Answer
Invalid by the converse error.
Work Step by Step
Converse Error: $\forall x$ if $P(x)$ then $Q(x)$.
$Q(a)$ for a particular $a$.
$\therefore P(a)$. (invalid conclusion)
In this case P(x) is: Let x be two rational numbers m and n.
Q(x) is: m+n is rational.
a is the two particular rational numbers r and s.
