Midterm — October 8th 2010
Mathematics 220
Page 1 of 5
This midterm has 4 questions on 5 pages, for a total of 25 points.
Duration: 50 minutes
• Read all the questions carefully before starting to work.
• With the exception of Q1, you should give complete arguments and explanations for
all your calculations; answers without justifications will not be marked.
• Continue on the back of the previous page if you run out of space.
• Attempt to answer all questions for partial credit.
• This is a closed-book examination. None of the following are allowed: documents,
cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.)
Full Name (including all middle names):
Student-No:
Signature:
Question:
1
2
3
4
Total
Points:
7
6
6
6
25
Score:
Mathematics 220
7 marks
Midterm — October 8th 2010
Page 2 of 5
1. (a) Write the negation of the following statement:
“For every (a, b) ∈ N × N, a > b or a + b ≤ ab.”
Solution: ∃(a, b) ∈ N × N s.t. (a ≤ b) and (a + b > ab).
(b) Write the converse and contrapositive of the following statement:
“If it is Tuesday then this is Belgium.”
Solution: Converse — “If this is Belgium then it is Tuesday.”
Contrapositive — “If it is not Belgium then it is not Tuesday.”
(c) Give a precise mathematical definitions of the following sets
[
\
B=
Sα
C=
Sα
α∈I
α∈I
Solution:
S
• α∈I Sα = {x|∃α ∈ I s.t. x ∈ Sα }
T
• α∈I Sα = {x|∀α ∈ I, x ∈ Sα }
(d) For any n ∈ N, let An =
B=
1
2
,
n
+
1
. Simplify the following sets
n
[
n∈N
An
C=
\
An
n∈N
Solution: Union is B = (0, ∞). Intersection is C = [1, 2].
(e) Give an example of two sets, A and B, such that A ⊆ P (A) and B 6⊆ P (B).
Solution: Put A = ∅. Then A is a subset of any set. Now let B = {1} then
P (B) = {∅, {1}}, so B 6⊆ P (B).
Mathematics 220
6 marks
Midterm — October 8th 2010
Page 3 of 5
2. (a) Prove or disprove the following statement
Let a, b, c, d ∈ R. If a ≤ c and b ≤ d then ab ≤ cd.
Solution: The statement is false. Let a = −1, b = −1, c = 0, d = 0. Then a < c
and b < d but ab = 1 and cd = 0.
(b) Determine whether the following four statements are true or false — explain your
answers (“true” or “false” is not sufficient).
(i)
(ii)
(iii)
(iv)
∀x ∈ R, ∀y ∈ R, xy = x + y.
∀x ∈ R, ∃y ∈ R s.t. xy = x + y.
∃x ∈ R s.t. ∀y ∈ R, xy = x + y.
∃x ∈ R s.t. ∃y ∈ R s.t. xy = x + y.
Solution:
(i) False — negation is ∃x ∈ R s.t. ∃y ∈ R s.t. xy 6= x + y. This statement is
true. Consider x = y = 0.1, then xy = 0.01 and x + y = 0.2.
(ii) False — negation is ∃x ∈ R s.t. ∀y ∈ R s.t. xy 6= x + y. Now pick x = 1,
this gives y 6= y + 1 which is true for any y.
(iii) False — negation is ∀x ∈ R, ∃y ∈ R s.t. xy 6= x + y. For any x pick y = 1,
then x 6= x + 1.
(iv) True — Pick x = y = 0.
Mathematics 220
6 marks
Midterm — October 8th 2010
Page 4 of 5
3. (a) Let n ∈ Z. Prove that n2 + 1 is even if and only if 7n + 3 is even.
Solution: The number n is either even or odd and so we consider the two cases.
• Assume n is even, so n = 2k for some k ∈ Z. Then n2 + 1 = 4k 2 + 1 =
2(2k 2 ) + 1 and so is odd. Similarly 7n + 3 = 14k + 3 = 2(7k + 1) + 1 and
so is odd.
• Now assume n is odd, so n = 2k+1 for some k ∈ Z. Then n2 +1 = 4k 2 +4k+
2 = 2(2k 2 + 2k + 1) and so is even. Similarly 7n + 3 = 14k + 10 = 2(7k + 5)
and so is even.
Thus we see that either n2 + 1 and 7n + 3 are both even or both odd.
(b) Let S = {(x, y) ∈ R × R s.t. y = 3x − 7}. Prove the following
If (a, b) ∈ S and (b, a) ∈ S then a = b.
Solution:
Proof. Let (a, b) ∈ S and (b, a) ∈ S. Thus b = 3a − 7 and a = 3b − 7. We must
solve these simultaneously. Substituting one of these equations into the other
− 7 = 27 . Thus if
gives a = 3(3a − 7) − 7 = 9a − 28. Hence a = 72 and so b = 21
2
7
(a, b) ∈ S and (b, a) ∈ S then a = b = 2 .
Midterm — October 8th 2010
Mathematics 220
6 marks
Page 5 of 5
4. Let A, B, C be sets.
(a) Prove that A ∪ B ⊆ A ∩ B.
Solution:
Proof. We could do this using a contrapositive.
• We must show that if x 6∈ A or x 6∈ B then x 6∈ A ∩ B.
• We prove the contrapositive of this statement, namely that if x ∈ A ∩ B
then x ∈ A and x ∈ B.
• Assume that x ∈ A ∩ B. By definition we have x ∈ A and x ∈ B. So we
are done.
Alternatively we could do cases.
• Let x ∈ A ∪ B and so x 6∈ A or x 6∈ B.
• This gives 2 cases.
– Assume x 6∈ A then x cannot be in A ∩ B and so x 6∈ A ∩ B.
– Assume x 6∈ B then x cannot be in B ∩ A and so x 6∈ A ∩ B.
• In either case we have that x ∈ A ∩ B.
(b) Prove that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C).
Solution:
Proof. Let x ∈ A ∩ (B ∪ C).
• So x ∈ A and x ∈ B ∪ C. Hence either x ∈ B or x ∈ C. This gives 2 cases
to check.
– Assume x ∈ B. Since x ∈ A and x ∈ B it follows that x ∈ A ∩ B.
Hence x ∈ (A ∩ B) ∪ (A ∩ C)
– Assume x ∈ C. Since x ∈ A and x ∈ C we know that x ∈ A ∩ C. It
follows that x ∈ (A ∩ C) ∪ (A ∩ B).
• In either case x ∈ (A ∩ B) ∪ (A ∩ C).