Binary Addition, Shifts & Two's Complement

Syllabus & Goals 3 min

Cambridge 1.1 · Number systems Paper 1 · Computer Systems

By the end of this lesson you can:

1. Add two positive 8-bit binary numbers, showing carries, and identify an overflow error.
2. Carry out a logical shift left or right and state its effect (× 2 or ÷ 2 per place) and when it goes wrong.
3. Use two's complement to convert positive and negative 8-bit numbers in both directions.

Textbook: Chapter 1, §1.1.4–1.1.6 (pp. 15–25)

Recap / Warm-Up 5 min

You can now convert between binary, denary and hex. Next you do arithmetic on binary numbers.

Quick starter

In denary, what is the largest number you can show with three digits? What happens if you add 1 to it?

Key Concept 14 min

1 · Binary addition

Add column by column from the right, just as in denary. In denary you carry when a column passes 9; in binary you carry when a column passes 1.

2 · Overflow

An 8-bit register holds 0 to 255 (2⁸ − 1). If an addition produces a 9th bit, the true answer is too big to store. The 9th bit is lost, so the register keeps a value 256 too small. This is an overflow error.

More bits allow bigger numbers: a 16-bit register holds up to 65 535 (2¹⁶ − 1) and a 32-bit register up to 4 294 967 295 (2³² − 1).

3 · Logical binary shifts

A logical shift moves every bit left or right by a number of places. Bits that fall off the end are lost. Empty positions are always filled with 0.

• Shift left 1 place = × 2. Left 2 = × 4, left 3 = × 8 (× 2ⁿ for n places).
• Shift right 1 place = ÷ 2. Right 2 = ÷ 4, right 3 = ÷ 8 (÷ 2ⁿ for n places).

There is a limit. If a 1 is pushed out of the register, the result is wrong. After enough shifts the register holds only 0s — as if 112 × 32 were 0.

4 · Two's complement — negative numbers

So far every number has been positive. To store negative integers, change one heading: the left-most bit (the most significant bit) is worth −128 instead of +128. All other headings stay the same.

This number is −128 + 16 + 2 + 1 = −109. The MSB tells you the sign: 1 = negative, 0 = positive. The range is −128 (10000000) to +127 (01111111).

Two ways to write a negative number such as −67:

• Method 1 — build it from −128. −67 = −128 + 61, and 61 = 32 + 16 + 8 + 4 + 1. So put 1s under −128, 32, 16, 8, 4 and 1: 10111101.
• Method 2 — invert then add 1. Write +67 (01000011), invert every bit (10111100), then add 1: 10111101.

Worked Example 12 min

Worked example 1 · add 01011011 + 00101110

1. Column 1 (right): 1 + 0 = 1, no carry.
2. Column 2: 1 + 1 = 0 carry 1.2 in binary is 10 — write the 0, carry the 1.
3. Column 3: 0 + 1 + 1 (carry) = 0 carry 1. Column 4: 1 + 1 + 1 = 1 carry 1.
4. Column 5: 1 + 0 + 1 = 0 carry 1. Column 6: 0 + 1 + 1 = 0 carry 1.
5. Column 7: 1 + 0 + 1 = 0 carry 1. Column 8: 0 + 0 + 1 = 1, no carry out.
6. Answer: 10001001 = 137. Check in denary: 91 + 46 = 137 ✓.converting both numbers to denary is a free check on every addition.

Worked example 2 · add 11001010 + 01011001 (overflow)

1. Add right to left with carries, exactly as before. The last column (128s) gives 1 + 0 + 1 = 0 carry 1.
2. That final carry has no column to go into. It is the 9th bit.
3. The 8 bits kept are 00100011 = 35. But 202 + 89 = 291.291 > 255, the largest 8-bit value.
4. Conclusion: an overflow error has occurred; the stored result is wrong by 256.

Worked example 3 · shift 10110000 right

Worked example 4 · write −94 in 8-bit two's complement

1. Write +94 in binary: 94 = 64 + 16 + 8 + 4 + 2 → 01011110.
2. Invert every bit (0 ↔ 1): 10100001.inverting on its own gives −95; the +1 corrects it.
3. Add 1: 10100001 + 1 = 10100010.
4. Check with the headings: −128 + 32 + 2 = −94 ✓. Method 1 agrees: −128 + 34 = −94, and 34 = 32 + 2.

Worked example 5 · two's complement 11011010 → denary

1. The MSB is 1, so the number is negative.
2. Add the headings: −128 + 64 + 16 + 8 + 2.
3. −128 + 64 = −64; −64 + 16 = −48; −48 + 8 = −40; −40 + 2 = −38.

Shifts as code

// A shift left multiplies by 2; a shift right divides by 2
DECLARE Value, Result : INTEGER
Value ← 27
Result ← Value * 4
OUTPUT "Shift left 2 places gives ", Result
Result ← Value DIV 2
OUTPUT "Shift right 1 place gives ", Result

value = 27
print("Left 2  =", (value << 2) & 255)
print("Left 4  =", (value << 4) & 255)
print("Right 1 =", value >> 1)

Left 2  = 108
Left 4  = 176
Right 1 = 13

Try It Yourself 12 min

📝 Exam Practice 10 min

Recap & Key Terms 3 min

Add binary right to left, carrying whenever a column passes 1; a 9th bit means overflow. A logical shift left multiplies by 2 per place and right divides by 2 per place; losing a 1 makes the answer wrong. Two's complement gives the MSB the value −128: invert then add 1 to negate.

Overflow error

The result of a calculation that is too large for the computer's allocated word size (e.g. a 9th bit in 8-bit addition).

Logical shift

Shifting bits left or right in a register; bits shifted out are lost and empty positions are filled with 0s.

Most significant bit (MSB)

The left-most bit; in two's complement it has the value −128 and shows the sign.

Least significant bit (LSB)

The right-most bit, with place value 1.

Two's complement

A method of representing negative numbers in binary; the MSB of an 8-bit number is worth −128.

Homework 1 min

Task (≤ 15 min): (a) Add 01111000 + 10011010 and comment on the result. (b) Convert −20 to 8-bit two's complement, and check your answer. [5]