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Epp - Discrete Mathematics with Applications, 5/e (Homework)

James Finch

Math - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 18 / 27

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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Question
Points
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2/5 3/3 7/8 2/6 1/2 1/1 2/2
Total
18/27 (66.7%)

  • Instructions

    Known for its accessible, precise approach, Susanna Epp's Discrete Mathematics With Applications, 5th Edition, published by Cengage Learning introduces discrete mathematics with clarity and precision. Coverage emphasizes the major themes of discrete mathematics as well as the reasoning that underlies mathematical thought. Students learn to think abstractly as they study the ideas of logic and proof. While learning about logic circuits and computer addition, algorithm analysis, recursive thinking, computability, automata, cryptography, and combinatorics, students discover that ideas of discrete mathematics are essential to and underlie today's science and technology. The author's emphasis on reasoning provides a foundation for computer science and upper-level mathematics courses.

    Question 1 has a Watch It, a video showing the solution of a similar problem.

    Question 2 assesses closure and parity of integers.

    Question 3 features a combination proof that scaffolds the writing experience and balances guidance with rigor. Students identify the first step, determine the final goal to develop a rough outline, complete the next few steps with drop-down multiple-choice menus and an open-ended algebraic blank, and finally provide the second half of the proof in essay form.

    Question 4 gives a scrambled list of nine statements, only six of which belong in the proof. Students must construct a proof with the correct statements and then explain why the remaining three statements do not belong in the proof.

    Question 5 presents an attempted proof to analyze. Students identify and explain both the mistakes and the correct components of the given proof, and then they use their answers to rewrite the proof.

    Question 6 provides feedback when an equivalent answer is entered but it is not in the required format.

    Question 7 is an Expanded Problem that requires the student to show intermediate answers necessary to complete the question.

    Click here for a list of all of the questions coded in WebAssign. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

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1. 2/5 points  |  Previous Answers EppDiscMath5 2.1.008. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5
0/1 0/1 1/1 1/1 0/1
5/5 5/5 3/5 3/5 5/5
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2/5
 
Write the statements in symbolic form using the symbols ~, ⋁, and ⋀ and the indicated letters to represent component statements.
Let h = "John is healthy," w = "John is wealthy," and s = "John is wise."
(a)
John is healthy and wealthy but not wise.
     Incorrect: Your answer is incorrect.
(b)
John is not wealthy, but he is healthy and wise.
     Incorrect: Your answer is incorrect.
(c)
John is neither healthy, wealthy, nor wise.
     Correct: Your answer is correct.
(d)
John is neither wealthy nor wise, but he is healthy.
     Correct: Your answer is correct.
(e)
John is wealthy, but he is not both healthy and wise.
     Incorrect: Your answer is incorrect.

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2. 3/3 points  |  Previous Answers EppDiscMath5 4.1.002. My Notes
Question Part
Points
Submissions Used
1 2 3
1/1 1/1 1/1
3/5 5/5 3/5
Total
3/3
 
Assume that c is a particular integer.
(a)
Is
−8c
 an even integer?
     Correct: Your answer is correct.
(b)
Is
6c + 5
 an odd integer?
     Correct: Your answer is correct.
(c)
Is
(c2 + 3) − (c2 − 3) − 6
 an even integer?
     Correct: Your answer is correct.
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3. 7/8 points  |  Previous Answers My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8
0/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1
5/5 2/5 3/5 1/5 1/5 1/5 3/5 4/5
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7/8
 
