First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 30
where
s → sign of x = ±1
e → exponent, with bounds L ≤ e ≤ U
(.a 1 ...a t ) β = a 1
β + a 2
β + ... + a t
; the mantissa
2 βt β → base
t → number of digits; the precision.
In the floating-point representation (1.1), if we specify e in such a way that a 1 ≠0, then
the representation will be unique. This is called the normalized floating-point representation.
For example if β =10, in the normalized floating-point we would write 0.012 as
0.12 × 10 −1 , instead of choices like 0.012 × 10 0 or 0.0012 × 10.
In most computers today, the base is β = 2. Bases 8 and 16 were used in old IBM
mainframes in the past. Some handheld calculators use base 10. An interesting historical
example is a short-lived computer named Setun developed at Moscow State University which
used base 3.
There are several choices to make in the general floating-point model (1.1) for the values
of s, β, t, e. The IEEE 64-bit floating-point representation is the specific model used in most
computers today:
x =(−1) s (1.a 2 a 3 ...a 53 ) 2 2 e−1023 . (1.2)
Some comments:
• Notice how s appears in different forms in equations (1.1) and(1.2). In (1.2), s is
either 0 or 1. If s =0,thenx is positive. If s =1,xis negative.
• Since β =2, in the normalized floating-point representation of x the first (nonzero)
digit after the decimal point has to be 1. Then we do not have to store this number.
That’s why we write x as a decimal number starting at 1 in (1.2). Even though precision
is t =52, we are able to access up to the 53rd digit a 53 .
• The bounds for the exponent are: 0 ≤ e ≤ 2047. We will discuss where 2047 comes
from shortly. But first, let’s discuss why we have e − 1023 as the exponent in the
representation (1.2), as opposed to simply e (which we had in the representation (1.1)).
If the smallest exponent possible was e =0, then the smallest positive number the
computer can generate would be (1.00...0) 2 =1: certainly we need the computer to