# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Even, Odd, Even, Odd… Topic: Algebra/S&S. Level: AMC/AIME. January 30th, 2007

Problem: Show that the sequence contains an infinte number of both even and odd integers. Furthermore, show that there can be at most two consecutive integers of the same parity.

Solution: Suppose their exists a finite number of either even or odd integers. Then we definitely have three integers of the same parity in a row. But since , they must each differ by . Hence

.

Adding the equalities together, we get

.

By the standard floor function inequalities, however, we know that when is not an integer. Hence

,

a contradiction. Hence we have proved both of the desired results. QED.

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Comment: An interesting result which can be simply generalized to any irrational and probably any positive irrational at all (see if you can prove this). It wasn't hard to reason out that because we really cannot have three integers of the same parity in a row in that sequence. Adding a touch of rigor to the proof turned out to be quite easy as well.

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Practice Problem: Prove or disprove that, given a positive irrational , the sequence contains an infinite number of both even and odd integers.

### 3 Responses to “Even, Odd, Even, Odd… Topic: Algebra/S&S. Level: AMC/AIME.”

1. Xuan Says:

would this be induction?

PP: um. duh. like isnt’ this obvious? plub in 1 for a and … POOF !
u got ur proof.
haha it rhymes