Theorem: the sum of any two even integers equals 4k for some integer k. "proof: suppose m and n are any two even integers. by definition of even, m = 2k for some integer k and n = 2k for some integer k. by substitution, m + n = 2k + 2k = 4k. this is what was to be shown." what's the mistakes in this proof?

Question

Jovan Stafford

Mathematics

19 June, 16:49

Theorem: the sum of any two even integers equals 4k for some integer k. "proof: suppose m and n are any two even integers. by definition of even, m = 2k for some integer k and n = 2k for some integer k. by substitution, m + n = 2k + 2k = 4k. this is what was to be shown." what's the mistakes in this proof?

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Cristal Norris · Answer 1

Remark

The proof is only true if m and n are equal. Make it more general.

m = 2k

n = 2v

m + n = 2k + 2v = 2 (k + v).

k and v can be equal but many times they are not. From that simple equation you cannot do anything for sure but divide by 2.

There are 4 combinations

m is divisible by 4 and n is not. The result will not be divisible by 4.

m is not divisible by 4 but n is. The result will not be divisible by 4.

But are divisible by 4 then the sum will be as well. Here's the really odd result

If both are even and not divisible by 4 then their sum is divisible by 4