This Problem Was a Challenge Problem on My Math Test

By the binomial expression, we get that -

$\sqrt[3]{n+1}-\sqrt[3]{n}= n^{1/3}(1+\frac{1}{n}^{1/3}-n^{1/3}=n^{1/3}(1+0(\frac{1}{n}))-n^{1/3}=O(n^{-2/3})$

so $\sqrt[3]{n+1}-\sqrt[3]{n} \rightarrow 0$ as $n \rightarrow \infty$ . If $m>r^3$ . Then, the numbers

$0=\sqrt[3]{m}-\sqrt[3]{m}<\sqrt[3]{m+1}-\sqrt[3]{m}<\dots<\sqrt[3]{m+7m}-\sqrt[3]{m}-\sqrt[3]{m}=\sqrt[3]{m}$ partition the interval $[0,\sqrt[3]{m}]$ , containing $r$ , in such a way that the largest subinterval is of size $O(m^{-2/3}$ . By taking $m$ sufficiently large, one can find a difference among these that is arbitrarily close to $r$ . Therefore, the answer is $\color{#20A900}\boxed{\text{Yes}}$ .