Download Solutions to Test 2 Mathematics 503 Foundations of Mathematics 1

Solutions to Test 2
Mathematics 503
Foundations of Mathematics
Spring 2014
1. (a) Some examples of McNugget numbers are 6, 9, 12 = 2·6, 15 = 6+9, and 18 = 3·6 = 2·9.
Some examples of non-McNugget numbers are 1, 2, 3, 4, and 5.
(b) There exists a McNugget number m such that for all natural numbers n ≥ m, n is also
a McNugget number. Namely, m = 44 is such a number.
Proof. We will prove this statement via Strong Induction. Note that 44 = 4 · 6 + 20,
45 = 5 · 9, 46 = 6 + 2 · 20, 47 = 3 · 9 + 20, 48 = 8 · 6, and 49 = 9 + 2 · 20 are all McNugget
numbers. These six numbers constitute the basis step.
For the inductive step, let k ≥ 44 be a natural number, and suppose that the integer n
is a McNugget number for every n such that 44 ≤ n ≤ k. We will prove that k + 1 must
also be a McNugget number. If 44 ≤ k ≤ 48, then 45 ≤ k + 1 ≤ 49, and it follows that
k + 1 is a McNugget number by the preceding paragraph. On the other hand, if k ≥ 49,
then 44 ≤ k − 5 < k. By assumption then, k − 5 is a McNugget number, i.e.,
k − 5 = 6a + 9b + 20c
for some nonnegative integers a, b, and c. It follows that
k + 1 = (k − 5) + 6 = (6a + 9b + 20c) + 6 = 6(a + 1) + 9b + 20c.
This shows that k + 1 is a McNugget number since a + 1, b, and c are all nonnegative
integers. This completes our inductive proof.
2. For any natural number n,
3 + 11 + · · · + (8n − 5) = 4n2 − n.
Proof. We will prove this statement by induction. We verify the basis step in which n = 1
by noting that
3 = 4 · 12 − 1.
in accordance with the stated formula.
For the inductive step, suppose that k is a natural number such that
3 + 11 + · · · + (8k − 5) = 4k 2 − k.
(1)
3 + 11 + · · · + (8(k + 1) − 5) = 4(k + 1)2 − (k + 1).
(2)
We will show that
To this end, we add 8(k + 1) − 5 to both sides of (1) to obtain
3 + 11 + · · · + (8k − 5) + (8(k + 1) − 5) = 4k 2 − k + (8(k + 1) − 5).
(3)
The left-hand side of (3) is clearly equal to the left-hand side of (2). Moreover, the right-hand
side of (3) is
4k 2 − k + (8(k + 1) − 5) = 4k 2 − k + 8k + 3
= 4k 2 + 8k + 4 − k − 1
= 4(k 2 + 2k + 1) − (k + 1)
= 4(k + 1)2 − (k + 1).
This verifies (2), and the result follows.
3. (a) A12 ∩ B12 consists of all the residents who were both born in December and on the 12th
of some month. In other words, all residents born on December 12. According to the
information given in the problem, only one resident was born on each day of the year.
Thus the number of elements in A12 ∩ B12 is 1.
(b) A4 ∪ B25 is the set of all residents born in April or born on the 25th of some month.
Note that |A4 | = 30 and |B25 | = 12. If we add these cardinalities, we obtain 42, but this
number counts the resident born on April 25 twice. Thus |A4 ∪ B25 | = |A4 | + |B25 | − 1 =
30 + 12 − 1 = 41.
(c) (A7 ∪ B30 ) ∩ (A2 ∪ B14 ) = (A7 ∩ A2 ) ∪ (A7 ∩ B14 ) ∪ (B30 ∩ A2 ) ∪ (B30 ∩ B14 ) = (A7 ∩ B14 ) ∪
(B30 ∩ A2 ). (A7 ∩ B14 ) is the set containing only the resident born on July 14. (B30 ∩ A2 )
is empty since February has only 28 days. Therefore, this set has only 1 element.
(d) A8 c ∩ B31 c = (A8 ∪ B31 )c . Now A8 , the set of residents born in August, has 31 elements.
Also, B31 , the set of residents born on the 31st of January, March, May, July, August,
October, or December, has 7 elements. Thus as in (c), A8 ∪ B31 has 31 + 7 − 1 = 37
elements. It follows that (A8 ∪ B31 )c has 365 − 37 = 328 elements.
(e) Ac1 ∩ · · · ∩ Ac11 = (A1 ∪ · · · ∩ A11 )c = A12 , which is the set of residents born in December.
Thus this set has 31 elements.
(f)
A2 = {(2, m) : m ∈ N, m ≤ 28}.
B31 = {(1, 31)(3, 31)(5, 31)(7, 31)(8, 31)(10, 31)(12, 31)}.
(g) If the maximum of m and n is 1, then both m and n must be 1 so only resident (1, 1)
has lucky number 1.
If the maximum of m and n is 2, then (m, n) is either (1, 2), (2, 1), or (2, 2). Thus 3
residents have lucky number 2.
If the maximum of m and n is 31, then m must be 31 so (m, n) is an element of B31 .
Hence as in (d), there are 7 residents with lucky number 31.
4. (a) The set {a, b, {a, b}, {a}} has 4 elements. Thus the set P({a, b, {a, b}, {a}}) has 24 = 16
elements, so the set P(P({a, b, {a, b}, {a}})) has 216 = 65536 elements.
(b) Suppose that A ∩ B = ∅, and let X ∈ P(B) − {∅}. Then X ∈ P(B) and X ∈
/ {∅}.
The statement X ∈ P(B) means that X ⊆ B, and the statement X ∈
/ {∅} means that
X 6= ∅. Thus X is a nonempty subset of B. We must show that X ∈ P(A ∪ B) − P(A).
This means that X ∈ P(A ∪ B) and X ∈
/ P(A). In other words, we must show that
X ⊆ A ∪ B but X 6⊆ A.
Since B ⊆ A ∪ B, it follows that X ⊆ A ∪ B. To show that X 6⊆ A, suppose to the
contrary that X ⊆ A. Then for each x ∈ X, we have x ∈ A. Moreover, since X ⊆ B,
for each x ∈ X, we have x ∈ B. Thus x ∈ A ∩ B = ∅, which is always false. It follows
that x ∈ X is always false, so X = ∅, which contradicts our original assumption about
X. Thus X 6⊆ A. This competes the proof.