You will need to use the denitions of even and odd integers for these two proofs

Question

You will need to use the denitions of even and odd integers for these two proofs. The integer n is even if n = 2k for some integer k. The integer n is odd if n = 2k + 1 for some integer k. Some facts you can use: (1) Every integer is either even or odd. (2) No integer is both even and odd.

(1) Give a proof by contradiction: If s is an even integer and t is an odd integer, then 4 does not divide s^2 + 2t^2. (Recall that an                 integer a divides an integer b means there is an integer c such that ac = b. In other words, we can multiply a by an integer to             get b.)

Explanation / Answer

Suppose (4) divides (s^2 + 2t^2). Since (s) is an even integer and (t) is an odd integer, then we can express them as (s = 2k) for some integer (k) and (t = 2l + 1) for some integer (l). By algebra, we have

[(2k)^2 + 2(2l + 1)^2 = 4k^2 + 2(4l^2 + 4l + 1) = 4k^2 + 8l^2 + 8l + 2]

But (4) can't divide (2), which shows that (4) does not divide (s^2 + 2t^2).

Either (n) is even or (n) is odd.

If (n) is even, then (n = 2m) for some integer (m). Then,

[n^2 + 27n - 18 = (2m)^2 + 27(2m) - 18]

is even since the product of integers with at least one even integer is even, the sum of all even integers is even, and the difference of all even integers is even.

If (n) is odd, then (n = 2p + 1) for some integer (p). Then,

[n^2 + 27n - 18 = (2p + 1)^2 + 27(2p + 1) - 18 = 4p^2 + 4p + 1 + 54p + 27 - 18 = 4p^2 + 58p + 8]

So by the same reasoning, (n^2 + 27n - 18) is even.

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