Math Problem Statement

  1. For each of the arguments below, define the propositions and state which rule of inference was used to conclude when the premise is True. If you cannot conclude based on the argument, state the reason. Argument Propositions Rule of Inference

  2. If 𝑛 is a real number such that 𝑛 > 3, then 2𝑛 > 6 Suppose that 𝑛 >= 2, can you conclude 2𝑛 > 6

2.If 𝑛 is a real number such that 𝑛 > 2, then 𝑛2 > 4. Suppose that 𝑛2 ≀ 4, then 𝑛 ≀ 2 not awarded.

Solution

Let's analyze each argument in terms of the propositions and rule of inference used.

Argument 1:

Statement: If nn is a real number such that n>3n > 3, then 2n>62n > 6. Suppose that nβ‰₯2n \geq 2, can you conclude 2n>62n > 6?

Propositions:

  • PP: n>3n > 3
  • QQ: 2n>62n > 6
  • Premise: Pβ†’QP \rightarrow Q
  • Additional assumption: nβ‰₯2n \geq 2

Analysis:

  • The premise is the implication Pβ†’QP \rightarrow Q, meaning if n>3n > 3, then 2n>62n > 6.
  • However, the assumption you are given is nβ‰₯2n \geq 2, which includes values of nn such as 2, 2.5, and 3 that do not satisfy n>3n > 3.
  • Since the premise only applies if n>3n > 3, and the new assumption doesn't guarantee n>3n > 3, you cannot conclude 2n>62n > 6. The condition for the premise to apply isn't met under the new assumption.

Conclusion:

  • Reason for no conclusion: The assumption nβ‰₯2n \geq 2 does not imply n>3n > 3, so you cannot apply the premise Pβ†’QP \rightarrow Q.
  • Rule of inference: No valid inference rule can be applied here due to the invalid assumption.

Argument 2:

Statement: If nn is a real number such that n>2n > 2, then n2>4n^2 > 4. Suppose that n2≀4n^2 \leq 4, then n≀2n \leq 2.

Propositions:

  • PP: n>2n > 2
  • QQ: n2>4n^2 > 4
  • Premise: Pβ†’QP \rightarrow Q
  • New assumption: n2≀4n^2 \leq 4

Analysis:

  • The premise states Pβ†’QP \rightarrow Q, meaning if n>2n > 2, then n2>4n^2 > 4.
  • The contrapositive of Pβ†’QP \rightarrow Q is Β¬Qβ†’Β¬P\neg Q \rightarrow \neg P, i.e., if n2≀4n^2 \leq 4, then n≀2n \leq 2.
  • Since the assumption is n2≀4n^2 \leq 4, you can apply the contrapositive of the original statement to conclude that n≀2n \leq 2.

Conclusion:

  • Valid conclusion: Using the contrapositive of Pβ†’QP \rightarrow Q, you can conclude that n≀2n \leq 2 when n2≀4n^2 \leq 4.
  • Rule of inference: Contrapositive rule.

Summary:

  1. First argument:

    • Propositions: Pβ†’QP \rightarrow Q where PP: n>3n > 3, QQ: 2n>62n > 6, but the assumption nβ‰₯2n \geq 2 does not satisfy PP, so no conclusion.
    • Reason: Invalid assumption.
    • Inference rule: None applicable.
  2. Second argument:

    • Propositions: Pβ†’QP \rightarrow Q where PP: n>2n > 2, QQ: n2>4n^2 > 4, and the contrapositive Β¬Qβ†’Β¬P\neg Q \rightarrow \neg P applies.
    • Conclusion: n≀2n \leq 2 (contrapositive rule).

Do you want further details or have any questions?

Related Questions:

  1. What is the difference between a contrapositive and converse in logic?
  2. Can you provide an example of Modus Ponens?
  3. How do you recognize when a rule of inference cannot be applied?
  4. What is the contrapositive of P→QP \rightarrow Q?
  5. How can you formally prove the contrapositive inference rule?

Tip:

Always check whether the premise or the contrapositive of a logical statement applies when given assumptions that seem to reverse the implication.

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Math Problem Analysis

Mathematical Concepts

Logic
Propositional Logic
Rules of Inference
Contrapositive

Formulas

If P β†’ Q, then contrapositive is Β¬Q β†’ Β¬P

Theorems

Contrapositive Rule
Implication in Logic

Suitable Grade Level

Grades 9-12