Comparing Square Root Differences: $\sqrt{7}-\sqrt{6}$ vs. $\sqrt{6}-\sqrt{5}$

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The discussion centers on comparing the differences between the square roots of consecutive integers: $\sqrt{7}-\sqrt{6}$ and $\sqrt{6}-\sqrt{5}$. The conclusion, provided by user eddybob123, establishes that $\sqrt{7}-\sqrt{6}$ is greater than $\sqrt{6}-\sqrt{5}$. This is derived from the fact that the function $\sqrt{x}$ is concave, leading to larger differences as the integers increase. The correct solutions were acknowledged from several forum members, including MarkFL and Pranav.

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Which is larger? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$? Answer without using a calculator.
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Congratulation to the following members for their correct solutions:

1) MarkFL
2) anemone
3) Pranav
4) eddybob123
5) kaliprasad

Solution (from eddybob123):
$$ 5041 > 5040$$
$$\Rightarrow 71>2\sqrt{1260}$$
$$\Rightarrow 1< 72-2\sqrt{1260}$$
$$\Rightarrow 1^2< (\sqrt{42}-\sqrt{30})^2$$
$$\Rightarrow 1< \sqrt{42}-\sqrt{30}$$
$$\Rightarrow 2<2(\sqrt{42}-\sqrt{30})$$
$$\Rightarrow 7-2\sqrt{42}+6<6-2\sqrt{30}+5$$
$$\Rightarrow (\sqrt{7}-\sqrt{6})^2<(\sqrt{6}-\sqrt{5})^2$$
$$\Rightarrow \sqrt{7}-\sqrt{6}<\sqrt{6}-\sqrt{5}$$
 

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