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If we have the function P(n): 2 | (2n-1), how could we prove that P(k) -> P(k+1) is a tautology?

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    I don't even know where to start. – 2017-02-08
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    $\;P(n)\;$ is false since $\;2 \not \mid 2n-1\;$ therefore $\;P(k) \implies \text{anything}\;$ is always true. – 2017-02-08
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    @dxiv does $n$ have to be an integer though? – 2017-02-08
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    n does have to be an integer, sorry I left that out. – 2017-02-08
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    @Archr Your answer makes the same assumption (how else do you define $\,2 \mid 2k-1\,$). – 2017-02-08
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    Well, I would have thought the function was true for $n=.5+z$ , $z \in \mathbb{N}$, and false for other inputs. But considering the revision, your answer is definitely cleaner and better than mine. – 2017-02-08

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Suppose $P(k)$. That is, $2|(2k-1)$. By the definition of | we have $2x = 2k-1$ for some integer $x$. We perform some algebraic manipulations: $$2x = 2k-1$$ $$2x +2 = 2k+1$$ $$2(x+1) = 2k+1$$ $$2(x+1) = 2k+1 +1 - 1$$ $$2(x+1) = 2k+2 - 1$$ $$2(x+1) = 2(k+1) - 1$$ $x$ is an integer, so $x+1$ is also an integer. Thus, we have shown that for some integer $y$, $2y=2(k+1)-1$. This is $P(k+1)$, so $P(k) \implies P(k+1)$.