Theorem: If 6|\(n\), then 3|\(n\).

  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

Theorem: If \(n\equiv7\) (mod 2), then \(n\equiv3\) (mod 2).

  1. Domain: The theorem deals with the modular congruence of an integer with respect to two others. Hypothesis\(n\) and 7 are congruent modulo 2. Conclusion: \(n\) and 3 are congruent modulo 2.
  2. If an integer and 7 are congruent modulo 2, then the integer and 3 are also congruent modulo 2.
  3. 7 and 7 are congruent modulo 2, and 7 and 3 are also congruent modulo 2, since 2 divides (7 - 3). 9 and 7 are congruent modulo 2, since 2 divides (9 - 7), and 9 and 3 are also congruent modulo 2, since 2 divides (9 - 3). Adding 4 to the difference does not affect its divisibility by 2.
  4. By definition, \(n\equiv7\) (mod 2) means 2 divides \(n-7\) and \(n-7=2k\). Letting \(k'=k+2\), we have \(n-3=2k'\), and, since \(k'\) is an integer, by definition, 2 divides \(n-3\) and n and 3 are congruent modulo 2.
  5. I don't feel like I worded this proof clearly. I have a lot to learn about organizing the thought and logic process behind proofs.

Theorem: If \(n\equiv7\) (mod 2), then \(n\equiv3\) (mod 2).