Answer
See the solution.
Work Step by Step
We are given that $n$ is a positive integer and must prove the biconditional sentence '$n$ is even $\leftrightarrow$ $7n+4$ is even'. To prove this, we will show the two conditionals
'$n$ is even $\rightarrow$ $7n+4$ is even' and '$7n+4$ is even $\rightarrow$ $n$ is even' are true.
$Proof.$
Let $n$ be a positive integer. To show '$n$ is even $\rightarrow$ $7n+4$ is even', we will use a direct proof. So suppose $n$ is even. Then by definition of 'even number', $n=2k$ for some integer $k$. Then by substitution, we have $7n+4=7(2k)+4=14k+2$. We factor out a two to get $2(7k+2)$. Thus $7n+4=2(7k+2)$, where $7k+2$ is an integer. Hence $7n+4$ is even, and, therefore, the conditional sentence '$n$ is even $\rightarrow$ $7n+4$ is even' is true.
Next, we show '$7n+4$ is even $\rightarrow$ $n$ is even' by contraposition. So suppose $n$ is not even. Then $n$ must be odd. Thus, by definition of 'odd', $n=2k+1$ for some integer $k$. Then by substitution, $$7n+4=7(2k+1)+4=14k+7+4=14k+10+1=2(7K+5)+1,$$
where $7k+5$ is an integer. Hence $7n+4$ is odd, and, therefore, not even. Thus, we have shown '$7n+4$ is even $\rightarrow$ $n$ is even' is true.
Now since '$n$ is even $\rightarrow$ $7n+4$ is even' and '$7n+4$ is even $\rightarrow$ $n$ is even' are both true, the biconditional, '$n$ is even $\leftrightarrow$ $7n+4$ is even', is true.$_\Box$\
