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Exam (elaborations)

MAT 230 Discrete Mathematics| SNHU | Module 5 Problem Set 5–3 with Solutions (Grade A)

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This document provides complete, clearly explained solutions for MAT 230 Discrete Mathematics Module 5 Problem Set (5–3) at Southern New Hampshire University (SNHU). It covers key Module 5 topics commonly tested in discrete mathematics, with step-by-step reasoning to support understanding and exam preparation. The assignment earned a Grade A and is suitable as a reliable study guide or reference for mastering the material.

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Discrete Mathematics
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Discrete Mathematics









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Institution
Discrete Mathematics
Course
Discrete Mathematics

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Uploaded on
December 22, 2025
Number of pages
13
Written in
2025/2026
Type
Exam (elaborations)
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MODULE FIVE PROBLEM SET


This document is proprietary to Southern New Hampshire University. It and
the problems within may not be posted on any non-SNHU website.




TESTBANKSNERD




1

, Directions: Type your solutions into this document and be sure to show all steps
for arriving at your solution. Just giving a final number may not receive full credit.



PROBLEM 1
Indicate whether the two functions are equal. If the two functions are not equal,
then give an element of the domain on which the two functions have different values.

(a)
f : Z → Z, where f (x) = x2.
g : Z → Z, where g(x) = |x|2.


If we square an integer, an non negative result is obtained.
Considering
x2 = |x|2 for allx ∈ Z
Hence both functions are equal.


(b)
f : Z × Z → Z, where f (x, y) = |x + y|.
g : Z × Z → Z, where g(x, y) = |x| + |y|.


Let us say x and y have the same sign
f (x, y) = |x + y| = |x| + |y| = g(x, y)
But if x and y have opposite signs
f (x, y) /= g(x, y)
If x = -1, y = 1
Then
f (−1, 1) = | − 1 + 1| = |0| = 0
And
g(−1, 1) = | − 1| + |1| = 1 + 1 = 2
Therefore, the two functions are not equal.

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