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HW #7 Solutions
August 26, 2008
Pg. 101, # 101
Theorem. Let p be prime, let n be a natural number, and let a1 , a2 , . . . , an be
integers such that p divides a1 a2 · · · an . Then there exists i, 1 ≤ i ≤ n, such that
p divides ai .
Proof. The proof is by induction. The statement clearly holds when n = 1; the
case where n = 2 is given in Corollary 3.13 of the text (prove this on your own).
Suppose the statement holds for some natural number n ≥ 2 and assume that p
is a prime number and a1 , a2 , . . . , an , an+1 are integers such that p|a1 a2 · · · an+1 .
Set q = a1 a2 · · · an , so that p|q · an+1 . By the case of the statement for n = 2,
p|q or p|an+1 . If p|an+1 , we are done. If not, then p must divide q = a1 a2 · · · an ,
and by the inductive hypothesis, p divides ai for some i, 1 ≤ i ≤ n. In either
case, p|ai for some i, 1 ≤ i ≤ n + 1, and the theorem now follows.
#104 A fraction ab is said to be in lowest terms provided gcd(a, b) = 1. Two
fractions ab and dc are said to be equivalent provided ad = bc.
Theorem. Every fraction is equivalent to a fraction in lowest terms.
Proof. Let gcd(a, b) = d; then there exists integers p and q such that a = pd and
b = qd. Since aq = (pd)q = (qd)p = bp, the fractions ab and pq are equivalent. By
Theorem 3.10, there exist integers x and y such that d = ax+by = (pd)x+(qd)y.
Canceling d from both sides gives px + qy = 1; by Theorem 3.10 again, we must
have gcd(p, q) = 1. So ab is equivalent to pq , and the latter fraction is in lowest
terms.
#105 We find a fraction equivalent to 1739/4042 that is written in lowest
terms. We are careful to show work so it doesn’t look like we just did it on a
1
1739/d
is
calculator! By the proof of #104 above, if gcd(1739/4042) = d, then 4042/d
the fraction we’re looking for. We use the Euclidean Algorithm to find d:
4042 = 2(1739) + 564
1739 = 3(564) + 47
564 = 12(47) + 0,
so d = 47 and 1739/4042 is equivalent to
is in lowest terms.
1739/47
4042/47
= 37/86, and the latter fraction
Theorem. Let a, b, c, d be natural numbers. If a/b and c/d are equivalent fractions that are in reduced form (lowest terms), then a = c and b = d.
Proof. Since the fractions are equivalent, ad = bc; so a divides bc. Because a/b
is in reduced form, a and b are relatively prime, so a divides c by Corollary 3.11
(you should prove this corollary for yourself). Likewise, ad = bc also shows that
c divides ad. Since c/d is in reduced form, c and d are relatively prime, so by
Corollary 3.11 again, c|ad implies that c|a. Since a|c and c|a, we must have that
a ≤ c and c ≤ a, and hence a = c. It then follows from the equation ad = bc
that b = d.
Pg. 107 #4, #5 For each of the following relations we state the domain
and range and compute S[1] and S[3].
1. S = {(x, y)√∈ R√
× R : x2 + y 2 = 16}.
Domain = Range = {x ∈ R : |x| ≤ 4}
√ √
S[1] = {− 15, 15}. S[3] = {− 7, 7}.
2. S = {(x, y)
R × R : y 2 = 2x}.
{x ∈ R : x ≥ 0}. Range = R.
√∈ √
√ Domain:
√
S[1] = {− 2, 2}. S[3] = {− 6, 6}.
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3. S =
= 2y 2√
}. Domain = Range = R. S[1] =
√ {(x,√y) ∈ R × R : x √
{− 2/2, 2/2}. S[3] = {−3/ 2, 3/ 2}.
4. S = {(x, y) ∈ R × R : x2 = y 2 }. Domain = Range = R. S[1] = {−1, 1}.
S[3] = {−3, 3}.
Pg. 120 #26 Let R = {(a, b) ∈ R × R : there is an integer k such that
a − b = 2kπ}.
