Problem 53
Question
Determine the truth value for each statement when \(p\) is false, \(q\) is true, and \(r\) is false. \(p \wedge(q \vee r)\)
Step-by-Step Solution
Verified Answer
The truth value of the statement \(p \wedge (q \vee r)\) when \(p\) is false, \(q\) is true, and \(r\) is false is false.
1Step 1: Evaluate \(q \vee r\)
Evaluate the operation within the brackets first according to BODMAS rules. As per the given condition, \(q\) is true and \(r\) is false. The OR (\(\vee\)) operation takes the value true if either of its operands is true. Therefore, \(q \vee r = true \vee false = true\).
2Step 2: Evaluate \(p \wedge (q \vee r)\)
Now, evaluate \(p \wedge (q \vee r)\) using the value of \(q \vee r\) and the given value of \(p\). The AND (\(\wedge\)) operation takes the value true only if both of its operands are true. In this case, \(p = false\) and \(q \vee r = true\), so \(p \wedge (q \vee r) = false \wedge true = false\).
Key Concepts
Logical ConjunctionLogical DisjunctionBODMAS RulesLogical Operands
Logical Conjunction
Logical conjunction, commonly represented by the symbol \( \wedge \) and known as the \( AND \) operator, is a fundamental concept in logical operations. It connects two statements (operands) in such a manner that the combined statement is true only if both of the operands are true.
For example, in the provided exercise, the conjunction is used between \( p \) and the result of \( q \vee r \). The truth value of \( p \) is false, and despite the truth value of \( q \vee r \) being true, the entire expression \( p \wedge (q \vee r) \) yields a false result because \( p \) is false, and both operands need to be true for a conjunction to be true.
For example, in the provided exercise, the conjunction is used between \( p \) and the result of \( q \vee r \). The truth value of \( p \) is false, and despite the truth value of \( q \vee r \) being true, the entire expression \( p \wedge (q \vee r) \) yields a false result because \( p \) is false, and both operands need to be true for a conjunction to be true.
- If \( p \) is true and \( q \) is true, then \( p \wedge q \) is true.
- If either \( p \) or \( q \) is false, then \( p \wedge q \) is false.
Logical Disjunction
Logical disjunction signifies the \( OR \) operation in logic, symbolized by \( \vee \) and it defines a statement that is true if at least one of the operands is true.
The exercise makes use of this principle when it evaluates \( q \vee r \) inside the parentheses. Since \( q \) is true and \( r \) is false, the disjunction \( true \vee false \) becomes true because at least one operand—\( q \)—is true.
Understanding disjunction is critical for analyzing complex logical statements and constructing truth tables in logic.
The exercise makes use of this principle when it evaluates \( q \vee r \) inside the parentheses. Since \( q \) is true and \( r \) is false, the disjunction \( true \vee false \) becomes true because at least one operand—\( q \)—is true.
- If \( p \) is true or \( q \) is true (or both), then \( p \vee q \) is true.
- If both \( p \) and \( q \) are false, then \( p \vee q \) is false.
Understanding disjunction is critical for analyzing complex logical statements and constructing truth tables in logic.
BODMAS Rules
BODMAS is an acronym representing an order of operations used to solve mathematical expressions: Brackets, Orders (powers and square roots, etc.), Division and Multiplication, Addition and Subtraction.
This rule dictates that operations inside brackets are to be performed first, as seen in the exercise \( p \wedge (q \vee r) \). Here, \( (q \vee r) \) indicates that the operation within the brackets, the logical disjunction \( q \vee r \), must be resolved before applying the logical conjunction \( p \wedge … \) to the result.
Carefully applying BODMAS ensures the correct order of operations in mathematics and logic and prevents errors in computation of complex expressions.
This rule dictates that operations inside brackets are to be performed first, as seen in the exercise \( p \wedge (q \vee r) \). Here, \( (q \vee r) \) indicates that the operation within the brackets, the logical disjunction \( q \vee r \), must be resolved before applying the logical conjunction \( p \wedge … \) to the result.
Carefully applying BODMAS ensures the correct order of operations in mathematics and logic and prevents errors in computation of complex expressions.
Logical Operands
In logic, operands are the elements that operators act upon. In the context of the exercise, \( p \) and \( q \) are instances of logical operands.
Each operand has a truth value: true or false. When logical operators like conjunction \( \wedge \) and disjunction \( \vee \) are applied to these values, they create new logical statements.
The truth value of an operand can change the outcome of a logical expression significantly. For example, altering the truth value of \( p \) from false to true would change the final output of the expression \( p \wedge (q \vee r) \) from false to true, showcasing the critical role of operands in logical expressions.
Each operand has a truth value: true or false. When logical operators like conjunction \( \wedge \) and disjunction \( \vee \) are applied to these values, they create new logical statements.
The truth value of an operand can change the outcome of a logical expression significantly. For example, altering the truth value of \( p \) from false to true would change the final output of the expression \( p \wedge (q \vee r) \) from false to true, showcasing the critical role of operands in logical expressions.
Other exercises in this chapter
Problem 53
Write the negation of each statement. Express each negation in a form such that the symbol \(\sim\) negates only simple statements. \(p \wedge(r \rightarrow \si
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Use a truth table to determine whether each statement is a tautology, a self- contradiction, or neither. \([(p \rightarrow q) \wedge(q \rightarrow r)] \rightarr
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Let \(p\) and \(q\) represent the following simple statements: p: Romeo loves Juliet. q: Juliet loves Romeo. Write each symbolic statement in words. \(\sim(q \v
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Translate each argument into symbolic form. Then determine whether the argument is valid or invalid. It is the case that \(x10\), but \(x \leq 10\), so \(x
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