Question

Prove the following statements using either direct or contrapositive proof. If \(a \in Z\) and \(a \equiv 1(\bmod 5)\), then \(a^{2} \equiv 1(\bmod 5)\).

Short answer

Given the assumption that \(a \) is congruent to 1 modulo 5, we are able to prove that \(a^{2} \) is also congruent to 1 modulo 5. This was achieved using direct proof method and the definition of modulo.

Step-by-step solution

- Identifying Type of Proof

The exercise does not specify any particular type of proof to use, but either direct or contrapositive proofs can be used. A direct proof will be used because it's simpler and more straightforward: If a number \(a\) is congruent to 1 modulo 5, then \((a-1) \) is a multiple of 5 - that's basically the definition of congruence modulo 5. We will use this fact for the proof

- Building the Proof

To show that \(a^{2} \) is congruent to 1 modulo 5, we need to show that \((a^{2}-1)\) is a multiple of 5. Using the definition of modulo and factoring (difference of squares), we get: \((a^{2}-1) = (a-1)(a+1)\), both a-1 and a+1 are either multiple of 5. Hence \(a^{2}-1\) is a multiple of 5. Therefore, \(a^{2} \equiv 1(\bmod 5)\)

- Checking the Conclusion

To ensure that our proof is accurate, we want to check if \(a^{2} \equiv 1(\bmod5)\) when \(a \equiv 1(\bmod5)\) holds true. We verified that if \(a-1\) is a multiple of 5, then \((a^{2}-1)\) is also a multiple of 5 and therefore, \(a^{2} \equiv 1(\bmod 5)\) as per the definition of modulo.