5.0 Data Representation - Real Numbers

Tags: #◇Computer Science #◇CCICOMP #◇Problem Solving #◇M2

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◇Overview

Real numbers contain both whole and fractional components. A fixed-point binary representation is generally impractical as it lacks the necessary range to represent very large numbers and the precision to represent very small numbers. Thus, computer systems utilize floating-point notation, based on binary scientific notation, to dynamically balance range and precision. Defined heavily by the IEEE 754 standard, floating-point representations often trade accuracy for range; truncation of exact mantissa bits results in stored values being close approximations rather than perfectly exact numbers.

◇Important Formulas

Fractional Binary Evaluation

The positional weight of bits to the right of the radix point uses negative powers of $2$.
$\dots + b_{-1}2^{-1} + b_{-2}2^{-2} + b_{-3}2^{-3} + \dots$
Where $2^{-1} = 0.5$, $2^{-2} = 0.25$, $2^{-3} = 0.125$, etc.

IEEE 754 Floating-Point Formats

Floating point numbers are encoded into three parts: a Sign bit ($S$), an Exponent ($E'$), and a Mantissa ($M$).
The value evaluated is given by:
$\text{Value} = (-1)^S \times 1.M \times 2^{E' - \text{bias}}$ Single Precision (binary32): $32$ bits total. $1$ sign bit, $8$ exponent bits (bias $127$), and $23$ mantissa bits. Double Precision (binary64): $64$ bits total. $1$ sign bit, $11$ exponent bits (bias $1023$), and $52$ mantissa bits.

Range Limitations Overflow: Occurs when a value is too large and its normalized exponent exceeds the maximum possible bits (e.g., an exponent $> 127$ for single precision). Underflow: Occurs when a very small value normalizes to an exponent smaller than the minimum possible bound (e.g., $< -126$ for single precision). Truncation: When a value has more significant digits than the mantissa capacity, the remaining least-significant bits are simply discarded, causing precision loss (e.g., $1/3 = 0.3333333\dots$).

◇Worked Examples

1. Binary to Decimal (Fractional)

Convert $101.101_2$ to decimal. Whole part: $101_2 = 1(2^2) + 0(2^1) + 1(2^0) = 4 + 0 + 1 = 5$ Fractional part: $.101_2 = 1(2^{-1}) + 0(2^{-2}) + 1(2^{-3}) = 0.5 + 0 + 0.125 = 0.625$
Result: $5.625_{10}$.

2. Decimal to Binary Float Conversion (IEEE Single Precision)

Convert $-12.125_{10}$ to IEEE Single Precision floating point format.

1. Determine Sign: Negative, so $S = 1$.
2. Convert to Binary:
   Whole: $12_{10} = 1100_2$
   Fractional: $0.125_{10} = 0.001_2$
   Combined: $1100.001_2$
3. Normalize (Scientific Notation):
   Shift the radix point left by $3$ to get $1.100001_2 \times 2^3$.
   The implicit leading $1$ is dropped. The mantissa $M = 10000100000000000000000_2$ (padded with zeros to $23$ bits).
4. Determine Exponent:
   Base exponent is $3$.
   Add bias ($127$): $3 + 127 = 130$.
   Convert $130$ to $8$-bit binary: $10000010_2$.
5. Final 32-bit String:
   1 10000010 10000100000000000000000
   In hexadecimal, this is $\text{C}1420000\text{h}$.

3. Binary Float to Decimal

Evaluate the IEEE single precision string 0 01111101 11010000000000000000000 to decimal.

1. Sign: $0 \implies$ Positive.
2. Exponent: $01111101_2 = 125_{10}$.
   Unbias: $125 - 127 = -2$.
3. Mantissa: $M = 1101000\dots$
   Add implicit $1$: $1.1101_2$.
4. Evaluate Value:
   $1.1101_2 \times 2^{-2} = 0.011101_2$
   Convert to decimal: $0(2^{-1}) + 1(2^{-2}) + 1(2^{-3}) + 1(2^{-4}) + 0(2^{-5}) + 1(2^{-6})$
   $= 0.25 + 0.125 + 0.0625 + 0.015625 = 0.453125_{10}$.

◇References & Navigation

• Previous Topic: ◇Data Representation - Integers
• Next Topic: ◇Data Representation - Characters and Boolean
• Related Notes: ◇Floating-Point Arithmetic Operations, ◇IEEE 754 Floating-Point Representation (Single & Double Precision)