Solutions for Assignment 2
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Restate the following in the form of if p then q.
1. Joey will pass the discrete mathematics exam if he studies hard.
If Joey studies hard, then he will pass the discrete mathematics exam.
3. A necessary condition for Fernando to buy a computer is that he obtains $2000.
If Fernando buys a computer, then he will obtain $2000.
4. A sufficient condition for Katrina to take the algorithms course is that she passes discrete
mathematics.
If Katrina passes discrete mathematics, then she will take the algorithms course.
7. You may inspect the aircraft only if you have the proper security clearance.
If you inspect the aircraft, then you have the proper security clearance.
Assume that p and r are false and q and s are true, find the truth value of each proposition below.
16. (𝑝 → 𝑞) ∧ (𝑞 → 𝑟)
Substitute p, q, and r by their values.
(𝐹 → 𝑇) ∧ (𝑇 → 𝐹) = 𝑇 ∧ 𝐹 = 𝐹
19. (𝑠 → (𝑝 ∧ ~𝑟)) ∧ ((𝑝 → (𝑟 ∨ 𝑞)) ∧ 𝑠)
Substitute p, q, r and s by their values.
(𝑇 → (𝐹 ∧ ~𝐹)) ∧ ((𝐹 → (𝐹 ∨ 𝑇)) ∧ 𝑇)
= (𝑇 → 𝐹) ∧ ((𝐹 → 𝑇) ∧ 𝑇)
= 𝐹 ∧ (𝑇 ∧ 𝑇) = 𝐹
Assume that p is true, q is false, and r’s status is unknown at this time. Tell whether each proposition is
true, is false, or has unknown status at this time.
30. (𝑞 ∨ 𝑟) ↔ 𝑟
q is false. So, (𝑞 ∨ 𝑟) = 𝑟.
We can get 𝑟 ↔ 𝑟. Therefore, it is true.
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Determine the truth value of each proposition below.
31.
If 3+5<2, then 1 + 3 = 4
𝐹→𝑇=𝑇
34.
If 3+5>2, then 1 + 3 ≠ 4
𝑇→𝐹=𝐹
37.
3+5<2 if and only if 1 + 3 ≠ 4
𝐹↔𝐹=𝑇
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Formulate the arguments below symbolically and determine whether each is valid.
Let p: I study hard.
q: I get A’s.
r: I get rich.
1. If I study hard, then I get A’s.
I study hard.
∴ I get A’s.
𝑝→𝑞
𝑝
∴ 𝑞
Valid
Since p is true, q cannot be false. If q is false, the statement 𝑝 → 𝑞 will be false which contradicts the
first statement. Therefore, q must be true.
4. If I study hard or I get rich, then I get A’s.
I get A’s.
∴ If I don’t study hard, then I get rich.
(𝑝 ∨ 𝑟) → 𝑞
𝑞
∴ ~𝑝 → 𝑟
Invalid
The fact of that the conclusion q is true does not guarantee that the hypotheses is true. That is,
(𝑝 ∨ 𝑟) could be true or false, not sure. In addition, nothing can be extended to ~𝑝 → 𝑟.
Therefore, it is invalid.
5. If I study hard, then I get A’s or I get rich.
I don’t get A’s and I don’t get rich.
∴ I don’t study hard.
𝑝→𝑞∨𝑟
~𝑞 ∧ ~𝑟
∴ ~𝑝
Valid
~𝑞 ∧ ~𝑟 = ~(𝑞 ∨ 𝑟) is true. That is, the conclusion of the first statement is false. So, the
hypothesis of the first statement must be false. The negation of false is true. ∴ ~𝑝 is true.
2
Write the given argument in words and determine whether each argument is valid. Let
p: 4 megabytes is better than no memory at all.
q: We will buy more memory.
r: We will buy a new computer.
6. 𝑝 → 𝑟
If 4 megabytes is better than no memory at all, then we will buy a new computer.
𝑝→𝑞
If 4 megabytes is better than no memory at all, then we will buy more memory.
∴ 𝑝 → (𝑟 ∧ 𝑞)
Therefore, if 4 megabytes is better than no memory at all, then we will buy a new computer and
we will buy more memory.
It is valid.
9. ~𝑟 → ~𝑝 If we will not buy a new computer, then 4 megabytes is not better than no memory at all
𝑟
We will buy a new computer.
