Question

Prove the following statements using either direct or contrapositive proof. If \(n\) is odd, then \(8 \mid\left(n^{2}-1\right)\).

Short answer

Through the direct proof method, we have shown that if \(n\) is an odd number, then \(8 \) divides \(n^{2}-1\). The proof was established by breaking down the odd number into the general format and demonstrating that \(n^{2}-1\) is always divisible by 8 for any odd number \(n\).

Step-by-step solution

Problem Interpretation

First, let's validate the structure of the 'if-then' statement we're attempting to verify: If \(n\) is odd (predicate A), then \(8 \) divides \(n^{2}-1\) (predicate B). We can approach this using a direct proof, showing that whenever \(n\) is odd, \(n^{2}-1\) is divisible by 8.

Assume Predicate A

Start by assuming that predicate A is true, i.e., \(n\) is an odd number. This infers that there exists an integer \(k\) such that \(n=2k+1\). This is the general format for odd numbers.

Finding Predicate B

Now, substitute \(n=2k+1\) in equation \(n^{2}-1\). This gives us \((2k+1)^{2}-1\) which simplifies to \(4k^{2}+4k+1-1 = 4k^{2}+4k = 4k(k + 1)\).

Proving Predicate B

Note that the integer \(k\) is either even or odd. If \(k\) is even, then \(k+1\) is odd, and vice versa. In either scenario, one of \(k\) or \(k+1\) is even, meaning it can be written as \(2m\) for some integer \(m\). So, \(4k(k + 1)\) can be written as \(8m\), which implies that \(8\) divides \(n^{2}-1\). This proves predicate B.

Final conclusion

Given that the statement 'If \(n\) is an odd number, then \(8\) divides \(n^{2}-1\)' is found to hold true, we have completed our direct proof.