First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 31
represent numbers less than 1! That’s why we use the shifted expression e − 1023,
called the biased exponent, in the representation (1.2). Note that the bounds for
the biased exponent are −1023 ≤ e − 1023 ≤ 1024.
Here is a schema that illustrates how the physical bits of a computer correspond to the
representation above. Each cell in the table below, numbered 1 through 64, correspond to
the physical bits in the computer memory.
1 2 3 ... 12 13 ... 64
• The first bit is the sign bit: it stores the value for s, 0 or 1.
• The blue bits 2 through 12 store the exponent e (not e − 1023). Using 11 bits, one can
generate the integers from 0 to 2 11 − 1 = 2047. Here is how you get the smallest and
largest values for e:
e =(00...0) 2 =0
e =(11...1) 2 =2 0 +2 1 + ... +2 10 = 211 − 1
2 − 1 = 2047.
• The red bits, and there are 52 of them, store the digits a 2 through a 53 .
Example 9. Find the floating-point representation of 10.375.
Solution. You can check that 10 = (1010) 2 and 0.375 = (.011) 2 by computing
10 = 0 × 2 0 + 1 × 2 1 + 0 × 2 2 + 1 × 2 3
0.375 = 0 × 2 −1 + 1 × 2 −2 + 1 × 2 −3 .
Then
10.375 = (1010.011) 2 =(1.010011) 2 × 2 3
where (1.010011) 2 × 2 3 is the normalized floating-point representation of the number. Now
we rewrite this in terms of the representation (1.2):
10.375 = (−1) 0 (1.010011) 2 × 2 1026−1023 .
Since 1026 = (10000000010) 2 , the bit by bit representation is:
0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 ... 0