Answer to Question #227145 in Discrete Mathematics for Aira

Question

Show that for any real number x, if x2
is odd, then x is odd.

Answer 1

If x is odd then x=2k+1.

$x^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1=2n+1.$

Thus, $x^2$ is odd.