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Short Proofs 3. Result: For every integer n, 4n + 7 is odd.
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Answer #1

If possible, let \left ( 4n+7 \right ) is not odd for some n .

Then it must be even and , it can be written as 4n+7=2m ,for some integer m .

This implies , 2\left ( m-2n \right )=7 .

This , shows that,  7 is an even number, which is not possible.

Hence, our assumption was wrong.

Therefore, \left ( 4n+7 \right ) is odd for every integer n .

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