2. 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.
3. 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.
4. 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.
5. Prove that (𝑛2 + 1) ≥ 2𝑛 where 𝑛 is a positive integer with 1 ≤ 𝑛 ≤ 4.
6. 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.

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:
1. 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)**.

2. **(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).

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

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



2. Show that for any positive integer n, n is odd if and only if 5n + 6 is odd.
3. Show that for any integer n, if n^3 + 5 is odd, then n is even. Provide 2 different proofs:
   a. Proof by contraposition.
   b. Proof by contradiction.
4. 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.
5. Prove that (n^2 + 1) ≥ 2n where n is a positive integer with 1 ≤ n ≤ 4.
6. 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.

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