∴𝑝
Therefore, 4 megabytes is better than no memory at all.
~𝑟 → ~𝑝 ≡ 𝑝 → 𝑟
So, this one is invalid.
The conclusion r is true. The hypothesis does not have to be true.
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Let p(n) be “n divides 77”. Write propositions below in words and tell whether it is true or false.
The domain of discourse is 𝑍 + .
7. p(11)
11 divides 77. It is true.
10. ∀𝑛 𝑝(𝑛) For every positive integer n, n divides 77. It is false since 2 cannot divide 77.
Let p(x) be “𝑥 ≥ 𝑥 2 ” Tell whether each proposition below is true or false. The domain of
discourse is R.
15. ∀𝑥 𝑝(𝑥)
False. The counterexample is that ∃𝑥, 𝑥 = 2 ∧ (2 ≱ 22 )
18. ~(∃𝑥 𝑝(𝑥)) ≡ ∀𝑥 ~𝑝(𝑥) False. The counterexample is that ∃𝑥, (𝑥 = 0.5 ) ∧ (0.5 ≥ 0.52 )
Let P(x) denote “x is a professional athlete.” Let Q(x) denote “x plays soccer.” The domain of
discourse is the set of all people. Write each proposition below in words. Determine the truth
value of each statement.
35. ∀𝑥 (𝑃(𝑥) → 𝑄(𝑥))
Every professional athlete plays soccer. False
38. ∃𝑥 (𝑄(𝑥) → 𝑃(𝑥))
True
There exists someone x such that if x plays soccer then x is a professional athlete.
It means that if x does not play soccer then x could be either a professional athlete or not.
The conclusion of the translation is followed.
Either someone does not play soccer or some soccer player is a professional athlete.
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41. ∀𝑥 (𝑃(𝑥) ∧ 𝑄(𝑥)) Everyone is a professional athlete and plays soccer. False
42. ∃𝑥 (𝑃(𝑥) ∧ 𝑄(𝑥)) Someone is a professional athlete and a soccer player. True
43. Write negation of each of #35, 38, 41, 42 in symbolically and in words.
Negation of #35: ∃𝑥 (𝑃(𝑥) ∧ ~𝑄(𝑥))
Someone is a professional athlete and does not play soccer.
Negation of #38: ∀𝑥 (𝑄(𝑥) ∧ ~𝑃(𝑥))
Every soccer player is a professional athlete.
Negation of #41: ∃𝑥 (~𝑃(𝑥) ∨ ~𝑄(𝑥))
Someone is not a professional athlete or not a soccer player.
Negation of #42: ∀𝑥 (~𝑃(𝑥) ∨ ~𝑄(𝑥))
Everyone is not a professional athlete or not a soccer player.
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Determine the truth value of each statement below. The domain of discourse is 𝑅 × 𝑅.
Justify your answers.
42. ∀𝑥∀𝑦(𝑥 2 < 𝑦 + 1) False. Counterexample: x=2, y=1.
45. ∃𝑥∃𝑦(𝑥 2 < 𝑦 + 1) True. x = 1, y = 1.
48. ∀𝑥∀𝑦(𝑥 2 + 𝑦 2 = 9) False. Counterexample: x=3, y=3.
51. ∃𝑥∃𝑦(𝑥 2 + 𝑦 2 = 9) True. x=0, y=3.
57. ∀𝑥∃𝑦((𝑥 < 𝑦) → (𝑥 2 < 𝑦 2 )) True.
If the hypothesis is false, no matter the conclusion is true or false, the statement is true.
Therefore, for any x, let y=x-1. That is, 𝑥 ≮ 𝑦.
60. Write the negation of each proposition of #42, 45, 48, 51 and 57.
Negation of #42: ∃𝑥 ∃𝑦 (𝑥 2 ≥ 𝑦 + 1)
Negation of #45: ∀𝑥 ∀𝑦 (𝑥 2 ≥ 𝑦 + 1)
Negation of #48: ∃𝑥 ∃𝑦 (𝑥 2 + 𝑦 2 ≠ 9)
Negation of #51: ∀𝑥 ∀𝑦 (𝑥 2 + 𝑦 2 ≠ 9)
Negation of #57: ∃𝑥 ∀𝑦 ((𝑥 < 𝑦) ∧ (𝑥 2 ≥ 𝑦 2 ))
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