MGMT 1000 Lecture 8: MGMT 1000 Lecture 8 Notes
MGMT 1000 Lecture 8 Notes – Performing arithmetic in different number bases
Introduction
• To do so, we break up the total number of bits into a sum that consists of values where
the range is easily figured out.
• The overall range is equal to the product of the sub ranges for each value.
• This method is best seen with examples.
• For example, if you need to know the range for 18 bits, you would break up the number
18 into the sum of 10 and 8, and then multiply the range for 10 bits to that for 8 bits.
• Since the range for 10 bits is approximately 1 K (1024, actually) and 8 bits is 256, the
range for 18 bits is approximately 256 K.
• Similarly, the range for 32 bits would be (10-bit range)×(10-bit range)×(10-bit range)×(2-
bit range)=1K×1K×1K×4=4 gigabytes.
• This technique becomes easy with a bit of practice.
• Notice that it takes 18 bits to represent a little more than five decimal digits.
• In general, approximately 3.3 bits are required for each equivalent decimal digit.
• This is true because 23.3 are approximately equal to 10.
• Next, we consider simple arithmetic operations in various number bases.
• Let us begin by looking at the simple base 10 addition table shown
• We add two numbers by finding one in the row and the other in the column.
• The table entry at the intersection is the result.
• For example, we have used the table to demonstrate that the sum of 3 and 6 is 9.
• Note that the extra digit sometimes required becomes a carry that gets added into the
next left column during the addition process.
• More fundamentally, we are interested in how the addition table is actually created.
• Each column (or row) represents an increase of 1 from the previous column (or row),
which is equivalent to counting.
• Thus, starting from the leftmost column in the table, it is only necessary to count up 1 to
find the next value.
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