Math Problem Statement

  1. Show that for any positive integer 𝑛, 𝑛 is odd if and only if 5𝑛 + 6 is odd. Hints: (i) π‘Ž ↔ 𝑏 ≑ (π‘Ž β†’ 𝑏) ∧ (𝑏 β†’ π‘Ž) (ii) To prove that π‘Ž ∧ 𝑏 is true, there is a need to prove that π‘Ž is true and to prove that 𝑏 is true.
  2. Show that for any integer 𝑛, if 𝑛3 + 5 is odd, then 𝑛 is even. Provide 2 different proofs: a. Proof by contraposition. b. Proof by contradiction.
  3. Prove that if π‘š + 𝑛 and 𝑛 + 𝑝 are even integers, where π‘š, 𝑛, 𝑝 are integers, then π‘š + 𝑝 is even. State the strategies used in the proof.
  4. Prove that (𝑛2 + 1) β‰₯ 2𝑛 where 𝑛 is a positive integer with 1 ≀ 𝑛 ≀ 4.
  5. Prove or disprove the following: a. βˆ€π‘›(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 β†’ 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛) b. βˆ€π‘›(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 β†’ 𝑛 𝑖𝑠 π‘œπ‘‘π‘‘) c. βˆƒπ‘›(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 β†’ 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛) d. βˆƒπ‘›(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 β†’ 𝑛 𝑖𝑠 π‘œπ‘‘π‘‘) e. βˆ€π‘›(𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 β†’ 6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛) f. βˆƒπ‘›(𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 β†’ 6𝑛 𝑖𝑠 π‘œπ‘‘π‘‘) where 𝑛 is in the domain of all integers.

Solution

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Math Problem Analysis

Mathematical Concepts

Parity (Odd and Even Numbers)
Logical Equivalence and Contraposition
Contradiction Proofs
Conditional Statements

Formulas

Odd and even integer definitions: n = 2k + 1 for odd, n = 2k for even
Expressions for checking odd or even results in composite functions (5n + 6, n^3 + 5)

Theorems

Contraposition
Contradiction
Conditional Statements in Logical Equivalence

Suitable Grade Level

Grades 11-12

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