Download Math 3000 Section 003 Intro to Abstract Math Homework 4

Math 3000 Section 003 Intro to Abstract Math Homework 4
Department of Mathematical and Statistical Sciences
University of Colorado Denver, Spring 2012
Solutions (February 25, 2012)
Please note that these solutions are only suggestions; different answers or proofs are also possible.
• Section 4.1: Proofs Involving Divisibility of Integers
1. Exercise 4.4: Let x, y ∈ Z. Prove that if 3 - x and 3 - y, then 3 | (x2 − y 2 ).
Solution: For the proofs of Exercises 4.4 and 4.16 (below), you can use that whenever a ≡ b
(mod n) and c ≡ d (mod n) for integers a, b, c, and d, then a±c ≡ b±d (mod n) and ac ≡ bd
(mod n) which you had proven already in class. In particular, if we choose a = c and b = d,
then a ≡ b (mod 3) only if a2 ≡ b2 (mod 3). This makes the following proofs very short.
Proof. Let 3 - x. Then either x ≡ 1 (mod 3) or x ≡ 2 (mod 3). In both cases x2 ≡ 1 (mod 3).
Similarly, if 3 - y, then y 2 ≡ 1 (mod 3). Hence x2 − y 2 ≡ 0 (mod 3) and 3 | (x2 − y 2 ).
• Section 4.2: Proofs Involving Congruence of Integers
2. Exercise 4.16: Let n ∈ Z. Prove each of the statements (a)–(f).
(a)
(b)
(c)
(d)
(e)
(f)
If n ≡ 0 (mod 7), then
If n ≡ 1 (mod 7), then
If n ≡ 2 (mod 7), then
If n ≡ 3 (mod 7), then
For every integer n, n2
For every integer n, n2
n2 ≡ 0 (mod 7).
n2 ≡ 1 (mod 7).
n2 ≡ 4 (mod 7).
n2 ≡ 2 (mod 7).
≡ (7 − n)2 (mod 7).
is congruent to exactly one of 0, 1, 2, or 4 modulo 7.
Solution: Parts (a)–(d) follow by observing that 02 = 0 ≡ 0 (mod 7), 12 = 1 ≡ 1 (mod 7),
22 = 4 ≡ 4 (mod 7), and 32 = 9 ≡ 2 (mod 7), respectively. Part (e) follows because (7−n)2 =
49 − 14n + n2 = 7(7 − 2n) + n2 ≡ n2 (mod 7). Finally, part (f) follows from combining parts
(a)-(d) which cover the congruence classes 0, 1, 2, and 3 with part (e) which can be used to
cover the remaining congruence classes 4 (42 ≡ 32 ≡ 3 (mod 3)), 5 (52 ≡ 22 ≡ 4 (mod 3))
and 6 (62 ≡ 12 ≡ 1 (mod 3)). The congruence class 7 is the same as the congruence class 0.
• Section 4.3: Proofs Involving Real Numbers
3. Exercise 4.20: Let x ∈ R. Prove that if 3x4 + 1 ≤ x7 + x3 , then x > 0.
Solution: We formulate the contrapositive and prove that 3x4 + 1 > x7 + x3 whenever x ≤ 0.
Proof. By contrapositive, let x ≤ 0. Because x4 is even, x4 ≥ 0 (for all x) but x7 ≤ 0 and
x3 ≤ 0. It follows that 3x4 + 1 > 0 ≥ x7 + x3 which we can reduce to 3x4 + 1 > x7 + x3 .
• Section 4.4: Proofs Involving Sets
4. Exercise 4.30:
(a) Give an example of three sets A, B, and C such that A ∩ B = A ∩ C but B =
6 C.
(b) Give an example of three sets A, B, and C such that A ∪ B = A ∪ C but B =
6 C.
Math 3000-003 Intro to Abstract Math Homework 4, UC Denver, Spring 2012
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(c) Let A, B, and C be sets. Prove that if A ∩ B = A ∩ C and A ∪ B = A ∪ C, then
B = C.
Solution: For (a), let A = ∅ be the empty set (so A ∩ B = A ∩ C = ∅ for any choice of B
and C), and B and C be any two sets that are not the same. For (b), let A = B ∪ C (so
A ∪ B = A ∪ C = B ∪ C for any choice of B and C), and B and C be any two sets that
are not the same. [These are two possibilities to define infinitely many generic examples; of
course there are other possibilities as well.] Finally, for part (c), consider the following proof.
Proof. Let A ∩ B = A ∩ C (*) and A ∪ B = A ∪ C (**). To prove that B = C, we need to
show that B ⊆ C and C ⊆ B. To strucure the proof, we will distinguish the two cases in
which both B and C are nonempty, and in which at least one of the two sets is empty.
Case 1 (B 6= ∅ and C 6= ∅). Let B and C be nonempty. Because of symmetry, it is sufficient
to only show that B ⊆ C; then C ⊆ B follows analogously (simply switch what B is and what
C is). Hence, now let x ∈ B (this only works because B is nonempty) and distinguish the
two subcases in which x ∈ A or x ∈
/ A. If x ∈ A, then x ∈ A ∩ B also and by (*) x ∈ A ∩ C
and thus x ∈ C. If x ∈
/ A, then still x ∈ A ∪ B and by (**) x ∈ A ∪ C also and thus x ∈ C.
Hence, in both subcases we find that x ∈ B implies that x ∈ C which shows that B ⊆ C.
Case 2 (B = ∅ or C = ∅). Without loss of generality, let B be empty (otherwise switch the
roles of B and C). By (**), then A ∪ B = A = A ∪ C and thus C ⊆ A and A ∩ C = C.
Furthermore, by (*), then A ∩ B = ∅ = A ∩ C = C so that B = ∅ implies that C = ∅ also.
