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I don't quite understand why I can not proof the following:

Assume that $n,m ∈ \mathbb{N}$.
Show: ($m$ is odd $∧$ $n$ is odd) $\Rightarrow$ $m+ n$ is even.
With this: Say $n, m$ are odd. Then the remains of $(m + n) / 2 $ is equal to zero.

Why is this not possible and why do I have to use (for instance) induction, to proof this?

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First observe that a natural number is odd if it can be written as: $2k + 1$ for some natural number $k$, and even if it can be written as $2k$, for some natural $k$. Then: $m = 2k + 1$, and $n = 2s + 1$. So: $m + n = 2k + 1 + 2s + 1 = 2(k + s + 1)$ is even.

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  • $\begingroup$ It actually makes a little bit more sense now. Thanks! $\endgroup$ – user149508 Jun 6 '14 at 11:20

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