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Chapter 1: Problem 19

Determine whether each of these arguments is valid. If an argument is correct, what rule of inference is being used? If it is not, what logical error occurs? a) If \(n\) is a real number such that \(n>1,\) then \(n^{2}>1\) Suppose that \(n^{2}>1 .\) Then \(n>1\) b) If \(n\) is a real number with \(n>3,\) then \(n^{2}>9\) . Suppose that \(n^{2} \leq 9 .\) Then \(n \leq 3\) . c) If \(n\) is a real number with \(n>2,\) then \(n^{2}>4\) . Suppose that \(n \leq 2 .\) Then \(n^{2} \leq 4 .\)

Short Answer

Expert verified
a) Invalid - affirming the consequent. b) Valid - contrapositive. c) Invalid - denying the antecedent.

Step by step solution

01

- Statement Analysis for (a)

Given: If \(n > 1\), then \(n^2 > 1\). Suppose \(n^2 > 1\), then \(n > 1\). We need to determine if the argument is valid.
02

- Determine the Validity for (a)

This argument is not valid. The argument commits the logical error known as 'affirming the consequent'. Explanation: The given implication is of the form \(p \rightarrow q\) (if \(n > 1\), then \(n^2 > 1\)). The statement \(n^2 > 1\) is the consequent \(q\), and just because the consequent \(q\) is true, it does not imply that the antecedent \(p\) (\(n > 1\)) is necessarily true.
03

- Statement Analysis for (b)

Given: If \(n > 3\), then \(n^2 > 9\). Suppose \(n^2 \leq 9\), then \(n \leq 3\). We need to determine if the argument is valid.
04

- Determine the Validity for (b)

This argument is valid. The rule of inference used here is 'contrapositive'. Explanation: The given implication is of the form \(p \rightarrow q\) (if \(n > 3\), then \(n^2 > 9\)). The contrapositive of \(p \rightarrow q\) is \(eg q \rightarrow eg p\) (if \(n^2 \leq 9\), then \(n \leq 3\)). Since the contrapositive is logically equivalent to the original implication, the argument is valid.
05

- Statement Analysis for (c)

Given: If \(n > 2\), then \(n^2 > 4\). Suppose \(n \leq 2\), then \(n^2 \leq 4\). We need to determine if the argument is valid.
06

- Determine the Validity for (c)

This argument is invalid. The argument commits the logical error known as 'denying the antecedent'. Explanation: The given implication is of the form \(p \rightarrow q\) (if \(n > 2\), then \(n^2 > 4\)). The statement \(n \leq 2\) is the negation of the antecedent \(eg p\), and just because the antecedent \(p\) is false (or \(eg p\) is true), it does not imply that the consequent \(q\) is false.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

validity of arguments
Understanding the validity of arguments is crucial when you are studying logical reasoning. An argument is valid if, assuming the premises are true, the conclusion must be true. Validity is about the structure of the argument rather than the actual truth of the premises or conclusion.

For example, in exercise (a), we looked at an argument and determined it was not valid. The structure did not allow for the conclusion to be guaranteed if the premises were true. Contrast this with exercise (b), where the argument was found to be valid because the conclusion was necessarily true if the premises were true. Being able to test the validity of an argument is a key skill in logical reasoning.
rules of inference
Rules of inference are logical rules that we use to deduce new statements from existing ones. They are foundational to constructing valid arguments and proofs. Some of the common rules include Modus Ponens, Modus Tollens, and the contrapositive rule.

Step-by-Step:
1. Modus Ponens: If 'If P then Q' and 'P' are true, then 'Q' must be true.
2. Modus Tollens: If 'If P then Q' is true and 'Q' is false, then 'P' must be false.
3. Contrapositive: If 'If P then Q' is true, then 'If not Q then not P' is also true.

In exercise (b), we utilized the contrapositive rule. The original statement was 'If n > 3, then n^2 > 9'. By taking its contrapositive, 'If n^2 鈮� 9, then n 鈮� 3', we preserved logical equivalence, ensuring the argument remains valid.
logical errors
Logical errors occur when the structure of an argument is flawed, leading to invalid conclusions even if the premises are true. Common logical errors include 'affirming the consequent' and 'denying the antecedent'.

Step-by-Step:
1. Affirming the Consequent: This error occurs in the form 'If P then Q' and 'Q', concluding 'P'. Just because Q is true, it does not prove that P is true.
2. Denying the Antecedent: This error happens in the form 'If P then Q' and 'Not P', concluding 'Not Q'. Just because P is false, it does not imply that Q is false.

For instance, in exercise (a), the error of affirming the consequent occurred. The argument assumed that since 'n^2 > 1' is true, 'n > 1' must also be true, which is not necessarily correct. In exercise (c), the error was denying the antecedent. The assumption was that because 'n 鈮� 2' was true, 'n^2 鈮� 4' must be true, which is another faulty conclusion. Identifying these errors helps you understand why certain arguments fail.

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Most popular questions from this chapter

Suppose that the domain of the propositional function \(P(x)\) consists of the integers \(1,2,3,4,\) and \(5 .\) Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. $$ \begin{array}{ll}{\text { a) } \quad \exists x P(x)} & {\text { b) } \forall x P(x)} \\ {\text { c) } \quad \neg \exists x P(x)} & {\text { d) } \neg \forall x P(x)}\end{array} $$ e) \(\quad \forall x((x \neq 3) \rightarrow P(x)) \vee \exists x \neg P(x)\)

Show that the equivalence \(p \wedge \neg p \equiv \mathbf{F}\) can be derived using resolution together with the fact that a conditional statement with a false hypothesis is true. [Hint: Let \(q=\) \(r=\mathbf{F}\) in resolution.

Show that the two statements \(\neg \exists x \forall y P(x, y)\) and \(\forall x \exists y \neg P(x, y),\) where both quantifiers over the first variable in \(P(x, y)\) have the same domain, and both quantifiers over the second variable in \(P(x, y)\) have the same domain, are logically equivalent.

Express the negations of these propositions using quantifiers, and in English. a) Every student in this class likes mathematics. b) There is a student in this class who has never seen a computer. c) There is a student in this class who has taken every mathematics course offered at this school. d) There is a student in this class who has been in at least one room of every building on campus.

Exercises \(48-51\) establish rules for null quantification that we can use when a quantified variable does not appear in part of a statement. Establish these logical equivalences, where \(x\) does not occur as a free variable in \(A\) . Assume that the domain is nonempty. $$ \begin{array}{l}{\text { a) }(\forall x P(x)) \vee A \equiv \forall x(P(x) \vee A)} \\ {\text { b) }(\exists x P(x)) \vee A \equiv \exists x(P(x) \vee A)}\end{array} $$
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