A knot in \(S^3\) is topologically slice if it bounds a locally flat disk in \(B^4\) . A knot in \(S^3\) is rationally slice if it bounds a smooth disk in a rational homology ball. We prove that the smooth concordance group of topologically and rationally slice knots admits a \(\mathbb {Z}^\infty \) subgroup. All previously known examples of knots that are both topologically and rationally slice were of order two. As a direct consequence, it follows that there are infinitely many topologically slice knots that are strongly rationally slice but not slice.
