$\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}$An example here seems best. How many digits in base 2 do I need to represent any odd integer from $1$ to $\sqrt{77}$, inclusive? It seems to be essentially half the digits required for $77$ --- in base 2. We can represent $77$ with $7$ digits in base 2. Half of that is $3.5$, so we need $4$ digits in base 2. So let's change "essentially half" to "exactly $\floor{n/2} + 1$, where $n$ is the number of digits in base 2 held by $N$.
Since I'm interested in the odd integers (from $1$ to $\sqrt{77}$), this means that the last digit must be $1$. But if $4$ is the number of bits, then I get the wrong list in base 2: $$0001_2 = 1_{10}, 0011_2 = 3_{10}, 0101_2 = 5_{10}, 0111_2 = 7_{10}, 1001_2 = 9_{10}, ...$$
That's wrong because $9 > \sqrt{77}$. So the number must be less than $4$ digits in base 2. When I try $3$, I get the correct list
$$001_2 = 1_{10}, 011_2 = 3_{10}, 101_2 = 5_{10}, 111_2 = 7_{10},$$
but I haven't found an argument to convince myself I'm right. (Will this always work? Why?)
The number of digits of a number $N$ in base $b$ is $\log_b N+1$. So the number of digits of $\sqrt{N}$ in base $2$ is
$$\log_2 \sqrt{N}+1 = \frac{1}{2}\log_2 N +1.$$
The $\frac{1}{2}$ shows why it takes half as many digits for $\sqrt{N}$ as $N$.