Find square root of non-rational fraction

If we have to compute this without using calculator, is there a quick way to find the answer approximately of the following problem:

which one is smaller ? $$A = \frac{\sqrt{2007}-\sqrt{2006}}{\sqrt{2008}-\sqrt{2007}}$$ or $$B = \frac{\sqrt{2010}-\sqrt{2009}}{\sqrt{2011}-\sqrt{2010}}$$

My thinking is to multiply A with $\displaystyle \frac{\sqrt{2008}+\sqrt{2007}}{\sqrt{2008}+\sqrt{2007}}$ and B with $\displaystyle \frac{\sqrt{2011}+\sqrt{2010}}{\sqrt{2011}+\sqrt{2010}}$, and simplify from fraction into multiplication and subtraction only to become:$A' = (\sqrt{2007}-\sqrt{2006})(\sqrt{2008}+\sqrt{2007})$, and $B' = (\sqrt{2010}-\sqrt{2009})({\sqrt{2011}+\sqrt{2010}})$.

This form is still not easy to calculate for me.

• you can use \sqrt{...} to get the root over the entire ... expression – Guest 86 Jan 27 '13 at 13:00
• So the idea in your last paragraph, how well does it work? Do you run into trouble? – Henning Makholm Jan 27 '13 at 13:07
• Sorry, now i completed my last paragraph. – kuskus Jan 27 '13 at 13:10
• @kusg1: But what happens when you do those multiplications? Does it solve the problem? Do you get stuck? If so, where? – Henning Makholm Jan 27 '13 at 13:22

This looks like a job for calculus. Basically the question is whether the function $$x \mapsto \frac{\sqrt{x}-\sqrt{x-1}}{\sqrt{x+1}-{\sqrt{x}}}$$ is increasing or decreasing around $x=2010$. Since the increases of $1$ are small compared to 2010 (which in this case means that none of the relevant derivatives show much relative change when $x$ varies by $1$) we can probably get away with setting $g(x)=\sqrt x$ and approximating $$\frac{g(x)-g(x-1)}{g(x+1)-g(x)} \approx \frac{g'(x-1)}{g'(x)} \approx \frac{g'(x)-g''(x)}{g'(x)} = 1 - \frac{g''(x)}{g'(x)} = 1 + \frac{1}{2x}$$
• So ? $x \mapsto 1-e^{-x}$ is also concave, but if you go and replace the square roots with it, you get $A=B=e$ – mercio Jan 27 '13 at 13:29