Why is $2^{\lfloor \log_{2}n \rfloor} \leq n$? Can someone explain to me why the following is true? $$2^{\lfloor \log_{2}n \rfloor} \leq n$$
In my calculator I can see that it's true but I don't know how to show mathimatically... 
 A: Note that $$\log_2 n -1<\lfloor \log_2 n \rfloor \le \log_2 n$$
Also, $2^n$ is an always increasing function since $(2^n)^{\prime}= \ln 2 \times 2^n >0$. 
$$\therefore 2^{\lfloor\log_2n\rfloor} \le 2^{\log_2 n}=n$$
A: Observe that $\lfloor x\rfloor\leq x$, then $2^{\lfloor x\rfloor}\leq 2^x$.
Since $2^{\log_2n}=n$ we have
$$2^{\lfloor\log_2n\rfloor}\leq 2^{\log_2n}=n$$ 
A: ${2^{\left\lfloor {{{\log }_2}n} \right\rfloor }} = {2^{{{\log }_2}n - \left\{ {{{\log }_2}n} \right\}}} = \frac{n}{{{2^{\left\{ {{{\log }_2}n} \right\}}}}} \leqslant n$ because $0 \le \left\{ {{{\log }_2}n} \right\} < 1 \Rightarrow {2^{\left\{ {{{\log }_2}n} \right\}}} \ge 1.$
A: The definition of $\lfloor x\rfloor$ is as follows:
$\lfloor x\rfloor = $the greatest integer less than or equal to $x$
So, logically we have $$\lfloor x\rfloor \le x$$
When $x=\log_2 n$, then we get that $$\lfloor \log_2 n\rfloor \le \log_2 n$$
Also $2^x$ is an increasing function for all real $x$. Hence we can conclude that,
$$2^{\lfloor \log_2 n\rfloor} \le 2^{\log_2 n} = n$$
