1. Domain: The theorem deals with the divisibility of an integer with respect to two others. Hypothesis: 6 divides \(n\)Conclusion: 3 divides \(n\).
  2. If an integer is divisible by 6, then it is also divisible by 3.
  3. 6 divides 6, since 6 = 1 * 6. 3 also divides 6, since 6 = 2 * 3. 6 divides 12, since 12 = 2 * 6. 3 also divides 12, since 12 = 4 * 3. In the notation of \(a=kd\), when \(d\) is 3 rather than 6, \(k\) is twice as big.
  4. Given \(6|n\), \(n=6k\) by definition. Letting \(k'=2k\), we have \(n=3k'\), and, by definition, \(3|n\).
  5. This proof was not too bad. I'm sure it will get harder soon.