Math 2112 Solutions
Assignment 5
5.1.10 Indicate which of the following relationships are true and which
are false:
a. Z+ ⊆ Q
b. R− ⊆ Q
c. Q ⊆ Z
d. Z− ∪ Z+ = Z
e. Q ∩ R = Q
f. Q ∪ Z = Q
g. Z+ ∩ R = Z+ h. Z ∪ Q = Z
a. True. Every positive integer is a rational number.
√
b. False. For example, − 2 is a negative real number, but is not rational.
c. False. For example,
2
3
is not an integer.
d. False. The integer 0 is in the set on the right, but not the one on the left.
e. True. Since Q ⊆ R, then Q ∩ R = Q.
f. True. Since Z ⊆ Q, then Q ∪ Z = Q.
g. True. Since Z+ ⊆ R, then Z+ ∩ R = Z+ .
h. False. For example,
right.
2
3
is in the set on the left, but not in the set on the
5.2.27 Prove by mathematical induction that for any integer n ≥ 1
and all sets A1 , A2 , ..., An and B,
(A1 − B) ∩ (A2 − B) ∩ · · · ∩ (An − B) = (A1 ∩ A2 ∩ · · · ∩ An ) − B.
Proof: Base Case: Let n = 1. Then LHS = A1 − B and RHS = A1 − B. So
LHS = RHS.
Inductive Step: Let k ≥ 1. Suppose that (A1 −B)∩(A2 −B)∩· · ·∩(Ak −B) =
(A1 ∩ A2 ∩ · · · ∩ Ak ) − B. Consider (A1 − B) ∩ (A2 − B) ∩ · · · ∩ (Ak+1 − B). By
our inductive hypothesis, we know that
(A1 − B) ∩ (A2 − B) ∩ · · · ∩ (Ak+1 − B)
= ((A1 ∩ A2 ∩ · · · ∩ Ak ) − B) ∩ (Ak+1 − B)
= ((A1 ∩ A2 ∩ · · · ∩ Ak ) ∩ B c ) ∩ (Ak+1 ∩ B c )
=
=
(A1 ∩ A2 ∩ · · · ∩ Ak ∩ Ak+1 ) ∩ B c ∩ B c
(A1 ∩ A2 ∩ · · · ∩ Ak ∩ Ak+1 ) ∩ B c
=
(A1 ∩ A2 ∩ · · · ∩ Ak ∩ Ak+1 ) − B
Thus the result holds by mathematical induction.
1
5.2.36 For all sets A and B, (B c ∪ (B c − A))c = B.
(B c ∪ (B c − A))c
= (B c ∪ (B c ∩ Ac ))c
Alternate Represenation for Set Difference
= ((B c )c ∩ (B c ∩ Ac )c )
De Morgan’s Laws
= ((B c )c ∩ ((B c )c ∪ (Ac )c )) De Morgan’s Laws
= (B ∩ (B ∪ A))
Double Complement law
= B
Absorption Laws
Proof:
5.3.33 Given any sets A and B, define the symmetric difference of A
and B, denoted A ⊕ B, as follows:
A ⊕ B = (A − B) ∪ (B − A).
Prove each of the following for all sets A, B, and C in a universal set
A.
a. A ⊕ B = B ⊕ A
b. A ⊕ (B ⊕ C) = (A ⊕ B) ⊕ C
c. A ⊕ ∅ = A
d. A ⊕ Ac = U
e. A ⊕ A = ∅
f. If A ⊕ C = B ⊕ C, then A = B.
Proof: We start our proof by noting that x ∈ A ⊕ B iff (x ∈ A and x 6∈ B) or
(x 6∈ A and x ∈ B) and x 6∈ A ⊕ B iff (x ∈ A and x ∈ B) or (x 6∈ A and x 6∈ B).
a. A ⊕ B = (A − B) ∪ (B − A) = (B − A) ∪ (A − B) = B ⊕ A.
x ∈ (A ⊕ B) ⊕ C
⇔
⇔
(x ∈ A ⊕ B and x 6∈ C) or (x ∈ C and x 6∈ A ⊕ B
([(x ∈ A and x 6∈ B) or (x ∈ B and x 6∈ A)] and x 6∈ C) or
(x ∈ C and [(x ∈ A and x ∈ B) or (x 6∈ A and x 6∈ B)])
⇔ [(x ∈ A and x 6∈ B and x 6∈ C) or (x ∈ B and x 6∈ A and x 6∈ C)] or
b.
[(x ∈ C and x 6∈ A and x 6∈ B) or (x ∈ C and x ∈ A and x ∈ B)]
⇔ x is in exactly one of the sets A, B, and C, or
x is in all three of the sets A, B, and C.
Similarly, it can be shown that
x ∈ A ⊕ (B ⊕ C) ⇔ x is in exactly one of the sets A, B, and C, or
Therex is in all three of the sets A, B, and C.
fore, A ⊕ (B ⊕ C) = (A ⊕ B) ⊕ C.
c.
