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 odd $\leftrightarrow$ $5n+6$ is odd'. To prove this, we will show the two conditionals
'$n$ is odd $\rightarrow$ $5n+6$ is odd' and '$5n+6$ is odd $\rightarrow$ $n$ is odd' are true.
$Proof.$
Let $n$ be a positive integer. To show '$n$ is odd $\rightarrow$ $5n+6$ is odd', we will use a direct proof. So suppose $n$ is odd. Then by definition of 'odd integer', $n=2k+1$ for some integer $k$. Then by substitution, we have $$5n+6=5(2k+1)+6=10k+5+6=10k+10+1=2(5k+5)+1,$$ where $5k+5$ is an integer. Hence $5n+6$ is odd, and, therefore, the conditional sentence '$n$ is odd $\rightarrow$ $5n+6$ is odd' is true.
Next, we show '$5n+6$ is odd $\rightarrow$ $n$ is odd' by contraposition. So suppose $n$ is not odd. Then $n$ must be even. Thus, by definition of 'even integer', $n=2k$ for some integer $k$. Then by substitution, we have $5n+6=5(2k)+6$. We factor out a two to get $2(5k+3)$, where $5k+3$ is an integer. Hence $5n+6$ is even, and, therefore, not odd. Thus, we have shown '$5n+6$ is odd $\rightarrow$ $n$ is odd' is true.
Now since '$n$ is odd $\rightarrow$ $5n+6$ is odd' and '$5n+6$ is odd $\rightarrow$ $n$ is odd' are both true, the biconditional, '$n$ is odd $\leftrightarrow$ $5n+6$ is odd', is true.$_\Box$
