Floating Point Numbers

In fixed point representation, the binary point (the fractional separator, like a decimal point) is in a fixed position. Bits to the left of the binary point represent integer values; bits to the right represent fractions.

For bit position $k$ relative to the binary point: value = $2^k$ (k can be negative for fractional positions).

Example — 4 integer bits + 4 fractional bits:

$\underset{13}{\underbrace{1101}}.\underset{0.375}{\underbrace{0110}}$

Bit	$2^3$	$2^2$	$2^1$	$2^0$	$.$	$2^{-1}$	$2^{-2}$	$2^{-3}$	$2^{-4}$
	8	4	2	1		0.5	0.25	0.125	0.0625
Digit	1	1	0	1		0	1	1	0
Value	8	4	0	1		0	0.25	0.125	0

$8+4+1+0.25+0.125=13.375_{10}$

Convert $2.75_{10}$ to fixed point (4 integer + 4 fractional bits):

• Integer part: $2=0010$
• Fractional part: $0.75$: multiply by 2 repeatedly — $0.75\times 2=1.5$ → 1; $0.5\times 2=1.0$ → 1 → $0.11_2$
• Result: $0010.1100$

Limitation: the binary point position is fixed — range and precision cannot change. The same format cannot represent both very large and very small numbers.

Floating point stores a number as:

$\text{value}=\text{mantissa}\times 2^{\text{exponent}}$

Both mantissa and exponent are stored as two's complement integers. This allows both large numbers (large positive exponent) and small numbers (large negative exponent).

AQA uses a simplified model — not the IEEE 754 standard. The IEEE standard is not required.

AQA format: the binary point in the mantissa is assumed to be after the sign bit (MSB). So a 4-bit mantissa represents a value in the range $[-1,+1)$.

Examples (4-bit mantissa, 4-bit exponent):

Mantissa bits	Mantissa value	Exponent bits	Exponent value	Represented value
0110	$+0.110_2=+0.75$	0010	$+2$	$0.75\times 4=3.0$
0111	$+0.111_2=+0.875$	0010	$+2$	$0.875\times 4=3.5$
1010	$-0.110_2=-0.75$	0000	$0$	$-0.75\times 1=-0.75$

Reading a mantissa: the sign bit is the MSB; the binary point follows it.

• $0110$: sign = 0 (positive); remaining bits $110_2$ after the binary point = $0.5+0.25=0.75$. Mantissa value = $+0.75$.
• $1010$: sign = 1 (negative). Apply two's complement: invert $1010$ → $0101$, add 1 → $0110=0.75$. So mantissa = $-0.75$.

Steps to convert a decimal number to floating point:

1. Convert the integer and fractional parts to binary
2. Write in the form $0.1\dots \times 2^e$ (normalised — see slide 5)
3. Fit the mantissa and exponent into the allocated bits; apply two's complement for negatives

Worked example — represent $3.5$ using a 4-bit mantissa and 4-bit exponent:

Step 1: $3.5_{10}=11.1_2$

Step 2: normalise → $0.111_2\times 2^2$ (shift right by 2 places; exponent = 2)

Step 3:

• Mantissa: $0.111_2$ → stored as $0111$ (sign=0, fractional bits=111)
• Exponent: $+2$ → stored as $0010$ (two's complement positive)

Floating point representation: mantissa = $0111$, exponent = $0010$

Verify: $0.111_2\times 2^2=0.111_2\times 4=(0.5+0.25+0.125)\times 4=0.875\times 4=3.5$ ✓

Given mantissa and exponent in bits, recover the decimal value:

1. Convert mantissa bits to a two's complement fractional value (binary point after MSB)
2. Convert exponent bits to a two's complement integer
3. Compute mantissa × $2^{\text{exponent}}$

Worked example — mantissa = $1100$, exponent = $1111$ (both 4-bit two's complement):

Mantissa $1100$: negative (MSB=1). Invert: $0011$, add 1: $0100=0.100_2=0.5$. So mantissa = $-0.5$.

Exponent $1111$: $-1$ in two's complement. ($-8+4+2+1=-1$, or invert+1: $0000+1=0001=1$ → exponent = $-1$.)

Value $=-0.5\times 2^{-1}=-0.5\times 0.5=-0.25_{10}$

Normalised form ensures maximum precision by using all available mantissa bits effectively. An un-normalised number has leading redundant zeros (positive) or ones (negative) in the mantissa.

Normalisation rules:

• Positive normalised mantissa starts with $0.1\dots$ (sign bit 0, next bit 1)
• Negative normalised mantissa starts with $1.0\dots$ (sign bit 1, next bit 0)

Un-normalised numbers can be normalised by shifting the mantissa and adjusting the exponent:

Example — normalise mantissa $0011$, exponent $0011$ (4-bit each):

Mantissa $0.011_2$: not normalised (starts with $0.0$, not $0.1$). Shift mantissa left by 1: $0.110_2$. Decrement exponent by 1: $3-1=2$. Stored as: mantissa = $0110$, exponent = $0010$. ✓

Example 2 — normalise mantissa $1110$, exponent $0010$:

Mantissa $1.110_2$: not normalised (starts with $1.1$, not $1.0$). Shift mantissa left by 1: $1.100_2$. Decrement exponent by 1: $2-1=1$. Stored as: mantissa = $1100$, exponent = $0001$. ✓

	Fixed point	Floating point
Binary point	Fixed position	Moves (via exponent)
Range	Limited	Much larger
Precision	Constant absolute precision	Relative precision (proportional to magnitude)
Speed	Faster (hardware simpler)	Slower (hardware more complex)
Best for	Known, bounded values	Scientific computing; very large or very small values

Precision vs range trade-off: adding more bits to the mantissa increases precision; adding more bits to the exponent increases range. With a fixed total bit count, more precision means less range and vice versa.

1. Placing the binary point in the wrong position for the mantissa

In the AQA model, the binary point is after the sign bit (MSB). A 4-bit mantissa $0110$ represents $0.110_2=0.75$, not $0110_2=6$.

2. Forgetting to adjust the exponent when normalising

When you shift the mantissa left by one position, you must decrement the exponent by one (and vice versa). Shifting without adjusting the exponent changes the value.

3. Confusing normalised and un-normalised

A normalised positive mantissa starts $0.1\dots$; a normalised negative starts $1.0\dots$. Starting $0.0\dots$ or $1.1\dots$ means the number is un-normalised and precision is being wasted.

4. Using IEEE 754 format

AQA uses a simplified two's complement model, not IEEE 754. Do not apply bias to the exponent or use the IEEE implicit leading 1 convention — the AQA format is different.