Math 201-B
Introduction to Proofs
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Fall 2015
Exam #2 (solutions)
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Duration of the exam 50 minutes.
The exam includes 5 regular questions and a bonus problem.
The total mark is 100 points for regular questions.
Please show all the work, not only the answers.
Calculators, textbooks, and help sheets are allowed.
1. [20 points]
(a) Prove that if A, B, and C are sets such that
A ∩ (B − C) = ∅
and
A ∩ (C − B) = ∅,
then
(B ∪ C) − (B ∩ C) ⊆ A.
(b) Prove that Z = {2m + 3n : m, n ∈ Z}.
Solution:
(a) Consider any element x such that
x ∈ (B ∪ C) − (B ∩ C) .
We must show that x ∈ A. To this end observe (drow the diagram) that
B ∪ C = (B − C) ∪ (C − B) ∪ (B ∩ C),
and that the three sets on the right-hand side are mutually disjoint. In other words, the union
of B and C can be partitioned into the disjoint union of the set of elements which are in B
but not in C, elements which are in C but not in B, and elements which are both in B and
C. It follows that
(B ∪ C) − (B ∩ C) = (B − C) ∪ (C − B).
Therefore, the condition
A ∩ (B − C) = ∅
and
A ∩ (C − B) = ∅
implies that (B ∪ C) − (B ∩ C) = (B − C) ∪ (C − B) is a subset of A.
(b) Clearly, {2m + 3n : m, n ∈ Z} ∈ Z. To prove the opposite inclusion we must show that for
any k ∈ Z there exist m, n ∈ Z such that k = 2m + 3n. There many ways to choose such a
pair (m, n). For instance, one can set m = −k and n = k. Another solution which appeared
frequently in your work is to represent k in the form k = 2i + j where i ∈ {o, 1} (thus to look
at the parity of k) and then set n = 1 and m = 2i − 2 if j = 1 (i. e. k is odd) and set n = 0
and m = 2i if j = 0 (i. e. k is even).
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2. [20 points] Prove or disprove:
(a) A × B = C × D if and only if A = C and B = D.
(b) (x, y) ∈ R2 : |x| + |y| ≤ 1 ⊆ (x, y) ∈ R2 : x2 + y 2 ≤ 1 .
Solution:
(a) This is a tricky one. The correct answer is that the statement is false. But the only counterexample is a situation when either A = D = ∅ or B = C = ∅ in which case both the
cross-products under consideration are empty sets. I gave full credit to students who wrongfully claimed that the statment is true and provided a correct argumet supporting their claim
in the case when neither of the cross-products is an empty set.
(b) Let A = (x, y) ∈ R2 : |x| + |y| ≤ 1 and B = (x, y) ∈ R2 : x2 + y 2 ≤ 1 . Notice first that if
(x, y) ∈ A then |x| ≤ 1 and |y| ≤ 1. This implies x2 = |x|2 ≤ |x| and, similarly, y 2 = |y|2 ≤ |y|.
Therefore, if (x, y) ∈ A then
x2 + y 2 ≤ |x| + |y| ≤ 1,
and hence (x, y) ∈ B. This yields A ⊆ B. The proof is complete.
3. [20 points] Using induction, show that (n3 + 2n) is a multiplyer of 3 for all n ∈ N.
Solution: The claim is clearly true for n = 0 (or n = 1, if you prefer). This provided the base of
the induction. Now assume that n3 + 2n = 3kn where kn ∈ N. This will serve us as the induction
hypothesis. Now write
(n + 1)3 + 2(n + 1) = (n3 + 3n2 + 3n + 1) + (2n + 2) =
= (n3 + 2n) + 3(n2 + n + 1) = 3(kn + n2 + n + 1).
This shows that (n + 1)3 + 2(n + 1) is a multiple of 3 whenever n3 + 2n is. The proof is therefore
complete.
4. [20 points]
(a) Let R be a relation on Z such that
xRy
if either x ≡ y (mod 3) or else x ≡ y (mod 5).
Is R reflexive? Symmetric? Transitive?
(b) Let R be a relation on Z such that
xRy
if 2x + 5y ≡ 0 (mod 3).
Is R an equivalence relation?
Solution:
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(a) Notice that both
x ≡ y (mod 3)
and
x ≡ y (mod 5)
are equivalence relations on Z. Thus R is a union of two equivalence relation. As we discussed
in the class, in general a union of two equivalence relation is not an equivalence relation, and
the problem is the transitivity property.
Specifically in our case, it is easy to see that R is both reflexive and symmetric. To see that
it is not transitive one can for instance observe that 3R15 since 3 ≡ 15 (mod 3) and 15R5
because 15 ≡ 5 (mod 5) but (3, 5) 6∈ R.
(b) The correct answer is that R is not an equivalence relation on Z. In fact, R is not reflexive
(consider x = y = 1), is symmetric (because 2x + 5y ≡ 2y + 5x (mod 3) due to the fact that
the difference (2x + 5y) − (2y + 5x) = 3(y − x) devises 3), and is not transitive. To check the
latter claim one can, for instance, observe that (1, 2) ∈ R, (2, 1) ∈ R, but as we have already
discussed (1, 1) 6∈ R.
5. [20 points] Exercise 12 from Section 12.2 of the textbook BP.
Solution: The function is injective. This wouldn’t be the case if a would take values in a reacher
set than two-point {0, 1}. So, from the first glance this reult might seem to be counterintuitive.
Nevertheless, we have:
1. First observation: θ(0, b1 ) = θ(0, b2 ) clearly implies b1 = b2 .
2. Second observation: θ(1, b1 ) = θ(1, b2 ) means
1 − 2b1 + b1 = 1 − 2b2 + b2 ,
and hence also implies b1 = b2 .
3. Third observation: θ(0, b1 ) = θ(1, b2 ) implies
b1 = 1 − 2b2 + b2 = 1 − b2 ,
and thus b1 +b2 = 1. Since b1 and b2 are both positive integers the latter equality is impossible.
The three observation we made show that the fuction θ is injective.
To see that θ is surjective, consider any n ∈ Z. If n > 0, that is n ∈ N, set
a=0
and
b = n.
If n ≤ 0 set
a=1
b = 1 − n.
and
Notice that in the latter case b ∈ N and
θ(a, b) = a − 2ab + b = 1 − b = n.
This completes the proof that θ is surgective. Since it is both injective and surjective, θ is a
bijection.
CONTINUED IN THE NEXT PAGE
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6. [Bonus] Is it possible to alternate just exactly one symbol/letter in the statement of Problem 4(b)
in such a way that the answer will change to the opposite one?
Solution: Let R be a relation on Z such that
xRy
if 2x−5y ≡ 0 (mod 3).
Then such defined R is an equivalence relation. Indeed, since 3y divides 3,
R = {(x, y) ∈ Z2 : 3|(2x − 5y)} = {(x, y) ∈ Z2 : 3|(2x − 2y)} = {(x, y) ∈ Z2 : 3|(x − y)}.
Thus R is in fact the equivalence mod 3 relation on Z.
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