Merge pull request #27 from spamegg1/2025-05-15

fix 3.3.41(g) again

Author: spamegg1

src/Epp.tex

@@ -9178,10 +9178,10 @@ \subsubsection{Exercise 41}
 \end{proof}
 
 (g)
-$\te x \in \R^+$ such that $\fa y \in \R^+$, $y < x$.
+$\fa x \in \R^+$, $\te y \in \R^+$ such that $y < x$.
 
 \begin{proof}
-False. If such an $x$ existed, then take $y = x+1$. So $y \in \R^+$ and $y > x$, a contradiction.
+True. For each $x$, we can take \(y = x / 2\) which satisfies the statement.
 \end{proof}
 
 (h)