Prove or disprove the following, Vx, y, z ≤ Z(((x+y) | (y+z)) → (x | z) V (y > 0))
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Title:** Logical Proof or Disproof of a Mathematical Proposition
**Objective:** Prove or disprove the given logical-mathematical statement.
**Proposition:**
\[
\forall x, y, z \in \mathbb{Z} \left( \left( (x + y) \mid (y + z) \right) \rightarrow \left( (x \mid z) \lor (y > 0) \right) \right)
\]
**Explanation:**
The statement is a universally quantified proposition involving integers \(x, y, z \in \mathbb{Z}\).
- **\( (x + y) \mid (y + z) \):** This denotes that \(x + y\) divides \(y + z\), meaning there exists an integer \(k\) such that \((y + z) = k(x + y)\).
- **\(\rightarrow\):** The logical "implies" operator; the entire expression on the left must lead to the expression on the right.
- **\( (x \mid z) \lor (y > 0) \):** This is the conclusion that \(x\) divides \(z\) or \(y\) is greater than 0.
**Approach:**
To prove this statement, we need to show that for all integers \(x, y, z\), the given implication holds true. Conversely, to disprove it, a single counterexample of integers showing the hypothesis is true and the conclusion false is sufficient. This exercise tests the understanding of divisibility over integers and the logical constructs of implication and disjunction within the realm of number theory.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F233b4b28-2634-4ce7-bb56-644fea0dac02%2Fd874d425-2c2c-4d01-ae8a-921d3bb6505b%2F7u1sp7m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title:** Logical Proof or Disproof of a Mathematical Proposition
**Objective:** Prove or disprove the given logical-mathematical statement.
**Proposition:**
\[
\forall x, y, z \in \mathbb{Z} \left( \left( (x + y) \mid (y + z) \right) \rightarrow \left( (x \mid z) \lor (y > 0) \right) \right)
\]
**Explanation:**
The statement is a universally quantified proposition involving integers \(x, y, z \in \mathbb{Z}\).
- **\( (x + y) \mid (y + z) \):** This denotes that \(x + y\) divides \(y + z\), meaning there exists an integer \(k\) such that \((y + z) = k(x + y)\).
- **\(\rightarrow\):** The logical "implies" operator; the entire expression on the left must lead to the expression on the right.
- **\( (x \mid z) \lor (y > 0) \):** This is the conclusion that \(x\) divides \(z\) or \(y\) is greater than 0.
**Approach:**
To prove this statement, we need to show that for all integers \(x, y, z\), the given implication holds true. Conversely, to disprove it, a single counterexample of integers showing the hypothesis is true and the conclusion false is sufficient. This exercise tests the understanding of divisibility over integers and the logical constructs of implication and disjunction within the realm of number theory.
Expert Solution
Step 1
Consider
x = 3, y = -2, z = 10
Then, x + y = 3 + (-2) = 3 - 2 = 1
y + z = -2 + 10 = 8
Step by step
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