Midterm 1 - MATH 2000
February 14, 2012
Last Name:
First Name:
Student ID:
Tutorial Section:
• No external aid is allowed. Cell phones should be turned off
before the start of this exam.
• This exam is 60 minutes long.
• There are 6 pages (including this page) and 8 questions in this
exam.
• Make sure your solutions are written clearly.
• Good luck.
Question Points
1
/8
2
/4
3
/3
4
/5
5
/5
6
/6
7
/6
8
/3
Total
/40
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Question 1: Write out each of the following sets by listing their
elements within braces, and write out their cardinality.
(a) {x ∈ Z | x2 < 3}
(b) {(x, y) ∈ R2 | x2 = 1, y = 2x}
(c) {{1, 2, . . . , n}|n ∈ {1, 2, 3}}
(d) {x ∈ Z | |x| = 1} × {y ∈ Z | y 2 = 3}
Question 2: Give an example of a set S such that
(a) S ⊆ P(N) and |S| = 3.
(b) S ∈ P(N) and |S| = 3
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Question 3: Give an example of two sets A and B such that |A −
B| = |A ∩ B| = |B − A| = 2. Draw the appropriate Venn diagram.
Question 4: For a real number r, define
Sr to be
S
T the interval [r −
1, r + 2]. Let A = {1, 3, 4}. Determine i∈A Si and i∈A Si .
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Question 5: For statements P , Q, and R, show that
P ⇒ (P ∨ Q ∨ R)
is a tautology.
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Question 6: True or False: for each statement below decide if the
statement is true or false, and give a brief explanation for each.
(a) Let G = {(x, y) ∈ R2 |x2 + y 2 = 1}. We have (1/2, 1/2) ∈ G.
(b) Let G = (4, 3) and H = [4, 3] be subsets of R. We have G ⊆ H.
(c) If Saint Valentine was a nun, then everyone will find true love.
(Note: Saint Valentine’s history is a bit of a mystery. For the
purposes of this question, you can assume that he was a religious
man in the third century.)
(d) For every z ∈ Z, if z 2 < 1 then z = 0.
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Question 7: Consider the open sentences:
P (x) : x = −2. and Q(x) : x2 = 4.
over the domain S = {−2, 0, 2}. State each of the following in words
and determine all values of x ∈ S for which the resulting statement is
true.
(a) ∼ P (x)
(b) P (x) ∧ Q(x)
(c) P (x) ⇒ Q(x)
(d) Q(x) ⇒ P (x)
Question 8: State the negation of the following statement: for
every x ∈ R, x > 0 or x < 0. (Your statement should be simplified.)
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Solutions:
(a) (a) {1, 0, −1}. Cardinality is 3.
(b) Note that x2 = 1 means x = ±1. When x = 1 then y = 2.
When x = −1, then y = −2. Therefore {(1, 2), (−1, −2)}
The cardinality is 2. (The other version was x = 2y, so the
set becomes {(1, 0.5), (−1, −0.5)}.)
(c) {{1}, {1, 2}, {1, 2, 3}}. Cardinality is 3.
(d) Note that {x ∈ Z | |x| = 1} = {1, −1} and {y ∈ Z | y 2 =
3} = {}. The product of these two sets is empty and the
cardinality is 0.
Note that the cardinality of the product of two sets is the
product of their cardinalities, which gives us a quick way
to test our answer: 2 × 0 = 0.
(b) (a) Note that {1}, {2, 3, 4}, and {2} are all subsets of N, hence
they are all elements in P(N). Therefore {{1}, {2, 3, 4}, {2}} ⊂
P(N). Furthermore this set has three elements, so this is
an example of the desired set.
(b) Note that {2, 3, 4} is a subset of N, hence {2, 3, 4} ∈ P(N).
Again since it has three element, it is an example of the
desired set.
(c) We can choose A = {1, 2, 3, 4} and B = {1, 2, 5, 6}. Then
A ∩ B = {1, 2},
A − B = {3, 4},
B − A = {5, 6},
which all have only two elements, as it was required.
(d) Note
\
Si = S1 ∩ S3 ∩ S4
i∈A
= [0, 3] ∩ [2, 5] ∩ [3, 6]
= {3},
since 3 is the only element in common in the three intervals
given.
Similarly note that
[
Si = S1 ∪ S3 ∪ S4
i∈A
= [0, 3] ∪ [2, 5] ∪ [3, 6]
= [0, 6].
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(e) Note that
P ⇒ (P ∨ Q ∨ R) ≡
≡
≡
≡
(∼ P ) ∨ (P ∨ Q ∨ R)
(∼ P ∨ P ) ∨ (Q ∨ R)
T ∨ (Q ∨ R)
T.
Therefore, for all possible truth values of P , Q, and R, we get
that P ⇒ (P ∨ Q ∨ R) is true, therefore this expression is a
tautology.
(f) (a) FALSE. If (1/2, 1/2) ∈ G then (1/2)2 + (1/2)2 = 1. However (1/2)2 + (1/2)2 = 1/2 6= 1.
(b) (G ⊆ H) TRUE. If x ∈ G = (3, 4), then 3 < x < 4. Since
H = [3, 4] = {t ∈ R | 3 ≤ t ≤ 4} we get that x ∈ H.
Therefore G ⊆ H.
(H ⊆ G) FALSE. Note that 3 ∈ H but 3 6∈ G.
(c) TRUE. Saint Valentine was not a nun, since he was male.
Therefore the hypothesis of the implication in question is
false. Therefore the whole sentence is true.
(d) (z = 0) TRUE. There are two cases to consider. If z = 0,
then z 2 < 1, so we have TRUE ⇒ TRUE, which is TRUE.
If z 6= 0, then z 2 ≥ 1. Therefore z 2 < 1 is FALSE, and we
have FALSE ⇒ FALSE, which is TRUE. Therefore for all
possible values of z ∈ Z the expression is TRUE.
(z is odd) FALSE. Note that when z = 0, z 2 < 1 and
z is even. Therefore we have TRUE ⇒ FALSE, which is
FALSE. Therefore the whole expression is FALSE.
(g) (a) (∼ Q(x)) “x squared is not equal to 4.” This is true for
x = 2 or x = −2 (but it is false for x = 0).
(∼ P (x)) “x is not equal to −2.” This is true for x = −2
(but it is false for x = 0 or x = 2).
(b) (P (x) ∨ Q(x)) “x equals −2 or x squared equals 4.” This
is true for x = 2 or x = −2.
(P (x) ∧ Q(x)) “x equals −2 and x squared equals 4.” This
is true for x = −2.
(c) “If x equals −2 then x squared is 4.” This is true for all
values of x ∈ S.
(d) “If x squared is 4 then x equals −2.” This is true for x = −2
or x = 0.
(h) Note that the negation of x > 0 is x ≤ 0. Similarly the negation
of x < 0 is x ≥ 0. Therefore, the negation of “x > 0 or x < 0”
is “x ≤ 0 and x ≥ 0”. (This can be simplified to “x = 0”.)
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Therefore the negation of “for every x ∈ R, x > 0 or x < 0”
becomes “there exists x ∈ R such that x ≤ 0 and x ≥ 0.”