In Discrete math and its app book from Kenneth H. Rosen says that
What is the symmetric closure of the relation $R=\{(a,b) | a >b\}$ on the set of positive integers ?
The answer of book following:
$R \cup R^{-1}=\{(a,b) | a >b\} \cup \{(b,a) | a >b\}= \{(a,b) | a \neq b\}$
I think that it should have been like
$R \cup R^{-1}=\{(a,b) | a >b\} \cup \{(b,a) | \color{blue}{b >a}\}= \{(a,b) | a \neq b\}$
The book gives the same answer for other editions,so i think that it is not just typo. Can you explain me why my answer is not correct ?
I think you're confused about the set-builder notation. Consider the pair $(1, 2)$. Clearly, $(1, 2) \in R^{-1}$. In the books solution, taking $b = 1$ and $a = 2$, we see that indeed $a > b$, so that $(1, 2) \in \{(b, a) \mid a > b\}$. In your solution, taking $b = 1$ and $a = 2$, we see that $b \not> a$, so that $(1, 2) \notin \{(b, a) \mid b > a\}$.