How can I write the numbers 5 and 7 as some sequence of operations on three 9s? I want to make the numbers $1, 2, ..., 9$ using exactly three copies of the number $9$ and the following actions: addition, subtraction, multiplication, division, squaring, taking square roots, and other action.
How can we make the numbers $5$ or $7$?
For example, we can make the below numbers using exactly three copies of the number 9.


*

*$1=\dfrac{\sqrt 9\times\sqrt9}{9}$

*$2=\dfrac{9+9}{9}$

*$3=\dfrac{\sqrt9\times9}{9}$

*$4=\dfrac{9}{9}+\sqrt9$

*$5=\,?$

*$6=\dfrac{9+9}{\sqrt9}$

*$7=\,?$

*$8=9-\dfrac{9}{9}$

*$9=9+9-9.$


Now, how can we make the numbers 5 and 7?
 A: $$\sqrt{9}+\sqrt{9}\pm \lfloor \sqrt{\sqrt{9}} \rfloor$$
A: $$5=\sqrt{9}! - \frac{9}{9}, \quad 7=\sqrt{9}! + \frac{9}{9}$$
A: $5 = \lfloor\sqrt{\sqrt{999}}\rfloor = \lfloor\sqrt{9+9+9}\rfloor$
$7 = \lceil\ln999\rceil = \lfloor\sqrt{9}\sqrt{9}\ln\ln9\rfloor$
It would be much more interesting if the target was a large number like $123456789$. Who can get it with the shortest $\LaTeX$ formula that only has 3 nines and no other digits?
A: $5 = \dfrac{\ln(9 \times \sqrt{\sqrt{9}})}{\ln(\sqrt{\sqrt{9}})}$
$7 = \dfrac{\ln\left(\dfrac{9}{\sqrt{\sqrt{\sqrt{9}}}}\right)}{\ln\left(\sqrt{\sqrt{\sqrt{9}}}\right)}$
A: Since the question did not say the number of actions must be finite,
$${\sqrt{\sqrt{\sqrt{...\sqrt9}}}}=1$$
Then $5=1+1+3$, $7=1+3+3$.
A: $$5=\log_{\sqrt9}9+\sqrt9,\qquad7=9-\log_{\sqrt9}9.$$
A: I just came up with one...
7 = 9P$\sqrt 9$ >> ($\sqrt 9$)!
where the >> is the bitwise shift operator (as in programming languages such as C).
9P$\sqrt 9$ is 9P3, or 504.
($\sqrt 9$)! is 3!, or 6.
And this gives 504 >> 6, or 7.
A: Just do a handstand and read:$$\frac{9}{9}\mp9$$ 
A: A couple in the classic vein for these, both taking stealthy advantage of being base-10: $\displaystyle 5=\frac{9}{.9+.9}$ and $\displaystyle 7=\frac{9}{.9}-\sqrt{9}$.
A: Try $\lceil\sqrt[9]{9!}\rceil$. If you insist on three 9s try $\lfloor\sqrt[9]{9!}\rfloor+0.\bar 9$.
A: Answering the question of the asker in his comment to the answer by fuglede:

Do we can make number 5 or 7 by only using of the actions Sum, Subtraction, multiplication, division, Square or root?

No, it is not possible. Assuming that the square root should only be applied once to each term (and not infinitely often, as proposed in another answer), there is no way of representing the numbers 5.0 and 7.0 under the given constaints. 
The following is a list of all terms that result in values between 1.0 and 10.0 that can be obtained with addition, subtraction, multiplication, division, squaring or square root, sorted by the actual result. It was computed with a program simiar to the one in this stackoverflow answer, which computes all combinations of terms with the given operations. The list here only contains the equivalence classes referring to the result (that is, there are many ways of obtaining "1.0" as a result, but only one is listed here). 
It can be seen that all whole numbers between 1.0 and 10.0 can be obtained, except for 5.0 and 7.0.
EDIT based on the request in the comments: The list now contains only the whole numbers that can be obtained, and, for 5 and 7, the next smaller/larger number, respectively. The full list can be seen in the previous revision


*

*$(9 \times (9/(9)^2)) = 1$

*$((9+9)/9) = 2$

*$((9+\sqrt{9})-9) = 3$

*$(\sqrt{9}+(9/9)) = 4$

*$\sqrt{(((9)^2-9)/\sqrt{9})} = 4.898979485566356$

*$\sqrt{(((9)^2-\sqrt{9})/\sqrt{9})} = 5.0990195135927845$

*$(9-(9/\sqrt{9})) = 6$

*$((9)^2/(9+\sqrt{9})) = 6.75$

*$((\sqrt{9}-(\sqrt{9}/9)))^2 = 7.111111111111111$

*$(9-(9/9)) = 8$

*$((9+9)-9) = 9$

*$(9+(9/9)) = 10$

A: How about 
$5 = \sqrt 9 + \sqrt 9 \ -\ .\bar{9}$ 
and 
$7 = \sqrt 9 + \sqrt 9 \ +\ .\bar{9}$ ?
A: Interestingly, if you allow for natural logarithms and cube roots in your "other action," you can actually express any natural number (including zero) using only three nines. Observe that
$$ n = -\frac{1}{\ln 9} \ln \left( \frac{\ln \left( \sqrt[3]{\sqrt[3]{\cdots \sqrt[3]{ 9}}}\right)}{\ln 9} \right) $$
where there are $2n$ cube roots applied to the $9$ on the upper-right.

To see why this is true, recall the following logarithm identity:
$$ \ln a^b = b \ln a.$$
Now, we can simplify:
\begin{align}
-\frac{1}{\ln 9} \ln \left( \frac{\ln \left( \sqrt[3]{\sqrt[3]{\cdots \sqrt[3]{ 9}}}\right)}{\ln 9} \right) &= -\frac{1}{\ln 9} \ln \left( \frac{\ln \left( 9^{9^{-n}} \right)}{\ln 9} \right) \\
&= -\frac{1}{\ln 9} \ln \left( \frac{9^{-n} \ln 9}{\ln 9} \right) \\
&= -\frac{1}{\ln 9} \ln 9^{-n} \\[1mm]
&= \frac{1}{\ln 9} n \ln 9 \\[2mm]
&= n
\end{align}
You see, all I've done is implicitly introduce a $9^{-n}$ by asking you to take a cube root $2n$ times and cleverly hide it behind a wall of logarithms.
