Miami Dade College MGF 1106 Math to Stats Exam 2 Solutions Spring 2019

This document provides solutions for the MGF 1106 Math to Stats Pathway (Blended) Exam 2, administered at Miami Dade College (MDC) in Spring 2019. The exam covers key concepts in logic and set theory, including translating symbolic statements into words and vice versa, writing negations of quantified statements, and constructing and interpreting truth tables. The solutions demonstrate how to determine the validity of arguments using truth tables, and how to find the inverse, converse, contrapositive, and negation of conditional statements. Specifically, the solutions address problems related to symbolic logic, truth tables, and argument validity, offering step-by-step explanations and detailed answers to each question in the exam, which includes determining whether two statements are equivalent.

MGF 1106Math to Stats Pathway (Blended)
MDC Spring 2019
Exam II Chapter 3
Prof. Charles
NAME: __________________ DATE: ____________
SCORE: _____________

Let p, q, and r represent the following simple statements:
p: I’m registered.
q: I’m a citizen.
r: I vote.
Write each of the symbolic statements in words:
1.
Sol.
Since, I am registered and I am a citizen, therefore, I vote.
2.
Sol.
If do not vote then either I am not registered or I am not a citizen.
Write each compound statement in symbolic form.
3. I am registered and a citizen, or I do not vote.
Sol.
𝑝 ∩ 𝑞 ∪ ~𝑟
4. If I am not registered or not a citizen, then I do not vote.
Sol.
(𝑝~ ∪ ~𝑞) → ~𝑟
Write the negation of each quantified statement. (The negation should begin with “all,”
“some,” or “no.”)
5. All numbers are divisible by 5.
Sol.
No numbers are divisible by 5.
6. Some people wear glasses.
Sol.
Some people do not wear glasses.
7. Not all dogs are playful.
Sol.

All dogs are playful.
Construct a truth table for the statement.
8.
Sol.
p q ~𝑞 𝑝 ∪ ~𝑞 ~(~𝑝 ∪ ~𝑞)
0 0 1 1 0
0 1 0 1 0
1 0 1 1 0
1 1 0 1 0
Construct a truth table for the given statement. Use the truth table to determine whether
the statement is a tautology.
9.
Sol.
p q r ~𝑞 𝑝 ∨ 𝑟 ∼ (𝑝
∨ 𝑟)
∼ 𝑞 ∧ 𝑟 ∼ (𝑝
∨ 𝑟) →
∼ 𝑞 ∧ 𝑟
0 0 0 1 0 1 0 0
0 0 1 1 1 0 1 1
0 1 0 0 0 1 0 0
0 1 1 0 1 0 0 1
1 0 0 1 1 0 0 1
1 0 1 1 1 0 1 1
1 1 0 0 1 0 0 1
1 1 1 0 1 0 0 1
From the truth table, we can see that this statement is not a tautology.

Use the truth table to determine whether the two statements are equivalent.
10. ~p q, q p
Sol.
p q ~𝑝 ~𝑝 → 𝑞 𝑞 → ~𝑝
0 0 1 0 1
0 1 1 1 1
1 0 0 1 1
1 1 0 1 1
From the above table, we can conclude that the above two statements are not equivalent to each
other.
11. Write the inverse, converse, and the contrapositive of: if it is August, it does not
snow.
a) Inverse
Sol.
If it is not August, it snows.
b) Converse
Sol.
If it does not rain, then it is August.
c) Contrapositive
Sol.
If it snows, then it is not August.
d) Negation
Sol.
It is August and it snows.

determine whether each argument in problem 11-14 is valid or invalid.
12. If a parrot talks, it is intelligent.
This parrot is intelligent.
Sol.
𝑝: 𝐴 𝑝𝑎𝑟𝑟𝑜𝑡 𝑡𝑎𝑙𝑘𝑠
𝑞: 𝑃𝑎𝑟𝑟𝑜𝑡 𝑖𝑠 𝑖𝑛𝑡𝑒𝑙𝑙𝑖𝑔𝑒𝑛𝑡
𝑆: 𝐼𝑓 𝑎 𝑝𝑎𝑟𝑟𝑜𝑡 𝑡𝑎𝑙𝑘𝑠, 𝑖𝑡 𝑖𝑠 𝑖𝑛𝑡𝑒𝑙𝑙𝑖𝑔𝑒𝑛𝑡 = 𝑝 → 𝑞
𝑇: 𝑇ℎ𝑖𝑠 𝑝𝑎𝑟𝑟𝑜𝑡 𝑖𝑠 𝑖𝑛𝑡𝑒𝑙𝑙𝑖𝑔𝑒𝑛𝑡, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑝𝑎𝑟𝑟𝑜𝑡 𝑡𝑎𝑙𝑘𝑠 = 𝑞 → 𝑝
p q 𝑆 = 𝑝 → 𝑞 𝑇 = 𝑞 → 𝑝
0 0 1 1
0 1 1 0
1 0 0 1
1 1 1 1
We can say that for all cases when S is true, T is not true. Therefore, it is not a valid
statement.
13. I am sick or I am tired.
I am not tired.
Sol.
𝑝: 𝐼 𝑎𝑚 𝑠𝑖𝑐𝑘
𝑞: 𝐼 𝑎𝑚 𝑡𝑖𝑟𝑒𝑑
𝑆: 𝐼 𝑎𝑚 𝑠𝑖𝑐𝑘 𝑜𝑟 𝐼 𝑎𝑚 𝑡𝑖𝑟𝑒𝑑 = 𝑝 ∪ 𝑞
𝑇: 𝐼 𝑎𝑚 𝑛𝑜𝑡 𝑡𝑖𝑟𝑒𝑑, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑖 𝑎𝑚 𝑠𝑖𝑐𝑘 = ~𝑞 → 𝑝
p q ~𝑞 𝑆 = 𝑝 ∪ 𝑞 𝑇 = ~𝑞 → 𝑝
0 0 1 0 0
0 1 0 1 1
1 0 1 1 1
1 1 0 1 0
We can say that for all cases when S is true, T is not true. Therefore, it is not a valid
statement.
14. All rabbis are Jewish.
Sol.
𝑝: 𝐴𝑙𝑙 𝑟𝑎𝑏𝑏𝑖𝑠 𝑎𝑟𝑒 𝐽𝑒𝑤𝑖𝑠ℎ
𝑞: 𝑆𝑜𝑚𝑒 𝐽𝑒𝑤𝑠 𝑜𝑏𝑠𝑒𝑟𝑣𝑒 𝑘𝑜𝑠ℎ𝑒𝑟 𝑑𝑖𝑒𝑡𝑎𝑟𝑦 𝑡𝑟𝑎𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑆: 𝑆𝑜𝑚𝑒 𝑟𝑎𝑏𝑏𝑖𝑠 𝑜𝑏𝑠𝑒𝑟𝑣𝑒 𝑘𝑜𝑠ℎ𝑒𝑟 𝑑𝑖𝑒𝑡𝑎𝑟𝑦 𝑡𝑟𝑎𝑑𝑖𝑡𝑖𝑜𝑛𝑠

Since it is given that p and q are always true, that All rabbis are Jewish and Some Jews observe
kosher dietary traditions. Hence this statement is also true that Some rabbis observe kosher
dietary traditions.