What does $23_4$ mean?

Question

I just saw this on a mathematical clock for $11$, i.e $23_4=11$:

$\qquad \qquad \qquad \qquad \qquad$

I guess it is some notation from algebra. But since algebra was never my favorite field of maths, I don't know this notation. Any explanations are welcome ;-))! Thanks

Answer 1

This denotes the number $11$ in base $4$. In everyday life, we write our numbers in base $10$.

$23_4$ is to be read as: $$2\cdot 4 + 3.$$ In general, $$(a_n...a_0) _ g = \sum_{i=0}^n a_i g^i = a_n g^n + a_{n-1}g^{n-1} + ... + a_1 g + a_0,$$ where the $a_i$ are chosen to lie in $\{0,...,g-1\}$.

EDIT: I have edited this post to write $2\cdot 4 +3$ rather than $3+2\cdot 4$. However, I still think that it is easier to decipher a (long) number such as $(2010221021)_3$ from right to left, simply by increasing the powers of $3$, rather than first checking that the highest occuring power of $3$ is $3^9$ and then going from left to right.

Answer 2

The base or radix of a number denotes how many unique digits are used by the numeral system that is representing the given number. Usually the radix is written as a subscript, as in your example $23_4$, where the radix is $4$.

Also note that when the radix is omitted, it is usually assumed to be $10$. So the number $23_4$ is equal to $11_{10}$ or simply $11$. Here is how we can convert $23_4$ to $11$ $$ 23_4=\left(2\cdot 4^1\right)+\left(3\cdot 4^0\right) =8+3=11 $$

In general, $$ \left(\alpha_{n-1}\dots\alpha_1\alpha_0\right)_{\beta}$$ $$= \left(\alpha_{n-1}\cdot\beta^{n-1}\right)+\cdots+ \left(\alpha_1\cdot\beta^1\right)+ \left(\alpha_0\cdot\beta^0\right) $$

Answer 3

$4$ is the base, $(23)_4$ means $2 \cdot 4^1 + 3 \cdot 4^0 = 11$. Similarly, if $10$ is the base, then $(23)_{10}$ means $2 \cdot 10^1 + 3 \cdot 10^0 = 23$.