Theorem What’s that? Any two consecutive integers have opposite parity. Jon Campbell – Introduction James Kendall – Proof Rob Chalhoub – Example, Questions, Conclusion

Let’s translate it! If a certain integer is even, the next integer in sequence is odd. If a certain integer is odd, the next integer in sequence is even.

Let’s prove it! Suppose that m and m+1 are consecutive integers. [we must show that either m or m+1 is even and that the other is odd] By the parity property, either m is even or m is odd. [we will break the proof into both of these cases]

PART I: We’re still provin’ it! CASE 1 (m is even): By the definition of even m=2k for some integer k Thus, m+1 = 2k+1 which is the definition of odd. Therefore, in the of m being even, m+1 will always be odd.

Part II: Electric Boogaloo! Case 2 (m is odd): By the definition of odd, m is equal to (2k + 1) for some value of k. Thus, m+1 = (2k+1)+1 = (2k +2) = 2(k+1). (If n = k+1) = 2n. Thus, by the definition of even, m+1 is even when m is odd.

I think we’ve got it! Part I proves that if m is even then m+1 is always odd. Similarly, Part II proves that if m is odd then m+1 is even. Thus, regardless of the case of m, m+1 is always of opposite parity. [this was what was to be proven]

I don’t believe it! Given two integers 4 and 5. (k = 2) 4 = 2k (thus 4 is even) 5 = 2k+1 (thus 5 is odd) Given two integers 5 and 6: (k = 2) 5 = 2k +1 (thus 5 is odd) 6 = 2k +2 or 2(k+1) (thus 6 is even)

Look out for this! (Exam question) Prove that any two consecutive integers have the same parity.

Try this out for yourself (Homework) Questions: Exercise Set 3.4 (P163-4) 24, 39

Scroll the credits! Everything – US Producer – Steve Spielberg