Prove that if n or m is an odd integer, then nm is an even integer. Proposed proof: Suppose that nm is odd. Then n*m = 2k + 1 for some integer k. Therefore, n or m must be odd.

Question

Eli Kline

Mathematics

15 January, 17:22

Prove that if n or m is an odd integer, then nm is an even integer. Proposed proof: Suppose that nm is odd. Then n*m = 2k + 1 for some integer k. Therefore, n or m must be odd.

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Trey Mcclure · Answer 1

Answer: Ok, we have two numbers, and one of them is an odd integer, and the other is even.

Lets call M to the odd integer and N the even.

We know that a even integer can be written as 2k, where k is a random integer, and a odd integer can be written as 2j + 1, where j is also a random integer.

then M = 2k, N = 2j+1

then the product of M and N is: M*N = 2*k * (2*j + 1) = 2 * (k*2*j + k)

is obvious to see that (k*2*J + k) is a integer, because k and j are integers.

then if we call g = (k*2*J + k), we can write M*N=2g, and we already know that this is an even number. So M*N is a even integer.

Prove that if n or m is an odd integer, then n*m is an even integer. Proposed proof: Suppose that n*m is odd. Then n*m = 2k + 1 for some integer k. Therefore, n or m must be odd.

Prove that if n or m is an odd integer, then nm is an even integer. Proposed proof: Suppose that nm is odd. Then n*m = 2k + 1 for some integer k. Therefore, n or m must be odd.