Nested floors and ceilings

$\lceil \lfloor \sqrt6 \rfloor \rceil=2$ .

The expression is equivalent to $\sqrt{2 +\sqrt{2+\sqrt{2 \ldots}}}$

Let $x=\sqrt{2 +\sqrt{2+\sqrt{2 \ldots}}}$ .

We have $\sqrt{2+x}=x$ or $x^2-x-2$ .

Solving the quadratic equation and taking only the positive root of $x$ (as the radical symbol of a number is always positive) we get $x=\boxed{2}$ .

$\left \lceil \left \lfloor \sqrt{6} \right \rfloor \right \rceil \\ =\left \lceil \left \lfloor 2.449\ldots \right \rfloor \right \rceil \\ =\left \lceil 2 \right \rceil \\ =2$

Substituting it into the nested radical, we get:

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}$

Now, let the value of this expression be $x$ .

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}=x \\ 2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}} = x^2\\ 2+x=x^2\\ x^2-x-2=0\\ (x-2)(x+1) = 0\\ x=2,\;-1$

Since we know that the square root of any number cannot result in negative numbers, we take the positive solution.

The value of the expression is $\boxed{2}$

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