Which is bigger: $\sqrt{1001} - \sqrt{1000}$, or $\frac{1}{10}$? 
Which is bigger: $\sqrt{1001} - \sqrt{1000}$, or $\frac{1}{10}$?

I can calculate the answer using a calculator, however I suspect to do so may be missing the point of the question.
The problem appears in a book immediately after a section called 'Rules for square roots'with $\sqrt{ab} = \sqrt{a}.\sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ as the given rules.
 A: Observe the following:
$$\frac{1}{10}+\sqrt{1000}=\frac{\sqrt{10}+1000}{\sqrt{1000}}>\frac{1001}{\sqrt{1000}}>\frac{1001}{\sqrt{1001}}=\sqrt{1001}$$
A: 
Is the pink border more or less than $\frac{1}{10}$cm wide?
A: Hint: $$\sqrt{n}-\sqrt{n-1}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}$$
To prove the above,multiply the numerator and denominator of the L.H.S by the conjugate.
A: Hint : Assume any one is bigger. write appropriate inequality sign. square. rearrange to put square root on one side. square.
A: This is not a proof but a way to "see" the answer, using just integer arithmetic.  Specifically, I'll use some perfect squares near 1000: 961 and 1024 have square roots equal to 31 and 32. So an increase of 63 (from 961 to 1024) in these numbers led to an increase of only 1 in their square roots. So, if (as one might reasonably expect) square roots increase in a rather regular fashion, increasing a number near 1000 by 1 should increase its square root by roughly 1/63. Although that's only a rough estimate, it's enough to convince me that 1/10 is way too big.
A: We want to find the sign of $\sqrt{1001} - \sqrt{1000} - \dfrac{1}{10}$.  Multiplying by the obviously positive $\sqrt{1001} + \sqrt{1000} + \dfrac{1}{10}$ we get $(\sqrt{1001})^2 - (\sqrt{1000} + \dfrac{1}{10})^2 = 1001 - 1000 - \dfrac{2\sqrt{1000}}{10} - \dfrac{1}{100} = \dfrac{99}{100} - \dfrac{2\sqrt{1000}}{10}$
Since $\sqrt{1000} > 30$ this quantity is obviously negative.  So $\dfrac{1}{10}$ is larger.
