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Number Bases

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Number Bases

Informatics INFO I101

February 9, 2004

John C. Paolillo, Instructor

- Last week
- Digital logic, Boolean algebra, and circuits
- Logic gates and truth tables

- This Week
- Numbers and bases
- Working with binary

Number Base Systems

…

…

b4

#

b3

#

b2

#

b1

#

b0

#

b-1

#

b-2

#

b-3

#

b-4

#

b-5

#

…

…

The number represented is the sum of all the products of the digit values and their respective place values

- Decimal (Base 10)
- Binary (Base 2)
- Octal (Base 8)
- Hexadecimal (Base 16)

- Identify each of the places in the new number base. These will correspond to the powers of the base, for example, with base 2, they are 1, 2, 4, 8, 16, 32, etc.
- Multiply the value for each place by the value of the digit appearing there;
- Add the results up, and you have the result in decimal
Note that if you divide and add correctly, you can reverse this procedure to convert decimal into another base. It’s harder, because you’re not used to using the appropriate addition and multiplication tables.

- What is 10011 Base 2 in decimal?
116+ 08 + 04 + 12 + 11 = 19

- What is 121 Base 8 in decimal?
164 + 28 + 11 = 81

- What is 247 Base 10 in Binary?
Here it helps to have a different procedure…

256

128

64

32

16

8

4

2

1

0

1

1

1

1

0

1

1

1

28

27

26

25

24

23

22

21

20

128

256

16

64

32

2

4

8

1

247

247

119

55

23

1

7

3

7

0

1

0

1

1

1

1

1

1

What we’re converting

- Sixteen digits:
0, 1, 2, 3, 4, 5, 6, 7

7 = 111two = 7eight

- Octal values are usually not specially indicated
Unix example:chmod 666 myfile.html

Decimal

0

1

2

3

4

5

6

7

Binary

000

001

010

011

100

101

110

111

Octal

0

1

2

3

4

5

6

7

- each octal digit corresponds to three binary digits (bits)
- convert binary to octal by parsing each group of three bits into one octal digit
- convert octal to binary by translating each digit into three bits
- Examples:
764eight =111101100two

011011101two= 335eight

- Sixteen digits:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

15 = 1111two = Fsixteen

- Hexadecimal (“hex”) values are usually indicated by a preceding base marker
HTML: #FFFFFF

JavaScript, C: 0xF1AD

Decimal

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Binary

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

Hex

0

1

2

3

4

5

6

7

8

9

A

B

C

D

E

F

- each hex digit corresponds to four binary digits (bits)
- convert binary to hex by parsing each group of four bits into one hex digit
- convert hex to binary by translating each digit into four bits
- Two hex digits make up one byte, a very common unit of memory

IP = “Internet Protocol”

IP Number

129.79.142.114

Host Name

dhcp-Memorial–142-114.memorial.indiana.edu

A Domain Name Server (DNS) has a database that matches IP and host name

Net

Subnet

Node

- Four Fields
- 0-255 in each field
- This is really base 256, but we use decimal numbers in each digit

129.79.142.114

Binary Addition

- Zero plus any other values leaves that value
Identity value for addition

No carry is generated

- One plus one leaves zero and causes a carry (one) to the next digit
Each successive digit must accept the carry from the previous

01001101 Base 2

+ 00011011 Base 2

01111111 Base 2

+ 00000001 Base 2

102 Base 8

+ 121 Base 8

01001101 Base 2

00000100 Base 2

0000101 Base 2

0000101 Base 2

01111111 Base 2

00011011 Base 2

CI

0

0

0

0

1

1

1

1

A

0

1

0

1

0

1

0

1

B

0

0

1

1

0

0

1

1

S

CO

S

0

1

1

0

1

0

0

1

CO

0

0

0

1

0

1

1

1

S

+

XOR

00

0

0

1

1

01

0

0

00

0

00

0

0

01

1

1

01

1

01

0

1

1

0

10

B

A

C

The half adder sends a carry, but can’t accept one

So we need another for the carry bit

- Two half-adders gives us a full adder
- two inputs plus carry

- Adders are cascaded to permit adding binary numbers
- Eight adders allows adding (0...256) + (0...256) in binary numbers
- Overflow can happen (200 + 100)

- Binary adders are used to do other computations as well...

Subtraction

Complement Representations

–

00

01

00

00

01

01

–01

00

- Subtraction is asymmetrical
- That makes it harder
- We have to borrow sometimes

827 Minuend

–223 Subtrahend

=604 Difference/remainder

456

–123

333

999

–123

876

- Subtraction is easy if you don’t have to borrow
- i.e. if all the digits of the minuend are greater than (or equal to) all those of the subtrahend

- This will always be true if the minuend is all 9’s: 999, or 999999, or 9999999999 etc.

- Subtract the subtrahend from 999 (or whatever we need) (easy)
- Add the result to the minuend (ordinary addition)
- Add 1 (easy)
- Subtract 1000 (drop highest digit)
Difference = Minuend + 999 – Subtrahend + 1 – 1000

This works for binary as well as decimal

- Subtract the subtrahend from 111 (or whatever we need) (easy)
- Add the result to the minuend (ordinary addition)
- Add 1 (easy)
- Subtract 1000 (drop highest digit)
Difference = Minuend + 111 – Subtrahend + 1 – 1000

10010101

–01101110

?????????

11111111

–01101110

10010001

This is the same as inverting each bit

00100111

+ 10010101

100100110

Regular addition

+1

100100111

Add one

–100000000

00100111

Now drop the highest bit (easy: it’s out of range)

Each of these steps is a simple operation we can perform using our logic circuits

Bitwise XOR

Invert each bit

Cascaded Adders

Regular addition

Add carry bit

Add one

Drop the highest bit ( it overflows)

Now drop the highest bit (easy: it’s out of range)

These steps make the negative of a number in twos-complement notation

Invert each bit

Add one

- Twos complements can be added to other numbers normally
- Positive numbers cannot use the highest bit (the sign bit)
- This is the normal representation of negative numbers in binary

000000000

000000011

000000102

000000113

000001004

000001015

000001106

000001117

000010008

000010019

etc.

11111111–1

11111110–2

11111101–3

11111100–4

11111011–5

11111010–6

11111001–7

11111000–8

11110111–9

11110110–10

etc.

- The number representation you use (encoding) affects the way you need to do arithmetic (procedure)
- This is true of all codes: encoding (representation) affects procedure (algorithm)
- Good binary codes make use of properties of binary numbers and digital logic

A computer program adds 20,000 and 20,000 and instead of 40,000, it reports –25,566

- No errors in encoding, decoding or addition
- How? Because the result is a negative number in twos-complement notation (highest bit = sign bit)

- 20,000 base ten is 0100111000100000 binary
010011100010000001001110001000001001110001000000

- Highest bit is set, so number is negative in twos complement notation: subtract one and invert to display
1001110001000000 – 1 = 1001110000111111

0110001111000000 = 25,566

Representations themselves, as we use them, have limits.

Interpretation depends on context

two procedures (encoding/decoding and addition) may be in and of themselves correct, but conflict in their application to specific examples

MER has landed