Consider the following statement.
The difference between any odd integer and any even integer is an odd integer.
Which of the following could begin a direct proof of the statement?
     Incorrect: Your answer is incorrect.
Plan Ahead!
Before continuing with the next steps, plan ahead and clarify the final goal. Fill in the blank.
It must be shown that Correct: Your answer is correct. at the end of the proof.
With an idea of where to start and where to finish, complete the next few steps that follow the first by filling in the blanks.
By definition, since m is an odd integer, there exists an Correct: Your answer is correct. r such that m = Correct: Your answer is correct. .
Similarly, by definition, since n is an even integer, there exists an Correct: Your answer is correct. such that n = Correct: Your answer is correct. .
Because the statement to be proved deals with the difference m − n, use the above information to rewrite the difference. Substitute the mathematical expressions found for m and n into the equation. (Enter your answer in terms of r and s.)
m − n =
2r+1−2s
Correct: Your answer is correct.
Consider how to algebraically manipulate the expression on the right side of the equation to reach the final goal. Finish the proof in essay form.

Score: 1 out of 1

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4. 2/6 points  |  Previous Answers EppDiscMath5 4.1.030. My Notes
Question Part
Points
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1 2 3 4 5 6
0/1 1/1 0/1 0/1 0/1 1/1
3/5 3/5 3/5 4/5 2/5 1/5
Total
2/6
 
Consider the following theorem.
The sum of any even integer and any odd integer is odd.
Only six statements in the following scrambled list belong in the proof.
By substitution,
m + n = 2t + 1,
 where t is an integer.
Let m + n be any odd integer.
So by definition of even, t is even.
Let t = r + s. Then t is an integer because it is a sum of integers.
By definition of even and odd, there are integers r and s such that
m = 2r
 and
n = 2s + 1.
Suppose m is any even integer and n is any odd integer.
Hence,
m + n
 is twice an integer plus one. So by definition of odd,
m + n
 is odd.
By substitution and algebra,
m + n = 2r + (2s + 1) = 2(r + s) + 1.
By definition of even and odd, there is an integer r such that
m = 2r
 and
n = 2r + 1.
Construct the proof by choosing the appropriate statements from the list and putting them in the correct order.
  1. Incorrect: Your answer is incorrect.
  2. Correct: Your answer is correct.
  3. Incorrect: Your answer is incorrect.
  4. Incorrect: Your answer is incorrect.
  5. Incorrect: Your answer is incorrect.
  6. Correct: Your answer is correct.
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5. 1/2 points  |  Previous Answers EppDiscMath5 4.2.018. My Notes
Question Part
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1 2
0/1 1/1
5/5 1/5
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1/2
 
Consider the following theorem.
The product of any even integer and any odd integer is even.
Here is a proof of the theorem with at least one incorrect step.
  1. Suppose m is any even integer and n is any odd integer.
  2. If
    m · n
     is even, then by definition of even there exists an integer r such that
    m · n = 2r.
  3. By definition of even and odd, there exist integers p and q such that
    m = 2p
     and
    n = 2q + 1.
  4. Therefore, by substitution, the product
    m · n = (2p)(2q + 1) = 2r.
  5. But r is an integer by the assumption above.
  6. Thus
    m · n
     is two times an integer, so by definition of even, the product is even.
Identify the mistakes in the proof. (Select all that apply.)
Incorrect: Your answer is incorrect.

Explain your reasoning.

Score: 1 out of 1

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6. 1/1 points  |  Previous Answers EppDiscMath5 4.3.002. My Notes
Question Part
Points
Submissions Used
1
1/1
3/5
Total
1/1
 
Write the rational number as a ratio of two integers. (Simplify your answer completely.)
5.6073
Correct: Your answer is correct.
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7. 2/2 points  |  Previous Answers EppDiscMath5 4.4.005.EP. My Notes
Question Part
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1 2
1/1 1/1
1/5 1/5
Total
2/2
 
Consider the expression below. Assume m is an integer.
6m(3m + 24)
Complete the equality. (Simplify your answer completely. Enter an expression in the variable m. If no expression exists, enter DNE.)
6m(3m + 24) = 9
2m(m+8)
Correct: Your answer is correct.
Given that m is an integer, is
6m(3m + 24)
 divisible by 9?
     Correct: Your answer is correct.
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Enter a reduced fraction or integer.