Answer:
It is proved that if [tex]n^3[/tex] is even the n is even.
Step-by-step explanation:
Given n is any integer.
To show [tex]n^3[/tex] is even then n is even.
Proving by contrapositive suppose [tex]n^3[/tex] is odd. Then we need to show n is odd.
Then, letting k is a ny integer,
[tex]n^3=2k+1\implies n=(2k+1)^{\frac{1}{3}}[/tex]
Now since (2k+1) is odd therefore n is odd.
Conversly let n is odd, then,
[tex]n=2k+1\implies n^3=(2k+1)^3[/tex]
since 2k+1 is odd so [tex]n^3[/tex] is odd.
This proves, if n is even then [tex]n^3[/tex] is even.
