- If \(a=kd\), then \(d\) and \(k\) both divide \(a\).
- The definition regards a property of a pair of integers (where order matters).
- 6 divides 12. 2 and 3 both divide 6. 2 and 3 both divide 12. 12 divides 60. 2, 3, 4, 5, and 6 all divide 60. 5 does not divide 12. 7 does not divide 60. 6 does not divide 3.
Definition: a and b are congruent modulo n
- If \(b=a+nk\) and \(n\) and \(k\) are both positive, then \(a\) and \(b\) are congruent modulo \(\)\(n\) and modulo \(k\).
- The definition regards a property of a pair of integers with respect to a third integer.
- \(1\equiv3\equiv5\equiv7\) (mod 2). \(0\equiv2\equiv4\equiv6\) (mod 2). \(a\equiv b\) (mod 1) for any two integers \(a\) and \(b\). \(1\not\equiv 2\) (mod 2). \(3\not\equiv 6\) (mod 2).
Theorem: If 6|\(n\), then 3|\(n\).
- Domain: The theorem deals with the divisibility of an integer with respect to two others. Hypothesis: 6 divides \(n\). Conclusion: 3 divides \(n\).
- If an integer is divisible by 6, then it is also divisible by 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.
- Given 6|\(n\), \(n\ =\ 6k\) by definition. Letting \(k'=2k\), we have \(n=3k'\), and, by definition, 3|\(\)\(n\).
- 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).
- 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.
- If an integer and 7 are congruent modulo 2, then the integer and 3 are also congruent modulo 2.
- 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.
- 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.
- 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).