1.
Theorem. R is an equivalence relation on R.
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Proof. Let a, b, c be real numbers. Since a − a = 2(0)π, (a, a) ∈ R and R
is reflexive. If (a, b) ∈ R then a − b = 2kπ, so b − a = −(a − b) = 2(−k)π.
Hence (b, a) ∈ R and R is symmetric. Suppose (a, b) and (b, c) are in R so
that there exists integers j and k such that a − b = 2jπ and b − c = 2kπ.
Then a − c = (a − b) + (b − c) = (2jπ) − (2kπ) = 2(j + k)π, so that
(a, c) ∈ R and R is transitive.
2. {−7π/4, π/4, 17π/4} ⊆ [π/4]
3. {1 − 20π, 1, 1 + 1000π} ⊆ [1]
4. Suppose x ∈ [1] ∩ [π/4], so that 1 − x = 2πj and π/4 − x = 2πk for some
integers j and k. Then 1 − π/4 = (1 − x) + (x − π/4) = 2π(j − k). But
this is a contradiction, since 1 − π/4 is clearly not an integer multiple of
2π. Hence [π/4] ∩ [1] = ∅.
#27 Define R = {(x, y) ∈ Q × Q : when x and y are represented by fractions
in lowest terms, these fractions have the same denominator}.
1.
Theorem. R is an equivalence relation on Q.
Proof. Let x, y, z be rational numbers. If x is represented in lowest terms
then it has the same denominator as the number x represented in lowest
terms, so (x, x) ∈ R and R is reflexive. Suppose (x, y) ∈ R, so that x and
y have the same denominator when represented in lowest terms; then y
and x have the same denominator when represented in lowest terms, so
(y, x) ∈ R and R is symmetric. If (x, y) and (y, z) are in R then in lowest
terms, x and y share the same denominator m, and y and z share the same
denominator n. We showed that the reduced form of a fraction is unique
(in the added HW problem above), so m = n and x and z have the same
denominator when represented in lowest terms. Hence (x, z) ∈ R and R
is transitive.
2. We show that [1/6] = [5/6]. Since 1/6 and 5/6 are both in lowest terms,
by definition we have [1/6] = {x ∈ Q : when 1/6 and x are represented
by fractions in lowest terms, these fractions have the same denominator
} = {x ∈ Q : when x is represented in lowest terms its denominator is
6} = {x ∈ Q : when 5/6 and x are represented by fractions in lowest
terms, these fractions have the same denominator } = [5/6].
3. We show that [4/6] and [5/6] are disjoint. If x ∈ [4/6] then it has a three
in the denominator when represented in lowest terms. If x ∈ [5/6] then it
has a six in the denominator when represented in lowest terms. Since the
denominator of a fraction in lowest terms is unique, no number can lie in
both [4/6] and [5/6], i.e. [4/6] ∩ [5/6] = ∅.
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#28 Define R = {(x, y) ∈ R × R : x2 − y 2 = 0}.
1.
Theorem. R is an equivalence relation on R.
Proof. Let x, y, z be real numbers. Since x2 − x2 = 0, (x, x) ∈ R and R
is reflexive. If (x, y) ∈ R then y 2 − x2 = −(x2 − y 2 ) = 0, so (y, x) ∈ R,
and R is symmetric. Suppose (x, y) and (y, z) are in R. Then x2 − z 2 =
(x2 − y 2 ) + (y 2 − z 2 ) = 0 + 0 = 0, so (x, z) ∈ R, and R is transitive.
2. [3] = {−3, 3}.
#29 Let R = {(3, 5)}. R is not reflexive on any set, otherwise R would
contain (3, 3) and (5, 5). R is not symmetric because (3, 5) ∈ R but (5, 3) 6∈ R.
However, R is transitive since there are not enough elements in R to violate the
transitive property. To put this another way, the statement “R contains pairs
of the form (a, b) and (b, c) implies that (a, c) is in R” is always true, since this
is a conditional statement whose hypothesis is never true.
