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.

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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.