meljun cortes handouts_the number system

8/8/2019 MELJUN CORTES HANDOUTS_The Number System

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TheNumber System: AReview

LogicDesignand Switching

* Property of STI

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Base or Radix 10 number ing sy stem

also known as the decimal numbering system

uses ten symbols: digits 0,1,2,3 up to 9

Binary number ing system uses binary digit (0 and 1) commonly called a

bit -- the bits correspond to the off and on

positions of the switches in the computer

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* Property of STI

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the position of the digit with respect to the

decimal point (radix point )

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Step 1: Make a listing of the position valuesabove each digit.

Step 2: For each column, multiply the position

value of the column by the digit

contained in that column.

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Step 3: Multiply .25 by 2.

Step 4: Multiply the fractional part again by 2.

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Step 3: Add the values listed in the decimal

equivalent row to get the decimal

answer.

Step 1: Separate the integer part from the

fractional part then; convert the decimalnumber 2110.

Step 2: Divide the integer 21 by 2 to get the

quotient 10 and the remainder 1.

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Separate the integer from the fractional part.

For the integer number 697 :

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Octal Number ing System

uses 0, 1, 2, 3, 4, 5, 6, 7 digits

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For the fractional part 0.78125 10 :

Step 11:

0.78125 * 2 = 1.5625 Answer: 0 . 1 ? ? ? ?

Step 12:

0.5625 * 2 = 1.125 Answer: 0 . 1 1 ? ? ?

Step 13:

0.125 * 2 = 0.25 Answer: 0 . 1 1 0 ? ?

Step 14:

0.25 * 2 = 0.5 Answer: 0 . 1 1 0 0 ?

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Hexadecimal Number ing System

uses the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

digits and the letters A, B, C, D, E, and F to

supplement for the values of 10 to 15

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binary digits are grouped into 3’s, right to left

before the binary point, left to right after thebinary point

Padding the number with 0:

add 0’s to the left of the most significant bit

add 0’s to the right of the least significant bit

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Step 1: Divide the binary number into groups of

3.

Step 2: Multiply each binary number to its place

value and add the products to get theoctal digit.

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binary digits are grouped into 4’s, right to left

before the binary point, left to right after the

binary point

the rules for adding 0’s remain the same

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Step 1: Divide the binary number into groups of4.

Step 2: Multiply each binary number to its placevalue and add the products to get the

hexadecimal digit.

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Solution:

1. 5.58 5 = 10125.58 = 101.1012

2. FACE16 F = 11112 A = 10102C = 11002E = 11102FACE16 = 1111 1010 1100

11102

3. 86416 8 = 100026 = 01102

4 = 0100286416 = 1000 0110 01002

4. 7001.68 7 = 11120 = 00021 = 00126 = 11027001.68 = 111 000 001 1102

5. 25.716 2 = 001025 = 010127 = 0111225.716 100101.0111 2(the two 0’s at the beginning wasdisregarded )

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Signed Magnitude Representation

padding the binary number with a sign bit at

the beginning to indicate its sign

0 - added at the start to indicate a positive

number

1 - stands for a negative number

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Complements

used in digital computers for simplifying the

subtraction operation and for logical operations

diminished radix complement of a number N of

base r and with n digits is defined as:

(r - 1)’s complement:

(r n - 1) - N

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Solution:

Formula:(r n - 1 ) -N

Given: n = 6; r = 10

Then: 106 = 1,000,000

r n - 1 = 1,000,000 -1 = 999,999

a. N = 5 678 94

9’s complement:

(106 - 1) - 567894 = 432105

b. N = 0 125 98

9’s complement:

(106 - 1) - 012598 = 987401

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1’s complement:

(2n - 1) – N

Solution:

a. 002 = 112

b. 01112 = 10002

c. 01001000102 = 10110111012

d. 1.10012 = 01.10012 = 10.01102

e. 1510 = 0011112 = 1100002

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Step 1: Find the least significant 1 bit and let

this be the anchor .

Step 2: Change each of the bits to the left of this

anchor while the 1 itself (or LSB) and all

other bits in the less significant positions

remain unchanged.Step 3: The result is the 2’s complement.

