What does the number and logic part of OCR section 1.4 cover?

It covers number systems (binary, denary and hexadecimal conversion, sign and magnitude and two's complement, binary addition and subtraction, fixed-point fractions and bitwise masks), floating-point representation with a mantissa and exponent and normalisation, Boolean algebra (the logic gates, truth tables, the laws of Boolean algebra, De Morgan's laws and Karnaugh maps), and logic circuits including the half adder, full adder and D-type flip-flop.

How do you represent a negative number in two's complement?

Write the positive value in binary, invert every bit, then add one. For example, in 8 bits, plus 5 is 0000 0101; inverting gives 1111 1010; adding one gives 1111 1011, which is minus 5. Two's complement has a single zero and an 8-bit range of minus 128 to plus 127, and it is preferred because ordinary binary addition and subtraction work directly on it.

What does normalising a floating-point number achieve?

Normalising shifts the mantissa so a positive value starts 0.1 and a negative value starts 1.0, with the exponent adjusted to keep the value unchanged. This removes leading zeros, so the maximum precision is kept for the available mantissa bits, and it gives each number a single unique representation. More mantissa bits increase precision, while more exponent bits increase range.

What are De Morgan's laws?

De Morgan's laws handle a NOT over a whole bracket. The complement of (A OR B) equals (NOT A) AND (NOT B), and the complement of (A AND B) equals (NOT A) OR (NOT B). In words, break the bar over the bracket and swap AND for OR. They are essential for simplifying Boolean expressions and for building circuits from only NAND or only NOR gates.

How should I revise the number and logic part of section 1.4?

Drill conversions and two's complement arithmetic until automatic, because no calculator is allowed. Practise evaluating and normalising floating-point numbers, simplifying Boolean expressions with the named laws and De Morgan's, and reading Karnaugh maps. Learn the half adder and full adder sum and carry expressions and how a D-type flip-flop uses a clock to store a bit.

EnglandComputer Science

OCR A-Level Computer Science Boolean algebra and logic: binary, floating point and logic circuits made exam-ready

A deep-dive OCR H446 guide to the data-representation and logic part of Component 01 section 1.4. Covers binary, hexadecimal and two's complement with binary arithmetic, floating-point representation and normalisation, Boolean algebra with logic gates, De Morgan's laws and Karnaugh maps, and logic circuits including half adders, full adders and D-type flip-flops.

Generated by Claude Opus 4.815 min readH446 1.4Updated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this section actually demands
Number representation
Boolean algebra and logic circuits
How this section is examined
Check your knowledge

What this section actually demands

The data-representation and logic part of section 1.4 is the most calculation-heavy of Component 01, and it is sat without a calculator. OCR rewards accurate, shown working: conversions, two's complement arithmetic, floating-point evaluation and normalisation, Boolean simplification with named laws, and gate-level circuits. Every answer should show the method, not just the result.

This guide walks through the topics in order and sets out the exam patterns OCR repeats. Each topic has a matching dot-point page with practice; this overview ties them together.

Number representation

Number systems covers binary, denary and hexadecimal conversion, sign and magnitude versus two's complement, binary addition and subtraction (by adding the two's complement), fixed-point fractions and bitwise masks. The recurring skill is representing a negative number in two's complement (invert and add 1) and performing a binary addition, discarding the final carry in fixed-width arithmetic.

Floating-point representation and normalisation covers storing a real number as a two's complement mantissa and exponent, shifting the binary point by the exponent, normalising the mantissa, and the range-versus-precision trade-off. The classic questions are evaluating a stored floating-point number and normalising one.

Boolean algebra and logic circuits

Boolean algebra and logic gates covers the six gates and their truth tables, building truth tables for expressions, and simplifying expressions with the laws of Boolean algebra, De Morgan's laws and Karnaugh maps. The recurring skill is simplifying an expression while naming each law and verifying with a truth table.

Logic circuits, adders and flip-flops covers the half adder ( $S = A \oplus B$ , $C = A \cdot B$ ), the full adder with its carry-in for cascading, and the D-type flip-flop as a clocked single-bit store that builds registers and counters.

How this section is examined

A typical OCR profile for the number and logic part of section 1.4:

Conversions and arithmetic. Binary/hex/denary conversion; two's complement negation; binary addition with carry.
Floating point. Evaluate a stored value; normalise a mantissa and adjust the exponent; explain range versus precision.
Boolean simplification. Simplify with named laws and De Morgan's; complete a truth table; group a Karnaugh map.
Circuits. Half adder and full adder sum/carry expressions; how a D-type flip-flop stores a bit using the clock.

Check your knowledge

A mix of recall and calculation questions covering the section. Attempt them under timed conditions, then check against the solutions.

Convert $1100\,1010$ to denary. (1 mark)
Represent $-12$ in 8-bit two's complement. (2 marks)
Convert the denary number $173$ to hexadecimal. (1 mark)
State the leading bits of a normalised positive mantissa. (1 mark)
Apply De Morgan's law to $\overline{A + B}$ . (1 mark)
Give the carry expression for a half adder. (1 mark)

Solutions

Six questions covering binary and hexadecimal conversion, two's complement, floating-point normalisation, De Morgan's law, and the half adder.

Step 1: Q1 - Convert $1100\,1010$ to denary

Each bit position represents a power of 2, with the leftmost bit being $2^7 = 128$ . Add the values at positions where a $1$ appears: the set bits here represent $128$ , $64$ , $8$ , and $2$ .

128 + 64 + 8 + 2 = 202

Final answer: $202$ .

Step 2: Q2 - Represent $-12$ in 8-bit two's complement

To form the two's complement negative, first write the positive value in 8-bit binary, then invert every bit, then add one. This procedure works because adding a number to its two's complement always gives zero in fixed-width arithmetic.

+12 = 0000\,1100

\text{Invert:}\ 1111\,0011

\text{Add 1:}\ 1111\,0100

Final answer: $1111\,0100$ .

Step 3: Q3 - Convert $173$ to hexadecimal

Dividing by 16 gives the hexadecimal digits from least significant to most significant. $173 \div 16 = 10$ remainder $13$ ; remainder $13$ is the digit D and quotient $10$ is the digit A.

173 = 10 \times 16 + 13 \to \text{AD}

Final answer: $\text{AD}_{16}$ .

Step 4: Q4 - State the leading bits of a normalised positive mantissa

Normalisation ensures the mantissa carries the maximum number of significant bits by removing any leading redundancy. For a positive value stored in two's complement, the sign bit (to the left of the binary point) is $0$ , and the first bit after the point must be $1$ to show no leading zeros have been left in.

Final answer: A normalised positive mantissa starts $0.1\ldots$ (the first bit after the point is $1$ ).

Step 5: Q5 - Apply De Morgan's law to $\overline{A + B}$

De Morgan's law for NOR (NOT OR) says: break the bar over the bracket and swap OR for AND. The complement of (A OR B) becomes (NOT A) AND (NOT B). This rule is the foundation for building circuits from universal NOR gates alone.

\overline{A + B} = \overline{A} \cdot \overline{B}

Final answer: $\overline{A} \cdot \overline{B}$ .

Step 6: Q6 - Give the carry expression for a half adder

A half adder adds two single bits. A carry of $1$ is produced only when both inputs are $1$ , which is exactly the condition described by the AND operation. The sum bit, by contrast, uses XOR because XOR gives $1$ when the inputs differ (one bit set, no carry).

Final answer: $C = A \cdot B$ (the carry is the AND of the two inputs; the sum is $A \oplus B$ ).

Sources & how we know this

OCR AS and A Level Computer Science (H046, H446) specification — OCR (2015)