GNED 1101 Chapter Notes - Chapter 2.2: 5,6,7,8, Hexadecimal

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2.2 Number Bases in Positional Systems
Changing Numerals in Bases Other then Ten to Base Ten
The base of a positional numeration system refers to the number of individual digit
symbols that can be used in that system as well as to the number whose powers define
the place value
o Example, the number 1001two
Is not read as one thousand one because that is of the base 10 system.
Going forward, when a numeral does not have a subscript attached to it
it is assumed to be of the base ten.
However. Since the two is subscripted to the base two, the number is
read as one zero zero one base two.
In any base system, the digit symbols begin at 0 and go up to one less than the base.
Table 4 (p.125)
Base
2Digit Symbols
Place Values
Two
0, 1
…, 4, 23, 22, 21, 1
Three
0, 1, 2
…, 4, 33, 32, 31, 1
Four
0, 1, 2, 3
…,4, 43, 42, 41, 1
Five
0, 1, 2, 3, 4
…,4, 53, 52, 51, 1
Six
0, 1, 2, 3, 4, 5
…, 4, 63, 62, 61, 1
Seven
0, 1, 2, 3, 4, 5, 6
…, 4, 73, 72, 71, 1
Eight
0, 1, 2, 3, 4, 5, 6, 7
…, 4, 83, 82, 81, 1
Nine
0, 1, 2, 3, 4, 5, 6, 7, 8
…, 4, 93, 92, 91, 1
Ten
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
…, 4, 103, 102, 101, 1
Follow these steps to change a base to base ten
o Find the place value for each digit in the numeral
o Multiply each digit in the numeral by its respective place value
o Find the sum of the products in step 2
Examples;
o Convert 4726eight to base 10
The numeral has 4 place values. From left to right the place values are 83,
82, 81, and 1
Multiply each digit in the numeral by its respective place value, then find
the sum of these products
4726eight = (4 x 83) + (7 x 82) + (2 x 81) + (6 x 1)
= (4 x 8 x 8 x8) + (7 x 8 x 8) + (2 x 8) + (6 x 1)
= (2048) + 448 + 16 + 6
= 2518
o Convert 100101two
o 100101two. = (1 x 25) + (0 x 24) + (0 x 23) + (1 x 22) + (0 x 21) + (1 x 1)
= (1 x 32) + (0 x 16) + (0 x8) + (1 x 4) + (0 x 2) + ( 1 x 1)
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