Math 220
Exam 2 Partial Solutions
November 9, 2012
S. Witherspoon
1. (a) f (O) = E, the set of even integers.
(b) f −1 (E) = Z since the image of each integer, under f , is even.
(c) No: For example, f (2) = f (3) but 2 6= 3.
2. (a) There are many possible examples. One is
f (1) = x,
f (2) = y,
f (3) = z.
(b) First we will prove that if f is injective, then f is surjective: Assume that f
is injective. Let b ∈ B. If b 6∈ Imf , then |Imf | < |B| = |A|. By the Pigeonhole
Principle, there are elements a1 , a2 ∈ A, a1 6= a2 , for which f (a1 ) = f (a2 ). That is, f
is not injective, contradicting the assumption that f is injective. Therefore b ∈ Imf ,
and f is surjective.
Next we will prove that if f is surjective, then f is injective: Assume that f is
surjective. Let a1 , a2 ∈ A be elements for which f (a1 ) = f (a2 ). If a1 6= a2 , then
|Imf | < |A| = |B|, so Imf cannot equal B, contradicting surjectivity of f . Therefore
a1 = a2 . It follows that f is injective.
3. (a) No: For example, 1 ∗ 2 = 8, 2 ∗ 1 = 7, so 1 ∗ 2 6= 2 ∗ 1.
(b) No: For example, (1 ∗ 2) ∗ 3 = 8 ∗ 3 = 25, 1 ∗ (2 ∗ 3) = 1 ∗ 13 = 41, so
(1 ∗ 2) ∗ 3 6= 1 ∗ (2 ∗ 3).
(c) Yes: Let a, b be even integers, so that a = 2x and b = 2y for some integers x, y.
Then
a ∗ b = 2(2x) + 3(2y) = 2(2x + 3y),
which is even since 2x + 3y is an integer.
(d) Yes: Let a, b be odd integers, so that a = 2x + 1 and b = 2y + 1 for some integers
x, y. Then
a ∗ b = 2(2x + 1) + 3(2y + 1) = 4x + 2 + 6y + 3
= 4x + 6y + 5
= 2(2x + 3y + 2) + 1,
which is odd since 2x + 3y + 2 is an integer.
4. We must prove that R is reflexive, symmetric, and transitive.
Reflexive: Let a ∈ R. Then a − a = 0, which is an element of Z, so aRa.
Symmetric: Let a, b ∈ R be elements such that aRb, that is, a − b ∈ Z. Then
b − a = −(a − b), which is an integer (as Z is closed under taking additive inverses),
so bRa.
Transitive: Let a, b, c ∈ R be elements such that aRb and bRc, that is, a − b ∈ Z
and b − c ∈ Z. Then
a − c = (a − b) + (b − c),
which is the sum of two integers, and so is an integer (as Z is closed under addition).
Therefore aRc.
1
2
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
5. We first check that the statement is true when n = 1: Since 4(1) − 3 = 1 and
1(2(1) − 1) = 1, it is true.
Now assume that the statement is true when n = k for some positive integer k,
that is, assume that
1 + 5 + 9 + 13 + · · · + (4k − 3) = k(2k − 1).
Then
1 + 5 + 9 + 13 + · · · + (4k − 3) + (4(k + 1) − 3) =
=
=
=
=
k(2k − 1) + (4(k + 1) − 3)
2k 2 − k + 4k + 1
2k 2 + 3k + 1
(k + 1)(2k + 1)
(k + 1)(2(k + 1) − 1),
which shows that the statement is true when n = k + 1. Therefore it is true for all
positive integers n.
6. (a) There are many possible examples. One is n = 4, a = 6, b = 10: 4 divides 60,
4 does not divide 6, and 4 does not divide 10.
(b) Let n be a positive integer greater than 1 with the property stated, that is, for
all a, b ∈ Z, if n divides ab, then n divides a or n divides b.
Assume that n is not prime, that is, assume n is composite. Then n = ab for some
integers a and b for which 1 < a < n and 1 < b < n. It follows that n divides the
product ab (since n = ab), n does not divide a (since a < n) and n does not divide
b (since b < n). This contradicts the hypothesis that n has the property stated.
Therefore n must be prime.