Given:
[tex]3n+2=6k+2[/tex]First we assume n is an even,
n=2k,
[tex]\begin{gathered} 3n+2=3(2k)+2 \\ =6k+2 \\ =2(3k+1) \end{gathered}[/tex]Multiple of 2 is always even.
Conversely assume that n is an odd
n=2k+1
[tex]\begin{gathered} 3n+2=3(2k+1)+2 \\ =6k+3+2 \\ =6k+5 \end{gathered}[/tex]Here 6k is even but 5 is an odd by adding even and odd we get an odd number.
Contradicts the proof .
