Floating Point Form of Real Numbers

Topics: Numerical Analysis - Rounding Error - Normalised Form of Real Numbers

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The floating point form of a number in its normalised form can be obtained by terminating the fractional part of the number at $k$ digits. There are two ways to do this:

• Truncate: the digits of the number up to the position that the word size allows (i.e. $k$)
• Round:
  • If $d^{k+1}\ge 5$, we add $1$ to $d^k$, ”rounding up”
  • If $d^{k+1}<5$, we ignore the digits from $k+1$ onward, ”rounding down”

$k$ is dependant on the size of the computer word.

Bit Assignation and Range §

Floating point numbers use specific bits of the computer word for specific purposes. Normally, there’s 1 bit dedicated to the sign, several bits for the exponent, and some more for the mantissa (the fractional part of the number).

Range and Exponent §

The size of the word defines the range of the exponent of the floating point number:

Total digits	Range of the exponent	Exponent
$N=p+q+1$	$-(2^{p-1}-1)$ to $2^{p-1}$	$c-(2^{p-1}-1)$

…where:

• $p$: number of digits for the exponent
• $q$: number of digits for the mantissa
• $c$: decimal form of the number in the exponent part of the word

Mantissa Bits

Please note that the mantissa section of the word must always have $1$ in its first position (left to right). The mantissa section cannot be $0$ in total.

Final Number §

The final number as represented on the computer is given by:

$$(-1{)}^s{B}^e(1+f)$$

…where:

• $s$: the bit corresponding to the sign
• $e$: the exponent (i.e. $c-(2^{p-1}-1)$)
• $f$: the decimal number in the mantissa section
• $B$: the base for the numbering system

Example

For instance, take the following number as represented in bits:

$s$	Exponent	Mantissa
1	1 0 0 1 1	1 0 0 1 0 1 0 0 0 0

In this case, we have that:

• $s=1$
• $e=19-15=4$
• $f=0.1001010000_2=0.578125_{10}$

Thus, the number is:

$$(-1{)}^1{2}^4(1+0.578125)$$

$$=-16(1.578125)=-25.25$$