First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 33
and the largest is
x =(−1) 0 (1.11 ...1) 2 × 2 1023 =
(
1+ 1 2 + 1 2 2 + ... + 1
2 52 )
× 2 1023
=(2− 2 −52 )2 1023
≈ 0.18 × 10 309 .
During a calculation, if a number less than the smallest floating-point number is obtained,
then we obtain the underflow error. A number greater than the largest gives overflow
error.
Exercise 1.3-1: Consider the following toy model for a normalized floating-point representation
in base 2: x =(−1) s (1.a 2 a 3 ) 2 × 2 e where −1 ≤ e ≤ 1. Find all positive machine
numbers (there are 12 of them) that can be represented in this model. Convert the numbers
to base 10, and then carefully plot them on the number line, by hand, and comment on how
the numbers are spaced.
Representation of integers
In the previous section, we discussed representing real numbers in a computer. Here we will
give a brief discussion of representing integers. How does a computer represent an integer n?
As in real numbers, we start with writing n in base 2. We have 64 bits to represent its digits
and sign. As in the floating-point representation, we can allocate one bit for the sign, and
use the rest, 63 bits, for the digits. This approach has some disadvantages when we start
adding integers. Another approach, known as the two’s complement, is more commonly
used, including in Julia.
For an example, assume we have 8 bits in our computer. To represent 12 in two’s
complement (or any positive integer), we simply write it in its base 2 expansion: (00001100) 2 .
To represent −12, we do the following: flip all digits, replacing 1 by 0, and 0 by 1, and
then add 1 to the result. When we flip digits for 12, we get (11110011) 2 , and adding 1 (in
binary), gives (11110100) 2 . Therefore −12 is represented as (11110100) 2 in two’s complement
approach. It may seem mysterious to go through all this trouble to represent −12, until you
add the representations of 12 and −12,
(00001100) 2 + (11110100) 2 =(100000000) 2