Which of the following is smallest?

A)\(10 - 3\sqrt{11} \qquad\) B) \(8-3\sqrt{7} \qquad\) C) \(5-2\sqrt{6} \qquad\) D) \(9-4\sqrt{5} \qquad\) E) \(7-4\sqrt{3}\)

We could note that \(10 - 3\sqrt{11} = 10-\sqrt{3^2\times 11} = \sqrt{100} - \sqrt{99}\).

Similarly, \[\begin{align*} 8-3\sqrt{7} &= \sqrt{64} - \sqrt{63}, \\ 5-2\sqrt{6} &= \sqrt{25}- \sqrt{24}, \\ 9-4\sqrt{5} &= \sqrt{81}-\sqrt{80}, \\ 7-4\sqrt{3} &= \sqrt{49}-\sqrt{48}. \end{align*}\]So all the options are of the form \(\sqrt{x} - \sqrt{x-1}\).

What does the graph of \(y = \sqrt{x} - \sqrt{x-1}\) look like? It seems intuitively obvious that as \(x\) gets bigger, \(y\) gets smaller.

For example, \(\sqrt{100}-\sqrt{99}\) is surely bigger than \(\sqrt{10\,000} - \sqrt{9\,999}\).

So we need to pick the largest value of \(x\) on offer, and the answer is (A).

We could note that \((\sqrt{x} - \sqrt{x-1} )(\sqrt{x} + \sqrt{x-1} ) =1\), and so \(\sqrt{x} - \sqrt{x-1} = \dfrac{1}{\sqrt{x} + \sqrt{x-1}}\), which certainly gets smaller as \(x\) increases.