 6⊂ S • {{2}} ⊂ S • {{3}} 6⊂ S • {{4}} ⊂
Chp.2. exp.1.10.1. Implication: If this book is interesting, then I am staying at home. • Converse: • Inverse: • Contrapositive:
• Converse: If I am staying at home, then this book is interesting. • Inverse: If this book is not interesting, then I am not staying at home. • Contrapositive: If I am not staying at home, then this book is not interesting.
exp. 2.3.2 Let f : N → R be defined by f(n) = √ n. • The domain is • The codomain is • The range is • The image of • The preimage of • The preimage of N is
• The domain is the set of natural numbers. • The codomain is the set of real numbers. • The range is {0, 1, √ 2, √ 3, 2, √ 5, . . . }. • The image of 5 is √ 5. • The preimage of 5 is 25. • The preimage of N is the set of all perfect squares in N.
exp. 2.3.1 f −1 (b) instead of f −1 ({b}). Regardless, f −1 (b) is still a subset of A. a ->z x b ->y y c ->z z d ->z • f(a) = • the image of d is • the domain of f is • the codomain is • f(A) = • f({c, d}) = • f −1 (y) = • f −1 (z) = • f −1 ({y, z}) = • f −1 (x) =
• f(a) = z • the image of d is z • the domain of f is A = {a, b, c, d} • the codomain is B = {x, y, z} • f(A) = {y, z} • f({c, d}) = {z} • f −1 (y) = {b}. • f −1 (z) = {a, c, d} • f −1 ({y, z}) = {a, b, c, d} • f −1 (x) = ∅.
