Proof:

Let m any odd integer.

The definition of even integer gives,

\(m = 2p\)

Here, p is also integer.

According to the question,

\(3m\ +\ 5 =3(2p)\ +\ 5\)

\(= 6p\ +\ 5\)

\(= 2(3p)\ +\ 2(2)\ +\ 1\)

\(= 2(3p\ +\ 4)\ +\ 1\)

Let, \((3p\ +\ 4) = a\)

Here, a is integer.

The above relation implies that,

\(3m\ +\ 5 = 2a\ +\ 1\)

The above relation \(3m\ +\ 5 = 2a\ +\ 1\) implies the definition of odd integer.

Therefore, for all integers m, if m is even then \(3m\ +\ 5\) is odd.

Let m any odd integer.

The definition of even integer gives,

\(m = 2p\)

Here, p is also integer.

According to the question,

\(3m\ +\ 5 =3(2p)\ +\ 5\)

\(= 6p\ +\ 5\)

\(= 2(3p)\ +\ 2(2)\ +\ 1\)

\(= 2(3p\ +\ 4)\ +\ 1\)

Let, \((3p\ +\ 4) = a\)

Here, a is integer.

The above relation implies that,

\(3m\ +\ 5 = 2a\ +\ 1\)

The above relation \(3m\ +\ 5 = 2a\ +\ 1\) implies the definition of odd integer.

Therefore, for all integers m, if m is even then \(3m\ +\ 5\) is odd.