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Chapter Six NUMBERING SYSTEMS

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Chapter SixNUMBERING SYSTEMS

- Way to count
- Types of numbering systems
- Non Systematic
- Old Egyptian numbers.
- Roman numbers

- Systematic (positional)
- Arabic numbers
- Indian numbers

- Non Systematic

- Symbols
- Phonetic
- Symbolic

- Base (radix)
- Number of the symbols is equal to the base

- Digits
- Relation between digits is that the one in the higher digit divided by the just lower digit is equal to the base.

- The value is determined by the face value multiplied by the base to the power of the position of that digit.

- Octal
- Symbols
- 0, 1, 2, 3, â€¦. ,7
- One, two , zero , â€¦., seven

- Base
- 8

- Digits
- (11)8 --- 1 in the higher digit / 1 in the lower digit = (8)10

- Symbols

- Start from left to right
- Pronounce each digit at a time
- At the end , pronounce in System so.
- Example (121)8
- One two one in octal

- Example (A98B)16
- A nine eight B in hexadecimal

- Example (1101.1)2
- One one zero one point one in binary

- There are many types, but we are going to study only some of them
- Binary system ( Base 2)
- Octal system ( Base 8)
- Decimal (denary) (base 10)
- Hexadecimal (base 16)

- We are familiar with decimal system.
- It is a good way to convert from/to decimal system
- Two ways to convert
- From any system to decimal
- From decimal to any system

- (dnâ€¦..d2 d1 d0 . d-1 d-2 â€¦â€¦) R = ( )10

- Place the face value with space doubling between digits
- Place under each digit the weight of it which is the power of the base of that number.
- Multiply the number by its weight into a third row.
- Sum all the numbers in the third row

- You do integer division with the whole part by the required base until you reach the zero value as the quotient.
- The following shape shows the processing of doing this.
- Integer division process is illustrated in the following shape.

- This method is for certain types of system which has a relation between.
- Suppose you have two numbering systems: R1 and R2. there is a conversion method those two systems if R1n1 = R2n2, where n1 and n2 are both integer values greater than 0 and one of them is 1.
- What is meant by that? It means that every digit from R1 is equivalent to n2 of digits from R2
- Examples: 23 = 81 and 24 = 161 , which means every 3 digits of the binary system is equivalent to one digits in octal and so on.

- Make a table for the two systems that have the relation discussed in previous charts.
- You didnâ€™t need to do that every time if you memorize that table.
- Group n2 of digits of the second system and replace it with one digit of first system starting from the right of the number.
- Or for every digit of first system, replace it with n2 digits of the second system

- Addition
- Subtraction
- Normal way
- Twoâ€™s complement

- Align the binary numbers based on the fraction point.
- Add leading zero to the fraction part for the number that has less digits and tailing zeros to the whole number for the number that has less digits.
- Use the table of the addition which is described below to do the addition.

- Align all the number beneath each other based on the fraction point.
- Strike through each two ones of one digit and replace them with one on the just higher digit.
- Each column (digit) which has a single one the result will be one, and if doesnâ€™t has any remaining ones then the result is zero.
- Repeat for all the digit until the last digit

- Subtraction using the borrow technique.
- Subtraction using the twoâ€™s complement
- Twoâ€™s complement. (oneâ€™s complement +1)
- Oneâ€™s complement

- Align the binary numbers based on the fraction point.
- Add leading zero to the fraction part for the number that has less digits and tailing zeros to the whole number for the number that has less digits.
- Use the table of the subtraction which is described below to do the subtraction.

- Align the two binary numbers starting from the fraction point.
- Do the following for each column of the numbers (four cases).
- If the two values of a single column are either zeroes or ones , then the result is zero with no borrow.
- Else if the higher value is one then the result is one with no borrow.
- Else, which means that (the top minuend is 0 and the subtrahand is one), we need to borrow. The borrow means we take one from the just higher bit leaving the one to be zero (cross it) . This borrowed one is interchanged to the lower bit into two ones. For this whole column cross out two ones and the remaining one is the result.

- Way to represent negative values.
- Change every one in binary to zero and every zero to 1

- Convert (1010011)2 oneâ€™s complement.
- (0101100) 1â€™s

- Convert (1000010011) which is oneâ€™s complement to its binary value.
- (0111101100)2

- It is a way to represent a number with both value and sign.
- The most important property is that the Most Significant Bit has a negative weight.
- To convert from twoâ€™s complement to decimal is an easy way. The only difference is that during computation the weight of the last bit is negative.

- We have two cases.
- If it is a positive decimal number. It is the same as normal binary except that you add 0 as MSB to the binary number.
- If it is negative decimal number.
- Treat the number as if it is positive.
- Convert it to binary number.
- Put Zeroes to the left of the number.
- Convert the zero to one and one to zero for all digits
- Add one to the whole number.

- Subtraction process requires two operands: minuend and subtrahend, normally the subtrahend is greater than the minuend.
- Make sure you have more digits that accommodate the value of the numbers by adding zeros to the left of the two numbers.
- Convert the subtrahend to twoâ€™s complement.
- Add the two numbers.
- If the number of digits of the result exceeds numbers of digits both numbers. Cancel that digit.

- Subtract 1001 from 101.
- Number of digits ( we need 5 digits for those numbers) so , we add trailing zeros
- Minuend becomes 00101 and the subtrahend becomes 01001

- Convert the subtrahend to twoâ€™s complement
- So it 01001 becomes 10111

- Add both numbers: 00101 to 10111
- 00101 +10111 = 11100

- As the number of digits of the result is that same as the number of the digits of both number. This is the result.
- (Note that as the last bit of the result is 1, it means the number is negative , to find its value. Follow the rules to convert from twoâ€™s complement to decimal)

- Number of digits ( we need 5 digits for those numbers) so , we add trailing zeros

- Subtract 110 from 1011.
- Number of digits ( we need 5 digits for those numbers) so , we add trailing zeros
- Minuend becomes 01011 and the subtrahend becomes 00110

- Convert the subtrahend to twoâ€™s complement
- So it 00110 becomes 11010

- Add both numbers: 01011 to 11010
- 01011+ 11010 = 100101

- Because we have six digits, the left most digits is discarded and keep the number of the digits to 5. the result will b 00101
- (Note that as the last bit of the result is 0, it means the number is positive, to find its value. Follow the rules to convert from twoâ€™s complement to decimal)

- Number of digits ( we need 5 digits for those numbers) so , we add trailing zeros