First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 34
and realize that the first 8 digits of the sum (from right to left), which is what the computer
can only represent (ignoring the red digit 1), is (00000000) 2 . So just like 12 + (−12) = 0 in
base 10, the sum of the representations of these numbers is also 0.
We can repeat these calculations with 64-bits, using Julia. The function bitstring outputs
the digits of an integer, using two’s complement for negative numbers:
In [1]: bitstring(12)
Out[1]: "0000000000000000000000000000000000000000000000000000000000001100"
In [2]: bitstring(-12)
Out[2]: "1111111111111111111111111111111111111111111111111111111111110100"
You can verify that the sum of these representations is 0, when truncated to 64-digits.
Here is another example illustrating the advantages of two’s complement. Consider −3
and 5, with representations,
−3 = (11111101) 2 and 5 = (00000101) 2 .
The sum of −3 and 5 is 2; what about the binary sum of their representations? We have
(11111101) 2 + (00000101) 2 =(100000010) 2
and if we ignore the ninth bit in red, the result is (10) 2 , which is indeed 2. Notice that if
we followed the same approach used in the floating-point representation and allocated the
leftmost bit to the sign of the integer, we would not have had this property.
We will not discuss integer representations and integer arithmetic any further. However
one useful fact to keep in mind is the following: in two’s complement, using 64
bits, one can represent integers between −2 63 = −9223372036854775808 and 2 63 − 1 =
9223372036854775807. Any integer below or above yields underflow or overflow error.
n
Example 10. From Calculus, we know that lim n
n−>∞ = ∞. Therefore, computing nn for
n! n!
large n will cause overflow at some point. Here is a Julia code for this calculation:
In [1]: f(n)=n^n/factorial(n)
Out[1]: f (generic function with 1 method)