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Math 310/ 2–13–
Solutions to Problem Set 6
- Premises:
A ↔ B
B∧ ∼ C
∼ A ∨ D
Prove: D.
- A ↔ B Premise
- B∧ ∼ C Premise
- ∼ A ∨ D Premise
- B Decomposing a conjunction (2)
- B → A Definition of biconditional
- A Modus ponens (4,5)
- D Disjunctive syllogism (3,6)
- Premises:
(A ∨ C) →∼ B
D →∼ A
Prove: A →∼ (B ∨ D).
- (A ∨ C) →∼ B Premise
- D →∼ A Premise
- A Premise for conditional proof
- A ∨ C Constructing a disjunction
- ∼ B Modus ponens (1,4)
- ∼ D Modus tollens (2,3)
- ∼ B∧ ∼ D Constructing a conjunction (5,6)
- ∼ (B ∨ D) DeMorgan (7)
- A →∼ (B ∨ D) Conditional proof (3,8)
- (a) Show that if ln x + ln(x − 1) = ln 2, then x = 2 or x = −1.
ln x + ln(x − 1) = ln 2 ln x(x − 1) = ln 2 eln^ x(x−1)^ = eln 2 x(x − 1) = 2 x^2 − x = 2 x^2 − x − 2 = 0 (x − 2)(x + 1) = 0
Therefore, x = 2 or x = −1.
(b) Show that it’s not true that if x = 2 or x = −1, then ln x + ln(x − 1) = ln 2.
Plugging x = 2 into ln x + ln(x − 1) = ln 2 gives ln 2 + ln 1 = ln 2, which is true. The answer checks. Plugging x = −1 into ln x + ln(x − 1) = ln 2 gives ln(−1) + ln(−2) = ln 2, which doesn’t make sense — the log of a negative number is undefined. The answer doesn’t check.
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- (a) Prove that if x > 3, then x^2 + 4x + 7 > 28.
x > 3 implies that x^2 > 9, and also that 4x > 12. Adding the last two inequalities, I get x^2 + 4x > 21. Adding 7 to both sides, I get x^2 + 4x + 7 > 28.
(b) Give a specific counterexample which shows that the converse is false.
The converse is: “If x^2 + 4x + 7 > 28, then x > 3.” If x = −10, then x^2 + 4x + 7 = 100 − 40 + 7 = 67 > 28. However, − 10 6 >3. Therefore, the converse is false.
- (a) Prove that if n is even, then n^2 + 3n + 16 is even.
Let n be an even integer. Then n = 2m for some m ∈ Z. So
n^2 + 3n + 16 = (2m)^2 + 3(2m) + 16 = 4m^2 + 6m + 16 = 2(2m^2 + 3m + 8).
Hence, n^2 + 3n + 16 is even.
(b) Give a specific counterexample to show that the converse is false.
The converse is: “If n^2 + 3n + 16 is even, then n is even.” However, if n = 1, then n^2 + 3n + 16 = 20, which is even, while n is odd. Thus, the converse is false.
- Suppose that f is a continuous function,
f (3) = 2, and f (6) = − 3.
Prove that x^2 f (x) has a root between x = 3 and x = 6.
g(x) = x^2 f (x) is continuous,
g(3) = 3^2 f (3) = 18, and g(6) = 6^2 f (6) = − 108.
By the Intermediate Value Theorem, g(c) = 0 for some c between 3 and 6 — that is, x^2 f (x) has a root between x = 3 and x = 6.
- Premises:
A →∼ B
B ↔ (A ∨ C)
Prove: B → (C∧ ∼ A).
- A →∼ B Premise
- B ↔ (A ∨ C) Premise
- B Premise for conditional proof
- (B → (A ∨ C)) ∧ ((A ∨ C) → B) Definition of biconditional
- B → (A ∨ C) Decomposing a conjunction (4)
- A ∨ C Modus ponens (3,5)
- ∼ A Modus tollens (1,3)
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