Question

. The number n is an even integer if and only if 3n + 2 = 6k + 2 for some integer k.

Answer 1

Given:

`3n+2=6k+2`

First we assume n is an even,

n=2k,

`\begin{gathered} 3n+2=3(2k)+2 \\ =6k+2 \\ =2(3k+1) \end{gathered}`

Multiple of 2 is always even.

Conversely assume that n is an odd

n=2k+1

`\begin{gathered} 3n+2=3(2k+1)+2 \\ =6k+3+2 \\ =6k+5 \end{gathered}`

Here 6k is even but 5 is an odd by adding even and odd we get an odd number.

Contradicts the proof .