• Section 4.5: Fundamental Properties of Set Operations
5. Exercise 4.36: Prove that A ∩ B = A ∪ B for every two sets A and B.
Solution: Note that this result is the set equivalent to De Morgan’s law in elementary logic.
Proof. Using the above observation, define the two statements A(x) : x ∈ A and B(x) : x ∈ B.
Then x ∈ A ∩ B is equivalent to the statement ∼ (A(x) ∧ B(x)) which by De Morgan’s law is
equivalent to (∼ A(x)) ∨ (∼ B(x)), or x ∈ A ∪ B - a very elegant proof for A ∩ B = A ∪ B.
• Section 4.6: Proofs Involving Cartesian Products of Sets
6. Exercise 4.44: It can be shown that if A, B, C, and D are sets such that A ⊆ C and
B ⊆ D, then A × B ⊆ C × D.
(a) Show that the converse of this result is false.
(b) Under what added hypothesis is the converse true? Prove your assertion.
Solution: For (a), let A = {“Amazing Spiderman”} and B = C = D = ∅. Then A × B =
C × D = ∅ but A * C. For (b), the converse becomes true if we add the hypothesis that
both A and B are nonempty. [Of course, you could also assume that all sets are nonempty;
moreover, if you prefer a trivial statement, you could also assume that all the sets are empty.]
Proof. Let A and B be nonempty and A × B ⊆ C × D (*). To show that A ⊆ C and B ⊆ D,
let x ∈ A and y ∈ B (this only works because A and B are nonempty). Then (x, y) ∈ A × B
and by (*) (x, y) ∈ C × D also, so x ∈ C and y ∈ D and thus A ⊆ C and B ⊆ D.
• Additional Exercises for Chapter 4
7. Exercise 4.58: Evaluate the proposed proof of the following result.
Result. Let x, y ∈ Z. If x ≡ 2 (mod 3) and y ≡ 2 (mod 3), then xy ≡ 1 (mod 3).
Math 3000-003 Intro to Abstract Math Homework 4, UC Denver, Spring 2012
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Proof. Let x ≡ 2 (mod 3) and y ≡ 2 (mod 3). Then x = 3k + 2 and y = 3k + 2 for some
integer k. Hence
xy = (3k + 2)(3k + 2) = 9k 2 + 12k + 4 = 9k 2 + 12k + 3 + 1 = 3(3k 2 + 4k + 1) + 1.
Since 3k 2 + 4k + 1 is an integer, xy ≡ 1 (mod 3).
Solution: This “proof” uses the same integer k for x = y = 3k + 2. Hence, it only shows that
if x ≡ 2 (mod 3), then x2 ≡ 1 (mod 3). Of course, this would also be true if x ≡ 1 (mod 3).
8. Exercise 4.60: A proof of the following result is given.
Result. Let n ∈ Z. If n4 is even, then 3n + 1 is odd.
2
Proof. Assume that n4 = n2 is even. Since n4 is even, n2 is even. Furthermore, since
n2 is even, n is even. Because n is even, n = 2k for some integer k. Then
3n + 1 = 3(2k) + 1 = 6k + 1 = 2(3k) + 1.
Since 3k is an integer, 3n + 1 is odd.
Answer the following questions.
(a)
(b)
(c)
(d)
Which proof technique is being used?
What is the starting assumption?
What must be shown to give a complete proof?
Give a reason for each of the following steps in the proof.
i. Since n4 is even, n2 is even.
ii. Furthermore, since n2 is even, n is even.
iii. Because n is even, n = 2k for some integer k.
iv. Then 3n + 1 = 3(2k) + 1 = 6k + 1 = 2(3k) + 1.
v. Since 3k is an integer, 3n + 1 is odd.
Solution: (a) Direct Proof. (b) The proofs starts out with an integer and assumes that its
fourth power is even. (c) The proof must conclude that the third multiple of that integer
increased by 1 is an odd number. (d) i.ii. We know (already have proven) that an integer and
its square have same parity. iii.v. These steps use the defintion of even and odd numbers. iv.
This step uses substitution and established rules and operations from mathematical algebra.
• Section 7.3: Testing Statements
9. Exercise 7.38 (Prove or disprove): For every positive rational numbers b, there exists an
irrational number a with 0 < a < b.
Solution: This statement is true. In particular, we will later prove that there are infinitely
many rational and irrational numbers, and that there is no smallest (positive) rational or
irrational number. Here, however, it suffices to produce (or “construct”) a single example.
√
Proof. Let b be a positive rational
number.
We
(assume
to)
know
that
2 is irrational and
√
greater than 1. Then a = b/ 2 < b remains positive and is also irrational.
10. Exercise 7.44 (Prove or disprove): For every two rational numbers a and b with a < b,
there exists a rational number r such that a < r < b.
Math 3000-003 Intro to Abstract Math Homework 4, UC Denver, Spring 2012
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Solution: This statement is also true, and most of you correctly used r = (a + b)/2 as the
averge (or mean) of a and b. Here is a slightly more general proof using a weighted average.
Proof. Let w ∈ (0, 1) be any rational number (the “weight”), e.g., w = 1/n for some positive
integer n, and define the weighted average r = a + w(b − a) = (1 − w)a + wb. Because
a, b, and w are rational numbers, and because the sum/difference and product of rational
numbers remain rational (a complete proof would need to show that this is true), it follows
that r is rational. Because a < b, w > 0, and 1 − w > 0, it follows that a = (1 − w)a + wa <
(1 − w)a + wb < (1 − w)b + wb = b which shows that a < r < b. In particular, if we choose
w = 1/2, then this weighted average reduces to the (equally weighted) average (or mean).
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