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Mathematics, 18.02.2021 21:00 perezriii4133

Use the following building blocks to assemble a proof that if m and n are integers, and mn is even, then m is even or n is even. Not all blocks belong in the proof. Before you start, you might want to write out such a proof on paper, and carefully consider whether a direct proof, a proof by contraposition or a proof by contradiction is the most appropriate. Suppose that m is odd and n is odd.
By definition of odd number, this means that nm is odd
We will give a proof by contraposition.
Suppose that nm is odd.
By definition of odd number, m = 2p + 1 for some integer p, and n 2q1 for some integer q.
We will give a direct proof.
Suppose that nm is even.
Therefore, nm-(2q+1(2p1) 4pq+ 2q+2p+1-2(2pq+ q+p)1
Therefore, we have finished our direct proof that one of the numbers n, m must be even.
By definition of odd number, m = 2k + 1 and n-2k + 1 for some integer k
Therefore, we have proved the contrapositive of the desired statement and thus completed the proof by contraposition.
By definition of even number, nm 2k for some integer k.
Therefore, n = 2 and m = korn.. k and m-2.
We will give a proof by contradiction
This contradicts our assumption that nm is even. Thus, we have proved by contradiction that m must be even, or n must De oven
Suppose that m is odd or n is odd.
Therefore, nms(2k + 1) (2k + 1) = 4ke + 4k + 1-2 (2k2 + 2k)

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