What's wrong with this proof that an even number plus an even number is even? Suppose n and m are even numbers. Then n is 2k and m is 2k as well. It follows that their sum is 2k plus 2k. By using the laws of algebra, we can show that 2k plus 2k is 4k, or 2 times 2k. Hence, n plus m is even by definition of even number.

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Mathematics

Pogo

5 May, 01:23

What's wrong with this proof that an even number plus an even number is even? Suppose n and m are even numbers. Then n is 2k and m is 2k as well. It follows that their sum is 2k plus 2k. By using the laws of algebra, we can show that 2k plus 2k is 4k, or 2 times 2k. Hence, n plus m is even by definition of even number.

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Mariyah Ayala · Answer 1

The proof assumes that n=m=2k, which is false in general.

Step-by-step explanation:

If n is an even number, then n=2k for some integer k. In the same way, if m is an even number, then m=2j for some integer j. It is important to write two different letters, k and j, because these integers are not necessarily equal.

For example, take n=10 and m=30. Then k=5 and j=15, so they are different. The fallacy of this proof is that it assumes k=j.

A correct proof would continue like this: by the usual laws of algebra we have that n+m=2k+2j=2 (k+j). Since k+j is an integer, n+m=2p for some integer p=k+j, hence n+m is even.