SELECTED SOLUTIONS TO HOMEWORK 5
1. 3.6
Exercise 12.
(1) Suppose P (x) is a statement with free variable x.
Find a formula meaning there are exactly two values for which
P (x) is true.
(2) Design a proof strategy for proving a statement of the form
“There are exactly two values for which P (x) is true”.
(3) Prove that there are exactly two solutions to the equation x3 =
x2 .
We’ll do parts (a) and (c). For part (a), we could use:
∃x∃y(¬x = y ∧ ∀z(P (z) ↔ (z = x ∨ z = y))).
Proof of (c). First, note that 0 and 1 each solve the equation x3 = x2 ,
because 0 · 0 = 0, and 1 · 1 = 1. Now suppose that x3 = x2 , and x 6= 0.
Then x2 6= 0, so multiplying by 1/x2 gives us x = 1, as we desired. 2. Section 3.7
Exercise 2. Suppose that a and B are sets. Waht can you say about
P(A\B)\(P(A)\P(B))?
Theorem. For any sets A, B, P(A\B)\(P(A)\P(B)) = {∅}.
Proof. First, note that because ∅ is a subset of any set, ∅ is an element
of every powerset. So, ∅ ∈ P(A\B) and ∅ ∈ P(A) ∩ P(B). Therefore
∅ 6∈ P(A)\P(B), so ∅ ∈ P(A\B)\(P(A)\P(B)), and we have that
{∅} ⊆ P(A\B)\(P(A)\P(B)).
For the other direction, suppose that C ∈ P(A\B)\(P(A)\P(B)).
Then C ⊆ A\B, and C 6∈ P(A)\P(B). But via the definition of
setminus and DeMorgan’s C 6∈ P(A)\P(B) if and only if C ∈ P(A)
implies C ∈ P(B).
So, for any C ∈ P(A\B)\(P(A)\P(B)), C ⊆ (A\B), and if C ⊆ A,
then C ⊆ B. Now, suppose for a contradiction that C contains an
element x. Then x ∈ A\B, so x ∈ A and x 6∈ B. But C ⊆ A\B ⊆ A,
therefore by our hypothesis that if C ⊆ A, then C ⊆ B we have
that C ⊆ B, so x ∈ B. This contradiction tells us that C = ∅, so
P(A\B)\(P(A)\P(B)) =⊆ ∅, and we are done.
1
2
SELECTED SOLUTIONS TO HOMEWORK 5
3. Additional Problems
Theorem 2. For all a, b ∈ Z, if there exist c, d ∈ Z so that ac+bd = 1,
then for all n ∈ Z, ab|n if and ony if a|n and b|n.
Proof. Suppose that ab|n. Then there exists k ∈ Z so that abk = n.
So a(bk) = n and b(ak) = n, and we have that a and b both divide n.
For the other direction, suppose that a|n and b|n, and that ac + bd =
1, where c, d ∈ Z. Then there exists integers k, l so that ak = n,
and bl = n. Multiplying both sides of ac + bd = 1 by n gives us
nac + nbd = n, so (bl)ac + (ak)bd = n. Factoring out ab gives us
(ab)(lc + kd) = n, so ab|n and we are done.
Theorem 3. For all n ∈ Z, 6|n if and only if 6|n2 .
Proof. We will use these three results that we have previously proved.
For all n ∈ Z:
(1) 2|n iff 2|n2 ,
(2) 3|n iff 3|n2 , and
(3) 6|n iff 2|n and 3|n.
First, suppose that 6|n. Then for some k ∈ Z, 6k = n, so 6(kn) = n2 ,
and 6|n2 .
For the other direction, suppose that 6|n2 . Then by (3), 2 and 3 each
divide n2 , so by (1) and (2) 2 and 3 each divide n. So, by (3) again
6|n, and we are done.