Two's complement, overflow, logical shift
Two's complement (signed 8-bit): the MSB place value is . For a negative value, write the positive in binary, invert every bit, then add 1, e.g. . Binary addition: add column by column from the right, showing the carries (denary is not accepted). Overflow: a sum needing a 9th bit loses that carry. Logical shift: left multiplies by , right divides by ; bits shifted off the end are lost.
Binary base 2; hex is shorthand for binary
A computer's switches are either on (1) or off (0), so all data is stored in binary (base 2). Denary is base 10 (digits 0-9); binary is base 2 (digits 0, 1); hexadecimal is base 16 (digits 0-9 then A-F, where A-F = 10-15). A single binary digit is a bit; 4 bits make a nibble; 8 bits make a byte. One hexadecimal digit represents exactly one nibble, because . The leftmost bit is the most significant bit (MSB), the rightmost the least significant bit (LSB).
Convert via place values and nibbles
Binary to denary: write the place values and add the values where there is a 1. Denary to binary: subtract each place value from 128 down if it fits (write 1), else write 0. Binary to hexadecimal: split into 4-bit nibbles from the right and write one hex digit per nibble, e.g. . Hexadecimal to binary: replace each hex digit with its nibble, e.g. .
Drawn from real examiner reports.
Converting to the wrong target base
Given a binary number and asked for its hexadecimal equivalent, a common error is to convert it to denary instead — the denary value earns no marks. Use nibble grouping: split the binary into 4-bit nibbles from the right and write one hex digit per nibble, going straight from binary to hex with no denary step. Always check which target base the question asks for.
November 2023 Paper 1 hexadecimal question: some candidates tried to convert the binary to denary instead of hexadecimal.
Vague definition of hexadecimal
A vague answer such as 'another type of number' scores zero when a question asks what hexadecimal is. The mark-scheme form is: a base-16 number system in which each digit represents 4 bits (a nibble), using the digits 0-9 and the letters A-F for the values 10-15. Hexadecimal is used because it is a much shorter, easier-to-read way of writing long binary numbers.
Binary addition done via denary
Converting the two binary numbers to denary, adding them, then converting the result back is not binary addition and is not credited. You must add the binary numbers directly, column by column from the right, carrying a 1 into the next column when a column sums to 2 (10) or 3 (11), and show the carries in your working.
November 2023 Paper 1: candidates were reminded that converting to denary, adding and converting back is not a valid method as the question requires binary addition.
Forgetting to identify overflow
When adding two 8-bit numbers gives a result that needs a 9th bit, the carry out of the most significant bit is an overflow: it cannot be stored in an 8-bit register, so it is lost and the stored answer is wrong. Candidates who write only the 8-bit result without stating that overflow has occurred lose the overflow mark. Always check for a 9th bit after adding.
November 2023 Paper 1: weaker candidates gave an 8-bit answer without identifying the overflow.
Shifting the wrong direction
A logical shift left multiplies the value (it gets bigger); a logical shift right divides it (it gets smaller). Reading 'shift left' but moving the bits right is a frequent slip that gives the wrong answer. Fix the direction first: left = larger ( per place), right = smaller ( per place). Any bits shifted off the end of the register are lost.
November 2023 Paper 1: the most common incorrect shift answer came from candidates who shifted right instead of left.
Reading a signed byte as if MSB is +128
When converting a two's-complement byte back to denary, the most significant bit has a place value of , not . Treating the MSB as positive turns every negative number into a large positive one. Give the MSB and add the positive place values of the remaining 1-bits, e.g. .
Group nibbles for hexadecimal
For hexadecimal, never go via denary. Split the binary into 4-bit nibbles from the right and convert each directly, e.g. . Working straight from binary to hex avoids the wrong-target-base error entirely.
Two's complement: flip and add one
To write a negative number, put the positive value in 8 bits, flip every bit, then add 1. To read a two's-complement byte back to denary, give the MSB a place value of and add the positive place values of the remaining 1-bits.
Add, then check for overflow
Stack the two bytes and add from the least significant bit, carrying into the next column. Then check whether a 9th bit (overflow) was produced and state it explicitly — omitting the overflow is a common slip that loses the overflow mark.
Write the place-value row first
For any conversion, write the row first, then mark the bits underneath. Add the place values where there is a 1 (binary to denary), or subtract from the largest downward (denary to binary). Showing the table earns method marks even after a slip.
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