Fall 2014
Math 3000-01 Quiz 1 Solution
Page 1 of 2
1. Solution:
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Fall 2014
Math 3000-01 Final Exam (Part 2)
Due by 4:00 p.m. Friday, 12/12/14
Page 1 of 1
Name:
Options of submission & (I) is PREFERRED:
(I) Arrange your work sheets in order and put the problem sheet on the top.
Indicate the total number of work sheets a
Fall 2014
Math 3000-01 Selected Homework Solution
Page 24
Section 2.4
1. Solution: Let S represent the given set in each part.
(a) For each natural number n 20, if n S, then n + 1 S. Thus, S has the
inductive property.
(c) Note that 2 S; however, its succ
Fall 2014
Math 3000-01 Quiz 1
Due: Monday, 10/20/2013 in class
Time spent on this quiz:
Page 1 of 1
Name:
Total number of work sheets:
Note: Read the following instructions carefully before you start the test.
1. This is a take-home exam. Do NOT get help
Fall 2014
Math 3000-01 Midterm 2 Partial Solution
Page 1 of 2
1. Proof:
(a) Prove by cases for n = 3q + 1 or n = 3q + 2, where q is some integer.
(b) Since n is a prime, its only natural number divisors are 1 and n. Since n = 3, n
is not divisible by 3. I
Fall 2014
Math 3000-01 Midterm 1 (Part 2)
Due: Wed. 11/5/2014 in class
Page 1 of 1
Name:
Note: Read the following instructions carefully before you start the test.
1. This is a take-home exam. Do NOT get help from anyone else and do NOT
share your work wi
Fall 2014
Math 3000-01 Quiz 2 Solution
Page 1 of 2
a
1. Proof: Assume on the contrary that 3 is rational. That is, 3 = for relatively
b
a2
prime integers a and b. Then, 3 = 2 or a2 = 3b2 . Since b2 is an integer, 3 divides
b
a2 . By Problem 2 in part 2 of
Fall 2014
Math 3000-01 Selected Homework Solution
Page 18
Section 2.1
1. Solution:
(b) cfw_x : x Z and x2 < 17.
(e) cfw_x : x R and 5 x < 1.
4. Solution:
1
5
2 < 2, < 2 < . This implies that 2 is an irrational number in
2
2
1 5
1 5
the interval , . But,
Fall 2014
Math 3000-01 Selected Homework Solution
Page 16
Section 1.6
4. (a) Solution: Lets consider positive integer x = 41. x2 + x + 41 = 412 + 41 + 41 =
41 (41 + 1 + 1) = 41 43 is a composite number. Therefore, it is false that for all
positive integer
Fall 2014
Math 3000-01 Selected Homework Solution
Page 15
Section 1.6
1. Proof:
(b) Lets choose m = 1 and n = 1. Then 15m + 12n = 15(1) + 12(1) = 3. Thus,
there exists integers m and n such that 15m + 12n = 3.
(h) Let m be an (arbitrary) odd integer. Then
Fall 2014
Math 3000-01 Selected Homework Solution
Page 12
Section 1.5
1. Solution:
(b) Assume that the determinant of a matrix B is zero.
.
.
.
Thus, B is a singular matrix.
Hence, its proven that if the determinant of a matrix B is zero, then B is a
sing
Fall 2014
Math 3000-01 Selected Homework Solution
Page 9
Section 1.4
1. Solution:
(c) Suppose that A, B, and C are sets.
Assume that A is a subset of B and B is a subset of C.
.
.
.
Thus, A is a subset of C.
Therefore, (its proven that) if A is a subset o
Fall 2014
Math 3000-01 Selected Homework Solution
Page 4
Section 1.2
1. Solution:
(c) Its antecedent is b divides 3. and consequent is b divides 9.
(e) Its antecedent is a is convergent. and consequent is a is bounded.
(g) Its antecedent is 1 + 1 = 2. and
Fall 2014
Math 3000-01 Selected Homework Solution
Page 7
Section 1.3
1. Solution:
(c) (x)(x is isosceles x is right).
(d) (x) (x is right x is isosceles) (x)(x is isosceles x is right) due to
the commutativity of . That is, (c) & (d) are denials of each o
Fall 2014
Math 3000-01 Selected Homework Solution
Page 1
Section 1.1
1. (d) Solution: Note that 4, 6, and 8 have a factor of 2 and 9 has a factor of 3.
That leaves 2, 3, 5, and 7 to be the only prime numbers (that are integers greater
than 1 whose only po