Chapter 4.7, Problem 10ES

Let S be the statement: For all positive real number r and s, $\sqrt{r+s}\ne \sqrt{r}+\sqrt{s}.$ Statement S is true, but the following “proof” is incorrect. Find the mistake.
“Proof by contradiction: Suppose not. That is, suppose that for all positive real numbers r and s, $\sqrt{r+s}=\sqrt{r}+\sqrt{s}.$ This means that the equation will be true no matter what positive real numbers are substituted for r and s. So let $r=9$ and $s=16.$ Then r and s are positive real numbers and $\sqrt{r+s}=\sqrt{9+16}=\sqrt{25}=5$ whereas $\sqrt{r}+\sqrt{s}=\sqrt{9}+\sqrt{16}=3+4=7.$ Since $5\ne 7,$ we have that $\sqrt{r+s}\ne \sqrt{r}+\sqrt{s},$ which contradicts the supposition that $\sqrt{r+s}=\sqrt{r}+\sqrt{s}.$ This contradiction shows that the supposition is false, and hence statement S is true.”