$\frac{\mathrm{n}}{\sqrt{\mathrm{n}}}=\sqrt{\mathrm{n}}$

Notice that $\frac{2}{\sqrt{2}}~1.41$ and $\sqrt{2}~1.41$ as well. In fact, $\frac{2}{\sqrt{2}}=\sqrt{2}$

Here is some proof of that specific case: $\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}$

And the same thing goes for any number, n, where n∊N: $\frac{n}{\sqrt{n}}=\frac{n}{\sqrt{n}}\times \frac{\sqrt{n}}{\sqrt{n}}=\frac{n\sqrt{n}}{\sqrt{n}\sqrt{n}}=\frac{n\sqrt{n}}{n}=\sqrt{n}$

Here's another proof: $\frac{n}{\sqrt{n}}=\frac{n^1}{n^{0.5}}=n^{\mathrm{1\; -\; 0.5}}=n^{0.5}=\sqrt{n}$

And another proof: $\sqrt{n}\sqrt{n}=n$ so $\frac{n}{\sqrt{n}}=\sqrt{n}$