contestada

Prove:
For all integers n, if n2 is odd, then n is odd.
Use a proof by contraposition, as in Lemma 1.1.
Let n be an integer. Suppose that n is even, i.e., n = for some integer k. Then n2 = = 2



is also even.

Respuesta :

LammettHash
Given the statement, "If [tex]n^2[/tex] is odd, then [tex]n[/tex] is odd," its contrapositive claims that, "If [tex]n[/tex] is not odd, then [tex]n^2[/tex] is not odd."

So assume [tex]n[/tex] is not odd, i.e. [tex]n[/tex] is even. This means there is an integer [tex]k[/tex] for which [tex]n=2k[/tex]. Squaring this gives [tex]n^2=(2k)^2=4k^2[/tex].

Well, we can write [tex]4k^2=2(2k^2)[/tex], and [tex]2k^2[/tex] is just another integer, which means [tex]4k^2=(2k)^2=n^2[/tex] must be even.

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