Binary — Edexcel GCSE Computer Science

Computers store and process everything as binary patterns of 0s and 1s, because each bit is a two-state signal that hardware can hold reliably. The same bit pattern can stand for different things depending on how it is interpreted: a number, a character of text, a sample of sound, a pixel colour in a graphic, or part of a program instruction. For example, the 8-bit pattern 01000001 can be the unsigned integer 65 or the character 'A', depending on context. The number of distinct patterns a group of n bits can form is 2ⁿ, so 4 bits give 2⁴ = 16 states, 8 bits give 2⁸ = 256 states, and 10 bits give 2¹⁰ = 1024 states. The maximum number of states is therefore found by raising 2 to the power of the bit length; the highest unsigned value is one less than that count.

Unsigned integers use every bit for magnitude, so an 8-bit unsigned pattern ranges from 00000000 (0) to 11111111 (255). Two's complement signed integers reserve the most significant bit as a sign indicator: 0 means non-negative and 1 means negative. To form the negative of a value, invert all bits and add 1; for example, +5 in 8 bits is 00000101, so −5 is 11111011. In 8-bit two's complement the range is −128 to +127, because 10000000 represents −128 and 01111111 represents +127. Addition and subtraction use the same binary circuits for both representations, and overflow occurs when a result falls outside the representable range.

This topic teaches conversion between denary and 8-bit binary in two ranges: unsigned 0 to 255 and signed -128 to +127. For unsigned conversion, use place values 128, 64, 32, 16, 8, 4, 2, 1. To convert denary to binary, subtract the largest place value that fits, set that bit to 1, and repeat with the remainder until it is zero; fill remaining bits with 0. For example, 200 = 128 + 64 + 8, giving 11001000. To convert binary to denary, add the place values where the bit is 1. For signed numbers, the leftmost bit has place value -128; the other bits keep 64, 32, 16, 8, 4, 2, 1. For example, 10000001 = -128 + 1 = -127. Always state whether the pattern is unsigned or signed, because the same 8 bits can mean different values.

This topic covers adding two positive binary patterns and applying logical and arithmetic shifts. To add, align the bits and work from right to left using 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 0 carry 1. For example, 00101100 (44) + 00010111 (23) = 01000011 (67). A logical left shift moves every bit one place left, puts 0 in the least significant bit, and multiplies the unsigned value by 2; a logical right shift moves every bit one place right, puts 0 in the most significant bit, and divides the unsigned value by 2, discarding any remainder. An arithmetic left shift works in the same way as a logical left shift for positive patterns. An arithmetic right shift moves every bit right but copies the original most significant bit into the new most significant bit, preserving the sign of a signed pattern.

Overflow happens when a value needs more bits than the storage location provides, so the extra most significant bits cannot be held and the stored result is wrong. With n bits you can represent 2ⁿ distinct patterns; for unsigned integers these are 0 to 2ⁿ − 1. An 8-bit register holds 0 to 255, so adding 200 + 100 gives a true value of 300, which needs 9 bits (1 0010 1100₂); the register keeps only the low 8 bits, 0010 1100₂ = 44, and the carry out of the most significant bit is lost. A 4-bit register holds 0 to 15, so 12 + 7 = 19 overflows and stores 3. Overflow is therefore about the number of bits available, not about a calculation being mathematically impossible.

Hexadecimal is base 16, using 0–9 then A–F for 10–15. It is used because one hex digit maps exactly to four binary bits, so long binary strings become shorter and less error-prone to read and write; 1011 1110₂ is BE₁₆. This makes hex convenient for memory addresses, colour codes and machine-level data. To convert binary to hex, group the bits into nibbles of four from the right, pad the leftmost group with leading zeros if needed, then replace each nibble with its hex digit: 1101 0110₂ becomes D6₁₆. To convert hex to binary, replace each hex digit with its four-bit pattern: 3F₁₆ becomes 0011 1111₂.