First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 32
Notice the first sign bit is 0 since the number is positive. The next 11 bits (in blue) represent
the exponent e = 1026, and the next group of red bits are the mantissa, filled with 0’s after
the last digit of the mantissa. In Julia, we can get this bit by bit representation by typing
bitstring(10.375):
In [1]: bitstring(10.375)
Out[1]: "0100000000100100110000000000000000000000000000000000000000000000"
Special cases: zero, infinity, NAN
In the floating-point arithmetic there are two zeros: +0.0 and −0.0, and they have special
representations. In the representation of zero, all exponent and mantissa bits are set to 0.
The sign bit is 0 for +0.0, and 1, for −0.0:
0.0 → 0 0 all zeros 0 0 all zeros 0
−0.0 → 1 0 all zeros 0 0 all zeros 0
When the exponent bits are set to zero, we have e =0and thus e − 1023 = −1023. This
arrangement, all exponents bits set to zero, is reserved for ±0.0 and subnormal numbers.
Subnormal numbers are an exception to our normalized floating-point representation, an
exception that is useful in various ways. For details see Goldberg [9].
Here is how plus and minus infinity is represented in the computer:
∞−→ 0 1 all ones 1 0 all zeros 0
−∞−→ 1 1 all ones 1 0 all zeros 0
When the exponent bits are set to one, we have e = 2047 and thus e − 1023 = 1024. This
arrangement is reserved for ±∞ as well as other special values such as NaN (not-a-number).
In conclusion, even though −1023 ≤ e − 1023 ≤ 1024 in (1.2), when it comes to representing
non-zero real numbers, we only have access to exponents in the following range:
−1022 ≤ e − 1023 ≤ 1023.
Therefore, the smallest positive real number that can be represented in a computer is
x =(−1) 0 (1.00 ...0) 2 × 2 −1022 =2 −1022 ≈ 0.2 × 10 −307