Midterm — February 14th 2012
Mathematics 220
Page 1 of 6
This midterm has 5 questions on 6 pages, for a total of 30 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
5
Total
Points:
8
8
4
6
4
30
Score:
Midterm — February 14th 2012
Mathematics 220
8 marks
Page 2 of 6
1. (a) Write the negation of the following statement:
“There exist (a, b) ∈ R × R s.t. (a + b)2 < (a − b)2 ”
Solution: ∀(a, b) ∈ R × R we have ((a + b)2 ≥ (a − b)2 ).
(b) Write the converse and contrapositive of the following statement:
“If this is Tuesday, then this test is hard.”
Solution: Converse — “If this test is hard, then it is Tuesday.”
Contrapositive — “If this test is not hard, 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
•
T
α∈I
Sα = {x|∃α ∈ I s.t. x ∈ Sα }
α∈I
Sα = {x|∀α ∈ I, x ∈ Sα }
(d) For any n ∈ N, let An = − n1 − 1, 2 −
B=
[
n∈N
An
1
n2
. Simplify the following sets
C=
\
An
n∈N
Solution: Union is B = [−2, 2). Intersection is C = [−1, 1).
(e) Let A, B be sets in some universal set U . Suppose that B̄ = {1, 7, 8}, A−B = {1, 7},
B − A = {3, 5} and A ∪ B = {1, 2, 3, 5, 6, 7} Determine A, B, U .
Solution: A = {1, 2, 6, 7}, B = {2, 3, 5, 6}, and U = {1, 2, 3, 5, 6, 7, 8} (use a
Venn diagram).
Mathematics 220
8 marks
Midterm — February 14th 2012
Page 3 of 6
2. (a) Prove or disprove the following statement
Let a, b, c, d ∈ R. If ab ≥ cd then a ≥ c and b ≥ d.
Solution: The statement is false. Let a = b = −1 and c = d = 0. Then
ab = 1 > 0 = cd, but a < c and b < d.
(b) Prove or disprove the following
Let a, b, c ∈ Z. If a + b ≡ 1 mod 3 and 2b + c ≡ 1 mod 3 then a + c ≡ 2
mod 3.
Solution: This is true. Assume a + b ≡ 1 mod 3 and 2b + c ≡ 1 mod 3. Then
(a + b) − 1 = 3k and (2b + c) − 1 = 3` for some integers k and `. Adding gives
a + 3b + c − 2 = 3(k + `) so a + c − 2 = 3(k + ` − b). Since k + ` − b is an integer,
a + c ≡ 2 mod 3.
(c) Repeat part (b), but replacing “mod 3” with “mod 4.”
Solution: This is now false. Consider a = 0, b = 1, c = 3. Then a + b = 1 ≡ 1
mod 4 and 2b + c = 5 ≡ 1 mod 4, but a + c = 3 is not congruent to 1 mod 4.
Mathematics 220
4 marks
Midterm — February 14th 2012
Page 4 of 6
3. Let n ∈ Z. Prove that 21n + 2 is even if and only if n2 + 220 is even.
Solution:
Proof. 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 + 220 = 4k 2 + 220 =
2(2k 2 + 110). Since 2k 2 + 110 is an integer, n2 + 220 is even. Similarly 21n + 2 =
42k + 2 = 2(21k + 1) and so is even, since 21k + 1 is an integer.
• Now assume n is odd, so n = 2k + 1 for some k ∈ Z. Then n2 + 220 =
4k 2 + 4k + 221 = 2(2k 2 + 2k + 110) + 1 and so is odd, since 2k 2 + 2k + 110 is
an integer. Similarly 21n + 2 = 42k + 23 = 2(21k + 11) + 1 and so is odd since
21k + 11 is an integer.
Thus we see that either n2 + 220 and 21n + 2 are both even or both odd.
Note that you can also solve this using a couple of lemmas. Namely
“If n2 + 220 is even then n is even.” and
“If 21n + 2 is even then n is even.”
Both of these are easy to prove using their contrapositives. The proof of the main
result then becomes
Proof. We must prove both the forward and backward implications.
• Assume that 21n + 2 is even, then by the previous lemma n is even and hence
n = 2k for some integer k. Then n2 + 220 = 4k 2 + 220 = 2(2k 2 + 110) and so
is even, since 2k 2 + 110 is an integer.
• Now assume that n2 + 220 is even, then by the above lemma, n is even and
hence n = 2k + 1 for some integer k. Then 21n + 2 = 2(21k + 11) + 1. Since
21k + 11 ∈ Z, it follows that 21n + 2 is even.
Mathematics 220
6 marks
Midterm — February 14th 2012
Page 5 of 6
4. Determine whether the following four statements are true or false — explain your answers
(“true” or “false” is not sufficient).
(i) ∀x ∈ R, ∀y ∈ R, if (x + y ≥ 0) then (xy ≥ 0).
(ii) ∀x ∈ R, ∃y ∈ R s.t. if (x + y ≥ 0) then (xy ≥ 0).
(iii) ∃x ∈ R s.t. ∃y ∈ R s.t. if (x + y ≥ 0) then (xy ≥ 0).
Solution:
(i) False — negation is ∃x s.t. ∃y s.t. (x + y ≥ 0) and (xy < 0). This is true; for
instance x = 2 and y = −1 works. Since the negation is true, the original
statement is false.
(ii) True — Pick any x, and let y = −x − 1. Then x + y = −1 < 0 and since the
hypothesis is false, the implication is true.
(iii) True — pick x = 0, y = −1. As above, the implication is true.
Mathematics 220
4 marks
Midterm — February 14th 2012
Page 6 of 6
5. Prove or disprove the following statement
Let n be an odd integer. If n is prime, then n2 ≡ 1 mod 8.
Solution: We will show that n2 ≡ 1 mod 8 whenever n is odd, so the hypothesis n
prime is irrelevant (this is an example of a “trivial” implication). Assume n is odd,
so n = 2k + 1 for some integer k. We need to show that 8 divides n2 − 1 = 4k 2 + 4k =
4k(k + 1). Now, since k and k + 1 are consecutive integers, one of them is even, so
one of k, k + 1 has a factor of 2 in it. Pulling out this 2 we see that n2 − 1 is 8 times
an integer, as required.