Question

Consider the following definition. Definition. A integer n is frumpable if n² + 6n is odd. Prove: All frumpable numbers are odd. Suppose n is an even integer, so n = for some integer k. Then n² + 6n = 2( ) is ______ 4 , so on A frumpable.

Answer 1

If a number n is frumpable (n² + 6n is odd), and n is even (n = 2k for some integer k), n² + 6n = 4(k² + 3k), which is even. Since frumpable numbers require n² + 6n to be odd, n cannot be even and must be odd, proving all frumpable numbers are odd.

Step-by-step explanation:

To prove that all frumpable numbers are odd, we will consider the definition provided: a number n is frumpable if n² + 6n is odd. The question provides a step in the proof for when n is an even integer; let's break it down further.

Let n be an even integer, which means n can be written as 2k for some integer k. Thus, n² + 6n = (2k)² + 6(2k) = 4k² + 12k = 4(k² + 3k). Since k² + 3k is an integer, we can denote it as m, so n² + 6n = 4m, which is clearly an even number because it is a multiple of 4.

Since the definition of a frumpable number requires n² + 6n to be odd and the expression 4m (where m is an integer) is even, it follows that n cannot be even and must therefore be odd. Thus, all frumpable numbers must be odd.