Answer to Question #249438 in Discrete Mathematics for Gayathri

Question

Prove by contradiction that for any integer n if n² is odd then n is odd.

Answer 1

Consider the next statement: for any integer if $n^{2}$ is odd then n is non-odd (if n is non-odd integer then n is even)

So, if $n^{2}$ is odd then n is even

Let $n^{2}$ = 2k+1, k $\in$ Z

n = 2m, m $\isin$ N ∪ {0}

$n^{2} = 2k+1 \to n=\sqrt{2k+1}$

$\sqrt{2k+1} = 2m \to 2k+1=4m^{2}$

Since 2k is even integer, then 2k+1 is odd. $4m^{2}$ is also even. We came to the conclusion that an odd number is equal to the even integer number, which is false. So, assumption that ' for any integer if $n^{2}$ is odd then n is non-odd ' is false, which means that for any integer if $n^{2}$ is odd then n is odd. The statement has been proved.