janmr blog

Let us consider a common way to represent non-negative integers. An integer $u \geq 0$ will be represented in radix $b \geq 2$ using the notation

$$u = (u_{n-1} \ldots u_1 u_0)_b = \sum_{k=0}^{n-1} u_k b^k, \quad 0 \leq u_k \leq b-1.$$

We will call $u$ an $n$-digit number and $u_0, u_1, \ldots$ its digits. Zero will be represented with no digits, $0 = ()_b$. Observe that

$$u \leq ((b-1) \ldots (b-1) (b-1))_b = \sum_{k=0}^{n-1} (b-1) b^k = b^n - 1$$

for any $n \geq 0$.

Unless stated otherwise we will always have that the most-significant digit is non-zero, that is, $u_{n-1} \neq 0$ for $n \geq 1$. This assumption has some important consequences. First, that

$$u \geq (1 0 \ldots 0)_b = b^{n-1}$$

for any $n \geq 1$. Secondly, that each non-negative integer has a unique representation, that is, to each number $u \geq 0$ and radix $b \geq 2$ corresponds exactly one $n \geq 0$ and digits $u_0, u_1, \ldots, u_{n-1}$ such that $u = (u_{n-1} \ldots u_1 u_0)_b$. Thirdly, that

$$b^{n-1} \leq u < b^n \quad \Leftrightarrow \quad n-1 \leq \log_b(u) < n \quad \Leftrightarrow \quad \lfloor \log_b(u) \rfloor = n-1.$$

This last relation can be quite useful since the number of needed digits can be found, given $u$ and $b$. For instance, the fact that $\lfloor \log_2(1317803400) \rfloor + 1 = 31$ means that the number 1317803400 can be represented using 31 binary digits.