Math 201-B
Introduction to Proofs
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Fall 2015
Exam #1 (solutions to the practice test)
October 7, 2015
Student name:
-
Student ID:
Duration of the exam 50 minutes.
The exam includes 5 questions.
The total mark is 100 points.
Please show all the work, not only the answers.
Calculators, textbooks, and help sheets are allowed.
1. [20 points] Exercise 2 from Section 1.5 of the textbook BP.
Solution:
(a) A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8}.
(b) A ∩ B = ∅.
(c) A − B = A = {0, 2, 4, 6, 8}.
(d) A − C = {0, 6}.
(e) B − A = B = {1, 3, 5, 7}.
(f) A ∩ C = C = {2, 4, 8}.
(g) B ∩ C = ∅.
(h) C − A = ∅.
(i) C − B = C = {2, 4, 8}.
2. [20 points] Verify the following set identities using Venn diagrams:
(a) A ∩ B = A − B.
(b) A ∪ B = A ∩ B.
Solution:
(a) Both sides of the identity A ∩ B = A − B are equal to A ∩ B.
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(b) Joint Venn diagram of A and B induces a partition of the plane into the following four
regions:
(I) A ∩ B.
(II) A ∩ B
(III) B ∩ A.
(IV) A ∩ B
The left hand side of the identity, namely the set A ∪ B is a complement to Region III
in the above list (which is exactly the right-hand side of the identity) because in order
not to belong to either A or B a point on the Venn diagram should clearly be inside
both A and B.
3. [20 points]
(a) Exercise 5 from Section 2.5 of the textbook BP.
(b) Exercise 5 from Section 2.6 of the textbook BP.
Solution:
(a) Let R = (P ∧ ∼ P ) ∨ Q. Since P ∧ ∼ P is always False, we obtain the following table:
P = T rue, Q = T rue yields R = T rue.
P = T rue, Q = F alse yield R = F alse.
P = F alse, Q = T rue yields R = T rue.
P = F alse, Q = F alse yields R = F alse.
(b) One can use truth tables directly. In common language, the identity says that “None
of the statements P, Q, and R is true if and only if all of them are wrong”.
4. [20 points]
(a) Exercise 4 from Section 2.10 of the textbook BP.
(b) Exercise 9 from Section 3.3 of the textbook BP.
(c) Exercise 2 from Section 3.5 of the textbook BP.
Solution:
(a) ∃ ε > 0 such that ∀ δ > 0 ∃ xδ ∈ (a − δ, a + δ) s. t. |f (xδ ) − f (a)| ≥ ε.
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(b) There are 62 = 15 possible locations for the ordered pair (D, A). For each choice, the
four remaining letters can be placed on the four remaining spots in 4! = 24 ways. Thus
the answer is 15 × 24 = 360.
(c) Denote:
A is the set of 4-digits positive integers for which there are no repeated digits,
B is the set of 4-digits positive integers with all digits odd.
Then, by inclusion-exclusion principle,
|A ∪ B| = |A| + |B| − A ∩ B|
= 9(all but zero) · 9(all but the fist one) · 8 · 7 + 54 − 5 · 4 · 3 · 2.
5. [20 points]
(a) Exercise A.6 from Chapter 5 of the textbook BP.
(b) Exercise A.13 from Chapter 6 of the textbook BP.
(c) Exercise 5 from Chapter 7 of the textbook BP.
Solution:
(a) Assume that x ≤ −1. Then
x3 − x = x x(x − 1)(x + 1) ≤
since all three factors in the brackets are non-positive:
x − 1 < x < x + 1 ≤,
where the last inequality is due to the assumption x ≤ −1 we made.
(b) Assume that ∃ x ∈ π/2, π] such that sin x − cos x < 1. Let y = π − x. Then y ∈ [0, π/2],
sin y = sin x ≥ 0, cos y = − cos x ≥ 0. In particular,
0 ≤ sin y + cos y < 1,
and hence
1 > (sin y + cos y)2 = 1 + 2 cos y sin y,
which is impossible since cos y ≥ 0, sin y ≥ 0.
(c) First, observe that if a is an odd integer it doesn’t contain 2 as a prime factors. Hence
a = a · a · a doesn’t contain 2 as a prime factor. Thus a3 is odd.
On the other hand, one can show that if a3 is an odd integer then a is also odd. Indeed,
to get a contradiction, assume that a is even. Then clearly, a · a · a divides 2 and thus
is even. This show that a must be odd whenever a3 is odd.
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