Math 373 - Quiz 1 Solutions
Thursday, 2 Sept 2010
1. Give an example of a universal set U and two sets A and B such that
|A B | = 4
|A B | = 3
|B A| = 2
AB =1 .
Find |A B |.
Let
U = cfw_0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
A = cfw_0, 1, 2, 3, 4, 5, 6,
B = cfw_4,
Math 373 - hw 9 solutions
8.16a, 8.26, 8.36, 8.38, 8.40;
Tues, 9 Nov 2010
8.16a Let R be a relation dened on Z by a R b if a + b is even. Show that R is an equivalence relation and determine
the distinct equivalence classes.
Let a Z. Then a + a = 2a and s
Math 373 - hw 11 solutions
9.30, 9.34, 9.48a, 10.6, 10.11;
Tuesday, 23 Nov 2010
9.30 Let A and B be nonempty sets. Prove that if f : A B, then f iA = f and iB f = f .
Let a A. Then
(f iA )(a) = f (iA (a) = f (a),
(iB f )(a) = iB (f (a) = f (a).
Hence, for
Math 373 - hw 8 solutions
6.14, 6.16, 6.24, 8.6, 8.20;
Tues, 1 Nov 2010
6.14 Prove that n! > 2n for every integer n 4.
Let P (n) denote this inequality (where n N). We use induction to prove: P (n) for every integer n 4.
Base step: For n = 4, we have
4! =
Math 373 Review sheet
Test I: Thurs, 23 Sept
19 Sept 2010 / A Kumjian
Review: Thurs, 21 Sept
Test I will cover sections Chapters 1 and 2. You may bring a formula sheet to the test; it should include denitions,
propositions etc., but please do not include
Math 373 - Test I Solutions
Thursday, 23 Sept 2010
Each problem is worth ten points. Please write clearly and neatly. For full credit please show all steps and give reasons.
1. Describe the following set as cfw_x N : p(x), where p(x) is some property on x
Math 373 Review sheet
Test II: Thurs, 28 Oct
Sun, 24 Oct 2010 / A Kumjian
Review: Tues, 26 Oct
Test I will cover sections 3.1 3.4, 4.1 4.5, 5.1 5.5, 6.1. You may bring a formula sheet to the test; it may include
denitions, results, propositions etc., but
Math 373 - Test II Solutions
Thursday, 28 Oct 2010
Each problem is worth ten points. Please write clearly and neatly. These problems all require proofs, which must be
written in complete English sentences. For full credit please show all steps and give re
Math 373 - hw 10 solutions
9.6abc, 9.12, 9.14, 9.20, 9.22;
Tues, 16 Nov 2010
9.6 In each of the following, a function f1 : Ai R (i = 1, 2, 3) is dened, where the domain Ai consists of all real
numbers x for which fi (x) is dened. In each case, determine t
Math 373 - hw 7 solutions
5.30, 5.36, 6.6, 6.8, 6.10;
Tuesday, 19 Oct 2010
5.30 Show that there exist a rational number a and an irrational number b such that ab is irrational.
By result 5.21, there exist irrational numbers x, y such that a = xy is ration
Math 373 - hw 6 solutions
4.38, 5.6, 5.8, 5.16, 5.22, 5.26;
Tuesday, 12 Oct 2010
4.38 Let A, B and C be sets. Prove that (A B) (A C) = A (B C).
The result can also be proven by combining Result 4.18 (p. 97), Theorem 4.21.4(b) (De Morgans Laws) and
Theorem
Math 373 - Quiz 10 solutions
Thursday, 23 Nov 2010
1. Let f : Z Z be the function dened by f (n) = 2n2 + 1.
(a) Determine whether f is one-to-one.
We will show that f is not one-to-one. Observe that
f (1) = 2 12 + 1 = 3 = 2(1)2 + 1 = f (1).
Hence, f is no
Math 373 - Quiz 5 solutions
Thursday, 7 Oct 2010
1. Let n Z. Prove that 3 | (n2 + 2) implies 3 n.
We prove the contrapositve. Suppose that 3 | n. Then n = 3k for some k Z. Then
n2 + 2 = (3k )2 + 2 = 3(3k 2 ) + 2.
Since m = 3k 2 Z, n2 + 2 = 3m + 2 is not d
Math 373 - Quiz 6 solutions
Thursday, 14 Oct 2010
1. Disprove the statement: For every x R, x2 + 3x + 2 0
3
We will disprove the statement by counterexample. Let x = 2 , then
x2 + 3x + 2 = (x + 1)(x + 2) =
3
+1
2
3
+2
2
=
1
< 0.
4
Thus, x2 + 3x + 2 0 for
Math 373 - Quiz 4 solutions
Thursday, 30 September 2010
1. Let x Z. Prove that if 3x + 7 is odd, then x is even.
We prove the contrapositive. Suppose that x is odd. Then x = 2k + 1 for some k Z. Hence,
3x + 7 = 3(2k + 1) + 7 = 6k + 3 + 7 = 2(3k + 5).
Sinc
Math 373 - Quiz 2 solutions
Thursday, 9 Sept 2010
1. For the open sentence P (A) : A cfw_1, 2 = over the domain S = P (cfw_1, 2, 3, 4), nd the set of all A S for
which P (A) is true.
Note that for A P (cfw_1, 2, 3, 4), we have A cfw_1, 2 = exactly when A
Math 373 - Quiz 11 solutions
Thursday, 2 Dec 2010
1. Show that the function f : R cfw_3 R cfw_0 dened by
2
,
for x R cfw_3,
f (x) =
x3
has an inverse and determine f 1 (x) for x R cfw_0.
To prove that f has an inverse, one could show that f is bijective i
Math 373 - Quiz 9 solutions
Tuesday, 16 Nov 2010
1. Let R be the relation dened on Z by a R b if a2 b2 (mod 4). Given that R is an equivalence relation, determine
the distinct equivalence classes. [Hint: For all a, b Z, if a b (mod 4) then a2 b2 (mod 4)].
Math 373 - Quiz 8 solutions
Thursday, 4 November 2010
1. Use mathematical induction to prove that 7n + 5 < 2n for every integer n 6.
Proof. Let P (n) denote this inequality (where n N). We use induction to prove: P (n) for every integer n 6.
Base step : F
Math 373 - Quiz 3 solutions
Thursday, 16 Sept 2010
1. Consider the following open sentences P (x) and Q(x) over the domain S = R.
P (x) : |x| 3;
Q(x) : x 2.
(a) Determine the truth values of the following statements: P (0) Q(0); P (3) Q(3); P (7) Q(7).
Fo
Math 373 - Quiz 7 solutions
Thursday, 21 Oct 2010
1. Use the Intermediate Value Theorem to prove that there exists x (0, 2) such that
x3 2x2 + 3x = 4.
Let f (x) = x3 2x2 + 3x. Since f is a polynomial, it is continuous on [0, 2]. Now
f (0) = 0 < 4 < 6 = f