Question #184438

Proof by contradiction that if n is a positive integer, then n is odd if and only 

if 5n + 6 is odd


1
Expert's answer
2021-04-26T05:59:12-0400

Let us prove by contradiction that if nn is a positive integer, then nn is odd if and only if 5n+65n + 6 is odd.


Let nn is not odd. Then nn is even, and hence n=2k, kN.n=2k,\ k\in\mathbb N. It follows that 5n+6=5(2k)+6=2(5k+3)5n+6=5(2k)+6=2(5k+3), and hence 5n+65n+6 is even, that is 5n+65n + 6 is not odd.


On the other hand, let 5n+65n+6 is not odd. Then 5n+65n+6 is even, and hence 5n+6=2t, tN.5n+6=2t,\ t\in\mathbb N.

It follows that 5n=2t6=2(t3)5n=2t-6=2(t-3), and hence 5n5n is even. Therefore, nn is also even, that is nn is not odd.


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