number of digits in the square root of a perfect square

Question

number of digits in the square root of a perfect square with n digits is ceil(n/2). We can get convinced of this result by following the division method. My question is, is there analytical proof of this result?

Answer 1

The number of digits of a number $n$ in base $b$ is given by

$$\lfloor\log_b (n)\rfloor + 1$$

Now we have

\begin{align*} \lfloor \log_b(\sqrt{n}) \rfloor + 1 &= \left\lfloor\frac{1}{2} \log_b(n) \right\rfloor + 1 \\ &= \begin{cases} \frac{1}{2}\lfloor\log_b(n)\rfloor +1 & \text{if } \lfloor\log_b(n)\rfloor \text{ is even} \\ \frac{1}{2}\lfloor\log_b(n)\rfloor + \frac{1}{2} & \text{if } \lfloor\log_b(n)\rfloor \text{ is odd} \end{cases} \\ &= \left\lceil \frac{1}{2} \left(\lfloor \log_b(n) \rfloor + 1 \right)\right\rceil. \end{align*}

Answer 2

An $m$-digit integer is in the interval $[10^{m-1},10^m)$ and has a square in the interval $[10^{2m-2},10^{2m})$ with $2m-1$ or $2m$ digits.

So to reverse, divide the number of digits in the square by two to give $\frac{2m-1}{2}= m-\frac12$ or $\frac{2m}{2}= m$ and round up to give the original integer $m$.

Answer 3

Claim: For a positive integer $a$, suppose $a$ has $m$ digits and $a^2$ has $n$ digits. Then $m = \left\lceil \frac{n}{2} \right\rceil$.

Proof: We have \begin{align*} 10^{m-1} \le a < 10^m \tag{1} \\ 10^{n-1} \le a^2 < 10^n \tag{2} \end{align*} Squaring (1), we get $$ 10^{2m-2} \le a^2 < 10^{2m} $$ Combining with (2), it follows that $$ 10^{n-1} < 10^{2m} \implies m > \frac{n-1}{2} \implies m \ge \frac{n}{2} $$ and $$ 10^{2m-2} < 10^n \implies m < \frac{n+2}{2} \implies m < \frac{n}{2} + 1. $$ Therefore $m$ is an integer greater than or equal to $\frac{n}{2}$ and less than $\frac{n}{2} + 1$. Thus by definition of the ceiling function, we must have $$ m = \left\lceil \frac{n}{2} \right\rceil. $$