Math 201
Introduction to Proofs
Instructor: Alex Roitershtein
Iowa State University
Fall 2015
Solutions to selected problems from Homework #2
Section 2.1.
9. The statement cos(x) = −1 can be interpreted as the claim that the function cos(x) is a
constant identically. This is of course a false statement.
10. (R × N) ∩ (N × R) = (R ∩ N) × (N ∩ R) = N × N. True.
11. Impossible to know as long as the value of x is not specified.
13. True.
14. Not a statement at all.
15. Depends on a point of view, but I guess the textbook is expecting the answer “impossible
to know”.
Section 2.2.
2. ∼ P, where P is the claim that A is invertible.
4. ∼ P, where P is the claim that x ≥ y.
9. P ∧ Q, where P claims that X ∈ A and Q says that x 6∈ B.
10. P ∧ Q, where P claims that x 6∈ A and Q says that x 6∈ B.
11. ∼ P, where P is the claim that either |A| = ∞ or A is at all not a subset of N.
Section 2.3.
1. IF the determinant of a matrix A is not zero, the matrix is invertible.
5. If an integer is divisible by 8, it is divisible by 4.
11. If you failed, it means that you had stopped writing.
Section 2.4.
1. A matrix A is invertible if and only if det(A) 6= 0.
1
3. xy = 0 if and only if either x = 0 or y = 0.
5. An occurrence becomes an adventure if and only if one recounts it.
Section 2.5.
1. The entire statement is false if and only if both P and (Q ⇒ R) are false. The table for
Q ⇒ R can be found on p. 49 of the textbook. The conclusion is that the statement is false
if and only if P = T, Q = T, and R = F.
9. The entire statement is false if and only if (∼ P ∨ ∼ Q) is true, if and only if either P or
Q are false.
11. Let A = (R ⇒ S) and B = (P ∧ Q). The table for A ⇔ B on p. 49 of the textbook
implies that A ⇔ B is true if and only if either A = B = T or A = B = F. Since P is false,
so is B. Thus A is false. This is only possible if R = T and S = F.
Section 2.6.
1. Let Pn be the set of polynomial of degree at least n. Then the statement reads:
f ∈ P3 ⇒ f 00 6≡ 0.
6. ∀ ε > 0 ∃ M > 0 s. t.
x > M ⇒ |f (x) − b| < ε.
Section 2.10.
3. There is a maximal prime number.
5. ∃ε > 0 s. t. ∀M > 0 the following is true:
∃ x > M : |f (x) − b| ≥ ε.
One can verify (this is beyond the scope of our present discussion) that the statement is
equivalent to the following one:
∃ ε > 0, (xn )n∈N : lim xn = ∞ and |f (xn ) − b| ≥ ε ∀ n ∈ N.
n∞
2