Number Systems

We count in decimal (base 10) because we have ten fingers. Computers use binary (base 2) because a transistor is either off (0) or on (1). Programmers also use octal (base 8) and hexadecimal (base 16) as compact ways to write binary.

#Place values

In base b, each position is worth a power of b, growing from right to left.

1011₂ = 1·2³ + 0·2² + 1·2¹ + 1·2⁰ = 8 + 0 + 2 + 1 = 11 2F₁₆ = 2·16¹ + 15·16⁰ = 32 + 15 = 47 157₈ = 1·8² + 5·8¹ + 7·8⁰ = 64 + 40 + 7 = 111

#Decimal to any base: repeated division

Divide by the base, record the remainder, repeat with the quotient until it reaches 0, then read the remainders bottom to top.

Convert 45 to binary:

Reading upwards: 45 = 101101₂.

#Try it

#Binary ↔ octal and hexadecimal: grouping

Because 8 = 2³ and 16 = 2⁴, each octal digit is exactly 3 bits and each hex digit exactly 4 bits. Group bits from the right:

1101 0110 1010₂ → D 6 A → D6A₁₆ 110 101 101 010₂ → 6 5 5 2 → 6552₈

That is why programmers love hex: one hex digit = 4 bits, two hex digits = one byte. Colours (#FF8800), memory addresses (0x7ffe…) and file signatures are all written in hex.

#Fractions in binary

Positions after the point are negative powers: 0.101₂ = 1/2 + 0/4 + 1/8 = 0.625. To convert a decimal fraction, multiply by 2 repeatedly and take the integer parts: 0.625 → 1.25 (1) → 0.5 (0) → 1.0 (1) → 0.101₂. Some fractions never terminate in binary — 0.1 is 0.000110011…₂ — which is why 0.1 + 0.2 != 0.3 in every language.

#Converting in code

#include <stdio.h>

void toBase(unsigned n, unsigned base) {
    const char *digits = "0123456789ABCDEF";
    char buf[40];
    int i = 0;
    if (n == 0) buf[i++] = '0';
    while (n > 0) { buf[i++] = digits[n % base]; n /= base; }   // remainders, last first
    while (i > 0) putchar(buf[--i]);                            // print them in reverse
}

int main(void) {
    unsigned n = 2026;
    printf("%u in decimal\n", n);
    printf("binary: "); toBase(n, 2); printf("\n");
    printf("octal:  %o\nhex:    %X\n", n, n);
    printf("base 5: "); toBase(n, 5); printf("\n");
    printf("0x7E2 = %d, 0b11111101010 = %d\n", 0x7E2, 0b11111101010);
    return 0;
}

2026 in decimal
binary: 11111101010
octal:  3752
hex:    7EA
base 5: 31101
0x7E2 = 2018, 0b11111101010 = 2026

n = 2026
print(bin(n), oct(n), hex(n))
print(int("11111101010", 2), int("3752", 8), int("7ea", 16), int("31101", 5))
print(f"{n:b} {n:o} {n:X} {n:#010b}")

0b11111101010 0o3752 0x7ea
2026 2026 2026 2026
11111101010 3752 7EA 0b11111101010