MATH 220.906
Examination 1 Solutions
October 7, 2013
1.
Determine if the following sentences are statements, open sentences, or neither. If a
sentence is open, incorporate it into a true statement.
(a) The quick brown fox jumped over the lazy dog.
(b) Z ⊆ R.
(c) sin(x) = 1.
(d) Let C be the unit circle.
c−b
.
a
Solution: (a) statement (b) statement (c) open sentence A possible way to incorporate
(e) For every real number x, if ax + b = c, then x =
this into a true statement: For all x ∈ R, if x = π2 , then sin(x) = 1. (d) neither (e)
open sentence A possible way to incorporate this into a true statement: For all a, b,
c ∈ R, if a 6= 0, then for every real number x, if ax + b = c, then x = c−b
a .
2.
Negate the following statements.
(a) For all x ∈ R there is some y ∈ R so that y 2 = x3 + 1.
π
π
or y = .
4
4
2
3
Solution: (a) There exists x ∈ R such that for every y ∈ R, y 6= x + 1. (b) There exist
x ∈ R, y ∈ R so that sin(y) = cos(x) and x 6= π4 and y 6= π4 .
(b) For all y ∈ R and for all x ∈ R, if sin(y) = cos(x), then x =
3.
Starting with the initial symbols provided, translate the given statement into an equivalent
statement made of mathematical symbols.
(a) If the product of two real numbers is zero, then one of the two factors in the product
must be zero.
∀x ∈ R, ∀y ∈ R, xy = 0 ⇒ (x = 0 ∨ y = 0) .
(b) Every positive real number has a square root.
∀x ∈ R, if x > 0, then ∃y ∈ R so that y 2 = x .
4.
For each of the following statements, determine if it is true or false. If a statement is false,
explain why it is false. (Notation: for a set A, we let P(A) denote its power set.)
(a) {0, 1} ∈ P({0, 1, 2}).
(b) {1} ⊆ P({0, 1, 2}).
(c) For all x ∈ R, x2 ≤ 10x if and only if x ≤ 10.
(d) For all x ∈ R, if x3 + 4x = 0, then x = 0 or x = 2.
Solution: (a) true (b) false ; it is false because {1} is an element of the power set
P({0, 1, 2}), not a subset of it. (c) false ; for a counterexample, take x = −1 (or any
negative x), and then x ≤ 10 but x2 = 1 6≤ −10 = 10x. (d) true ; the key thing here is
that x = 0 is the only real solution of the equation x3 + 4x = 0, so the statement does hold
(the x = 2 part is a red herring).
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5.
Consider the following 10 statement forms.
(A)
¬(Q ∧ (P ⇒ Q))
(C)
Q ∧ ¬Q
(B)
(P ⇒ Q) ⇒ Q
(D)
P ∨ ¬P
(E)
P ⇒Q
(F)
Q⇒P
(G)
¬Q
(H)
¬P
(I)
P ∧Q
(J)
P ∨Q
Match up each statement on the left (A–B) with its logically equivalent statement on the
right (one of C–J). Explain your answers.
Solution: For (A) we consider the truth table of ¬(Q ∧ (P ⇒ Q)):
P
T
T
F
F
Q P ⇒ Q Q ∧ (P ⇒ Q) ¬(Q ∧ (P ⇒ Q)) ¬Q
T
T
T
F
F
F
F
F
T
T
T
T
T
F
F
F
T
F
T
T
From this we see that (A) is logically equivalent to (G) .
For (B) we consider the truth table of (P ⇒ Q) ⇒ Q:
P
T
T
F
F
Q P ⇒ Q (P ⇒ Q) ⇒ Q P ∨ Q
T
T
T
T
F
F
T
T
T
T
T
T
F
T
F
F
From this we see that (B) is logically equivalent to (J) .
6.
For i ≥ 1, consider the intervals
Ai =
1 i+2
,
2i i + 1
.
(a) Express A1 ∩ A2 as an interval or a union of intervals.
∞
[
(b) Express
Ai as an interval or a union of intervals.
i=1
Solution: (a) A1 = [ 12 , 23 ) and A2 = [ 14 , 43 ), so A1 ∩ A2 = [ 21 , 43 ) . (b) As i → ∞, we
have
S∞
1
2i
i=1 Ai
→ 0 from above. As i → ∞, it follows that
= (0, 32 ) .
2
i+2
i+1
→ 1, also from above. Thus
7.
Alice, Bob, Carrie, and Dave ran a race, and the following statements were true about their
finishes:
(i) Dave did not finish last.
(ii) Alice was faster than Dave, and Carrie was faster than Bob.
(iii) Carrie did not finish first.
Who finished first, and who finished last? Explain your reasoning.
Solution: Alice finished first, and Bob finished last. There are many possible ways to
show this, but here is one.
By (ii), Alice was faster than Dave, so Dave did not finish first. By (ii), Carrie was faster
than Bob, so Bob did not finish first. By (iii), Carrie did not finish first, so by process of
elimination Alice finished first.
We already know that Alice and Dave did not finish last (Alice, because she was first;
Dave by (i)). By (ii), Carrie was faster than Bob, so she did not finish last. By process of
elimination, Bob finished last.
8.
Let A, B, and C be sets in a universe U . Consider the statement
S = “If (A ∩ C) ⊆ B, then A ∩ B ∩ C is the empty set.”
(a) What is the contrapositive of S?
(b) What is the converse of S?
(c) Prove that S is true.
Solution: (a) “If A ∩ B ∩ C is not the empty set, then (A ∩ C) 6⊆ B.”
(b) “If A ∩ B ∩ C is the empty set, then (A ∩ C) ⊆ B.”
(c) There were many possible answers here, but the main thing was to write a valid proof.
We will prove that the contrapositive of S is true. (Since the contrapositive of S is logically
equivalent to S, proving one is equivalent to proving the other.) Suppose that A∩B ∩C 6= ∅.
Then we can pick an element z ∈ A ∩ B ∩ C. We need to show that (A ∩ C) 6⊆ B. Suppose
on the contrary that (A ∩ C) ⊆ B. Well, then z ∈ A ∩ C, since A ∩ B ∩ C ⊆ A ∩ C, so then
z ∈ B. But then z ∈ B and z ∈ B, which is a contradiction. Therefore our assumption
that (A ∩ C) ⊆ B must have been false, and so (A ∩ C) 6⊆ B.
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