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MAT 230 - Discrete Mathematics |Problem Set 4-3- Exam One | Grade A

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This document includes fully worked solutions for MAT 230 Discrete Mathematics Problem Set 4–3, aligned with Exam One at Southern New Hampshire University (SNHU). The material focuses on key exam topics such as logic, proofs, sets, relations, functions, and foundational discrete mathematics concepts, with clear explanations for each solution. The assignment received a Grade A and is suitable for exam preparation and concept review. If LaTeX formatting is required for professionally structured mathematical notation, this can be offered as an additional service.

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MAT 230 EXAM ONE


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
(a) The domain for all variables in the expressions below is the set of real numbers. Determine
whether each statement is true or false.
(i) ∀ x ∃ y (x + y ≥ 0)


∀x ∃y (x + y ≥ 0)
∀x means “for every real number x” ∃y means there exists some real number y depending
on x. We need to find at least one y for each x such that
x+y ≥0
Expressing y in terms of x
Choose
y = −x
Then
x + y = x + (−x) = 0
This means
0≥ 0
hence the inequality holds. Alternatively, any y ≥ −x which satisfies x + y ≥ 0 Therefore,
since for every real x we can find such a y, the statement is TRUE.


(ii) ∃ x ∀ y (x · y > 0)


∃x ∀y (x · y > 0)
For any x ∈ R, let y = 0. Then
x · y = x · 0 = 0 /> 0
Hence, no x exists that makes x · y > 0 true for all y Therefore, the statement is FALSE.


(b) Translate each of the following English statements into logical expressions.
(i) There are two numbers whose ratio is less than 1.

Let x and y be numbers. Since there are two numbers, it implies an existential quantifier
and the ratio must be defined (y /= 0), the statement become
∃x ∃y (y /= 0 ∧ x/y < 1)




(ii) The reciprocal of every positive number is also positive.

Let x be a positive number (x > 0). Its reciprocal is 1/x. Since the statement asserts
that this reciprocal is positive (1/x > 0) for all x > 0.
Therefore, the logical expression will be
∀x (x > 0 → 1/x > 0)

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