#30 Let A = {1, 2, 3, 4}. We construct relations R on A that have various
combinations of the reflexive, symmetric, and transitive properties, which we
abbreviate by r, s, and t.
1. r but not s or t: R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 3)}.
2. s but not r or t: R = {(1, 2), (2, 3), (2, 1), (3, 2)}.
3. t but not s or r: R = {(1, 2)}.
4. r and s but not t: R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 3), (2, 1), (3, 2)}.
5. r and t but not s: R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2)}.
6. s and t but not r: R = ∅.
#31 Let R = {(a, b) ∈ N × N : a divides b}. R is reflexive since for any
natural number a, a = (1)(a), so that a divides a. R is not symmetric since, for
example, two divides four but four does not divide two. Let a, b, c be natural
numbers, and suppose a divides b and b divides c. Then there exist integers x
and y such that b = xa and c = yb, so that c = yb = yxa and therefore a divides
c. It follows that R is transitive.
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#36 For any two points (a, b) and (c, d) in the plane, define (a, b) ' (c, d)
provided that a2 + b2 = c2 + d2 .
1.
Theorem. ∼ is an equivalence relation on R × R.
Proof. Let (a, b), (c, d), (e, f ) be points in the plane. (a, b) ∼ (a, b) since
a2 +b2 = a2 +b2 , so ∼ is reflexive. Suppose (a, b) ∼ (c, d), so that a2 +b2 =
c2 + d2 . Then c2 + d2 = a2 + b2 , so (c, d) ∼ (a, b) and ∼ is symmetric. If
(a, b) ∼ (c, d) and (c, d) ∼ (e, f ), then a2 + b2 = c2 + d2 = e2 + f 2 . Hence
(a, b) ∼ (e, f ), and ∼ is transitive.
2. [(0, 0)] = {(0, 0)}.
2
2
2
2
3. [(5, 11)] = {points (c, d) in the plane
√ : 5 + 11 = c + d } = the circle
centered at the origin with radius 52 + 112 .
Pg. 127 # 48
Theorem. Let ∼ be an equivalence relation on a nonempty set S, and let S/ ∼
denote the family of all equivalence classes on S. The S/ ∼ is a partition of S.
Proof. This is Theorem 4.4 from the text; part of the proof has already been
given there. Namely, it has been shown that each member of S lies in some
member of S/ ∼, S/ ∼ is a family of nonempty sets, and if p, x, and y are
members of S such that p ∈ [x]∩[y] then [x] ⊆ [y]. Under the same assumptions,
let q ∈ [y]. Then y ∼ q, and since p ∈ [x] and p ∈ [y] we have x ∼ p and y ∼ p.
By symmetry, q ∼ y and p ∼ x. By transitivity, q ∼ y and y ∼ p imply that
q ∼ p. Applying transitivity again, q ∼ p and p ∼ x imply that q ∼ x. By
symmetry, x ∼ q, i.e. q ∈ [x]. Hence [y] ⊆ [x], and it follows that [x] = [y], i.e.
the sets in S/ ∼ are pairwise disjoint. The theorem now follows.
Pg. 127 #49
Theorem. Let A be a partition of a nonempty set S. If x and y are members
of S, define x ∼ y provided that some member of A contains both x and y. Then
∼ is an equivalence relation on S.
Proof. Let x, y, and z be members of S. Since A is a partition, x lies in some
member of A, call it X. Since x and x both lie in X, x ∼ x, and ∼ is reflexive.
Suppose x ∼ y; then there exists and element Y ∈ A such that x ∈ Y and
y ∈ Y . Since Y contains both y and x, y ∼ x and ∼ is symmetric. If x ∼ y and
y ∼ z then there are elements A and B in A such that x, y ∈ A and y, z ∈ B;
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in particular, y ∈ A ∩ B. Since S is a partition, A and B are either disjoint or
equal, and since they are not disjoint we must have A = B. So x and z both
lie in A, which means that x ∼ z and ∼ is therefore transitive. Hence ∼ is an
equivalence relation.
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