Answer to Question #273406 in Discrete Mathematics for haji

Question #273406

Use a proof by contraposition to show that if 𝑛𝑛2 + 1 is even, then 𝑛𝑛 is odd.


1
Expert's answer
2021-11-30T18:21:38-0500

The first step in a proof by contraposition is to assume that the conclusion of the conditional statement "If n2+1n^2 + 1 is even, then nn is odd" is false; namely, assume that nn is even.

Then, by the definition of an even integer, n=2k,k∈Z.n = 2k, k\in \Z.

Substituting 2k2k for n,n, we find that 

n2+1=(2k)2+1=4k2+1=2(2k2)+1n^2+1=(2k)^2+1=4k^2+1=2(2k^2)+1

This tells us that n2+1n^2+1is odd, and therefore not even. This is the negation of the premise of the theorem.

Because the negation of the conclusion of the conditional statement implies that the hypothesis is false, the original conditional statement is true.

Our proof by contraposition succeeded; we have proved

the theorem " If n2+1n^2 + 1 is even, then nn is odd". 


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