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radix complement of a number N of base r and

with n digits is defined as r’s complement:

r n - N , for N not 0

0, for N = 0

for values of N not 0:

r’s complement = (r - 1)’s complement +1

since r n - N = (r n - 1) - N + 1

10’s complement:

(106 - 1) - 567894 + 1 = 432106

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Solution:

a. 002 = 0 (the special caseN = 0)

b. 011.12 = 101.12

c. 11001000102 = 00110111102

d. 01000112 = 10111012

e. Convert the decimal number to binary and pad

leading or trailing 0’s as necessary to get the

fixed 16-bit size.

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* Property of STI

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may represent the numbers, letters, symbols, or

any information or characters

Solution:

2n > = number of information to be

represented

where n

= the number of binary digits required to

represent the information

n = 1

= 2n > = 21 = 2 number of information

that can be r epresented

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Solution:

for 5 different colors of hair, choose n = 3 to satisfythe formula:

2n > = number of information to berepresented

where n = 3

23 > = 8 = 6 number of information that can

be represented

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Solution:

2n > = number of information to be

represented

where n = 3

23 > = 8 = 6 number of information that can

be represented

the following combinations can be used:

000, 001, 010, 011, 100, 110, and 111

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the direct translation of the 10 decimal digits

into its 4-bit binary counterparts

a form of coding that which is numeric in nature

since it encodes numeric characters

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1 = 0001

6 = 0110

5 = 0101

BCD equivalent = 000101100101.01012

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American Standard Code for Information

Interchange

uses 7 bits to represent the text data

has 128 combinations or 128 values which

includes 94 printable characters:

26 uppercase letters

26 lowercase letters

10 decimal digits/numbers

32 special printable characters like @, *, &, etc.

34 special non-printable characters called

control characters

Control Characters

correspond to those keys and/or commands

that include formatting of data (e.g. BS for

Backspace), sending information (e.g. CR for

Carriage Return or the Enter key)

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Solution:

Step 1: Write the numbers aligning the radixpoint then start evaluating from the

rightmost column.

001.1

011.1

+ 101.1

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Step 4: Continue evaluating until all the columns

have been evaluated.

Step 5: The final carry can be immediatelycarried down as the most significant bit.

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Step 2: Cross out each pair of 1’s and generate a

carry every time a pair is crossed out.

Step 3: Two carries cannot be represented by

the numeral 2. However, it can be

represented as shown below.

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Solution:

Step 1: Write the numbers aligning the radix

point then start evaluating from the

rightmost column. Note: You need toborrow to evaluate the rightmost column.

1 0 0 1 0 . 0

- 1 0 1 1 . 1

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* Property of STI

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Step 2: If borrowing is required, borrow from the

nearest 1 at the left digit. Add this to thenext digit to its right until we arrive at

the column you are evaluating.

Step 3: Perform rules 1 to 3 since you havealready borrowed from 1.

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Step 1: Align the multiplicand with the

multiplier.

Step 2: Remember the two equations in

performing binary multiplication.

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Step 4: Perform borrowing from the nearest 1 at

the left digit. Repeat Step 2 .

Step 5: Perform rules 1 to 3 until you arrive at theleftmost column.

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Note that you are to align the product with the

multiplier digit to which you are multiplying the

multiplicand.

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* Property of STI

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Step 3: Add the resulting products as aligned

above. The decimal point is placed onthe fourth place from the right.

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Step 3:

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Solution:

Step 1:

Step 2:

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Dividend

the number to be divided

Divisor

the quantity by which the dividend will be

divided by

Quotient

the resulting quantity when the dividend has

been divided by the divisor

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Solution:

Step 1: Determine the number of digits (n) of thedivisor.

101101 - dividend

101 - divisor

n = 3

the first n digits of the dividend is equal

to the divisor

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Step 4: Repeat step 3 until all the difference is 0

and/or the bits of the dividend has beenbrought down.

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Step 2: Write a 1 aligned with the resulting

number’s place value. Subtract the

divisor from the corresponding number.

Step 3: Bring down the next bit from the

dividend. NOTE: Place 0 if the divisor is> the dividend. Place 1 if the divisor is


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Step 2:

Step 3:

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Step 4:

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