A⊕∅ =
=
(A − ∅) ∪ (∅ − A)
(A ∩ ∅c ) ∪ (∅ ∩ Ac )
=
=
(A ∩ U ) ∪ (Ac ∩ ∅)
A∪∅
=
A.
2
d.
A ⊕ Ac
=
=
=
(A − Ac ) ∪ (Ac − A)
(A ∩ (Ac )c ) ∪ (Ac ∩ Ac )
(A ∩ A) ∪ Ac
=
=
A ∪ Ac
U.
e.
A⊕A
=
(A − A) ∪ (A − A)
=
A−A
=
=
(A ∩ Ac )
∅.
f. Let A, B, and C be any sets with A ⊕ C B ⊕ C.
A ⊆ B : Suppose x ∈ A. Either x ∈ C or x 6∈ C. If x ∈ C, then x 6∈ A⊕C.
But A ⊕ C = B ⊕ C. Thus x 6∈ B ⊕ C. Since x ∈ C and x 6∈ B ⊕ C, then
x ∈ B. On the other hand, if x 6∈ C, then since x ∈ A, x ∈ A ⊕ C. But
A ⊕ C = B ⊕ C. So, since x 6∈ C and x ∈ B ⊕ B, then x ∈ B. Hence, in
either case, x ∈ B.
B ⊆ A : The proof is the same as A ⊆ B, but with A and B reversed.
Therefore, since A ⊆ B and B ⊆ A, then A = B.
2.7 (CE) Explain why the Fibonacci seqeunce appears on the shallow
diagonals of Pascal’s triangle.
Proof: The Fibonacci sequence is defined
F0 =
by n−1
1, F1 = 1 and Fn = Fn−1 +
Fn−2 , n ≥ 2. Recall also that nk = n−1
+
k
k−1 . We conjecture that
2n
2n − 1
n
F2n =
+
+ ... +
0
1
n
2n + 1
2n
n+1
F2n+1 =
+
+ ... +
0
n
n
for all n ≥ 0.
Proof of conjecture: We proceed
by strong induction.
Base Cases: F0 = 1 = 00 and F1 = 1 = 10 , so the result is true for n = 0, 1.
Inductive Step: We will only prove the even case. The odd case follows analgously. Let k > 0. Assume that the formula holds true for all i such that
3
0 ≤ i < 2k. Consider
F2k
=
=
=
=
=
F2k−1 + F2k−2
2k − 1
2k − 2
k
2k − 2
2k − 3
k−1
+
+ ... +
+
+
+ ... +
0
1
k−1
0
1
k−1
2k − 1
2k − 2
2k − 2
2k − 3
2k − 3
+
+
+
+
+ ...
0
1
0
2
1
k
k
k−1
+
+
+
k−1
k−2
k−1
2k − 1
2k − 1
2k − 2
k+1
k−1
+
+
+ ... +
+
0
1
2
k−1
k−1
2k
2k − 1
k+1
k
+
+ ... +
+
0
1
k−1
k
As stated above, the odd case follows anagously. Thus the result holds by strong
mathematical induction.
2.8.3 (CE) Consider n lightbulbs in a room, numbered 1 to n, Determine the number of ways the lightbulbs can be turned on or off. By
solving this problem in two different ways, prove that
n
n
n
n
+
+
+ ... +
= 2n
0
1
2
n
Proof: Light 1 can be in one of two positions, namely on or off. Light 2 can
be in one of two positions, on or off... Thus, there are a total of 2n different
positions the n lightbulbs can be in. On the other hand, there are n0 ways
0 lights can be on. There are n1 ways 1 light can be on... Since every light
position is counted
by nexactly
one of thenbinomial coefficients, we can deduce
n
n
that
there
are
+
+
1 2 + ... + n different positions. Thus, we have
0
n
n
n
n
n
+
+
+
...
+
=
2
, as desired.
0
1
2
n
2.10.1 (CE) Prove, using a routes argument, that
2n
2 2n − 2
2 2n − 2
2 2n − 2
=
+
+
n
0
n
1
n−1
2
n−2
Proof: Consider all possible routes on an n×n grid. We label the start position
(0, 0) and the end position (n, n). To get fromthe start to the end, we must
make n rights and n ups. Thus, there are 2n
different possible routes. On
n
4
the other hand, consider the checkpoints (2, 0), (1, 1) and (0, 2). Clearly, every
route must pass through exactly one of these checkpoints. There are 20 ways
to get to (2, 0) (we choose which of the two steps are rights), and 2n−2
ways
n
to get from (2, 0) to (n, n) (again,
we
choose which of the remaining steps are
rights). Similarly, there are 21 2n−2
n−1 routes going through the checkpoint (1, 1)
and 22 2n−2
routes
going
through
the checkpoint (0, 2). Thus there are a total
n−2
2 2n−2
2 2n−2
2 2n−2
+ 1 n−1 + 2 n−2 different routes.
of 0
n
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