Proofs on Odd and Even Integer Properties and Conditional Statements

Math Problem Statement

Show that for any positive integer 𝑛, 𝑛 is odd if and only if 5𝑛 + 6 is odd. Hints: (i) 𝑎 ↔ 𝑏 ≡ (𝑎 → 𝑏) ∧ (𝑏 → 𝑎) (ii) To prove that 𝑎 ∧ 𝑏 is true, there is a need to prove that 𝑎 is true and to prove that 𝑏 is true.
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.
Prove that if 𝑚 + 𝑛 and 𝑛 + 𝑝 are even integers, where 𝑚, 𝑛, 𝑝 are integers, then 𝑚 + 𝑝 is even. State the strategies used in the proof.
Prove that (𝑛2 + 1) ≥ 2𝑛 where 𝑛 is a positive integer with 1 ≤ 𝑛 ≤ 4.
Prove or disprove the following: a. ∀𝑛(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 → 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛) b. ∀𝑛(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 → 𝑛 𝑖𝑠 𝑜𝑑𝑑) c. ∃𝑛(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 → 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛) d. ∃𝑛(6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 → 𝑛 𝑖𝑠 𝑜𝑑𝑑) e. ∀𝑛(𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 → 6𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛) f. ∃𝑛(𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 → 6𝑛 𝑖𝑠 𝑜𝑑𝑑) where 𝑛 is in the domain of all integers.

Solution

Let's go through each question systematically.

Question 2:

Statement: Show that for any positive integer $n$ , $n$ is odd if and only if $5n + 6$ is odd.

Proof:

To prove "if and only if" ( $\iff$ ), we need to prove both directions:
- (If $n$ is odd, then $5n + 6$ is odd) and
- (If $5n + 6$ is odd, then $n$ is odd).
(Forward direction: If $n$ is odd, then $5n + 6$ is odd):
- If $n$ is odd, we can write $n = 2k + 1$ for some integer $k$ .
- Then $5n + 6 = 5(2k + 1) + 6 = 10k + 5 + 6 = 10k + 11 = 2(5k + 5) + 1$ , which is of the form $2m + 1$ (odd).
(Reverse direction: If $5n + 6$ is odd, then $n$ is odd):
- Suppose $5n + 6$ is odd. Then $5n$ must be odd, since 6 is even.
- For $5n$ to be odd, $n$ must be odd because the product of an odd number with an even number is even.

Thus, $n$ is odd if and only if $5n + 6$ is odd.

Question 3:

Statement: Show that for any integer $n$ , if $n^3 + 5$ is odd, then $n$ is even. Provide two different proofs.

(a) Proof by Contraposition:

Contrapositive Statement: If $n$ is odd, then $n^3 + 5$ is even.
If $n$ is odd, we can write $n = 2k + 1$ for some integer $k$ .
Then $n^3 = (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1$ , which is odd.
So $n^3 + 5 =$ odd + even = even.

Thus, by contraposition, if $n^3 + 5$ is odd, $n$ must be even.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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

Related Recommendation

Proof Exercises: Odd Integers and Proof by Contradiction

Prove by Contradiction: No Integer Exists Where n^2 is Even and n^3 is Odd

Proof by Contradiction: Proving n is Even if n^3 + 5 is Odd

Proof: If n^2 is Odd, then n is Odd

Proof Techniques for Odd and Even Properties in Integer Sets