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Number Systems

Decimal, Binary, and Hexadecimal

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Base-N Number System

• Base N• N Digits: 0, 1, 2, 3, 4, 5, …, N-1• Example: 1045N

• Positional Number System •

• Digit do is the least significant digit (LSD).• Digit dn-1 is the most significant digit (MSD).

1 4 3 2 1 0

1 4 3 2 1 0

n

n

N N N N N N

d d d d d d

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Decimal Number System• Base 10• Ten Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9• Example: 104510

• Positional Number System

• Digit d0 is the least significant digit (LSD).• Digit dn-1 is the most significant digit (MSD).

1 4 3 2 1 0

1 4 3 2 1 0

10 10 10 10 10 10n

nd d d d d d

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Binary Number System• Base 2• Two Digits: 0, 1• Example: 10101102

• Positional Number System

• Binary Digits are called Bits• Bit bo is the least significant bit (LSB).• Bit bn-1 is the most significant bit (MSB).

1 4 3 2 1 0

1 4 3 2 1 0

2 2 2 2 2 2n

nb b b b b b

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Definitions• nybble = 4 bits• byte = 8 bits• (short) word = 2 bytes = 16 bits• (double) word = 4 bytes = 32 bits• (long) word = 8 bytes = 64 bits• 1K (kilo or “kibi”) = 1,024• 1M (mega or “mebi”) = (1K)*(1K) = 1,048,576• 1G (giga or “gibi”) = (1K)*(1M) = 1,073,741,824

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Hexadecimal Number System

• Base 16• Sixteen Digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F• Example: EF5616

• Positional Number System •

0000 0

0001 1

0010 2

0011 3

0100 4

0101 5

0110 6

0111 7

1000 8

1001 9

1010 A

1011 B

1100 C

1101 D

1110 E

1111 F

1 4 3 2 1 016 16 16 16 16 16n

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Collaborative Learning

Learning methodology in which students are not only responsible for their own learning but for the learning of other members of the group.

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Think - Pair - Share (TPS) Quizzes

• Think – Pair – Share– Think individually for one time units– Pair with partner for two time units– Share with group for one and half time units– Report results

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Quiz 1-A (Practice)

• Assemble in groups of 4

• Question: Convert the following binary number into its decimal equivalent:

110102

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Quiz 1-A (Practice)

THINKOne Unit

(e.g. 30 Seconds)

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Quiz 1-A (Practice)

PAIRTwo Units

(e.g. 60 Seconds)

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Quiz 1-A (Practice)

SHARE1.5 units

(e.g. 45 Seconds)

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Quiz 1-A (Practice)

Report

• Write names of all group members and the consensus answer on one sheet of paper.• All sheets will be collected.• One will be picked at random to read to the class.• All papers will be graded!

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Quiz 1-A Solution

• Convert the following number into base 10 decimal:

4 3 12 1011010 2 2 2 16 8 2 26

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Quiz 1-B

• Convert the following number into base 10 decimal:

1010116

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Collaborative Learning

• Think for 30 seconds

• Pair for 1 minute

• Share for 45 seconds

• Report

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Quiz 1-B Solution

• Convert the following number into base 10 decimal:

1010116 = 1·164 + 0·163 + 1·162 + 0·161 + 1·160

= 164 + 162 + 160

= 65,536 + 256 + 1= 65,793

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TPS Quiz 2

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Binary Addition

•Single Bit Addition Table

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 10 Note “carry”

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Hex Addition

• 4-bit Addition

4 + 4 = 8

4 + 8 = C

8 + 7 = F

F + E = 1D Note “carry”

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Hex Digit Addition Table+ 0 1 2 3 4 5 6 7 8 9 A B C D E F0 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 1 2 3 4 5 6 7 8 9 A B C D E F 10

2 2 3 4 5 6 7 8 9 A B C D E F 10 113 3 4 5 6 7 8 9 A B C D E F 10 11 124 4 5 6 7 8 9 A B C D E F 10 11 12 13

5 5 6 7 8 9 A B C D E F 10 11 12 13 14

6 6 7 8 9 A B C D E F 10 11 12 13 14 157 7 8 9 A B C D E F 10 11 12 13 14 15 168 8 9 A B C D E F 10 11 12 13 14 15 16 17

9 9 A B C D E F 10 11 12 13 14 15 16 17 18

A A B C D E F 10 11 12 13 14 15 16 17 18 19B B C D E F 10 11 12 13 14 15 16 17 18 19 1AC C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B

D D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C

E E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1DF F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E

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TPS Quiz 3

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Complements

• 1’s complement– To calculate the 1’s complement of a binary

number just “flip” each bit of the original binary number.

– E.g. 0 1 , 1 0– 01010100100 10101011011

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Complements

• 2’s complement– To calculate the 2’s complement just calculate

the 1’s complement, then add 1.

01010100100 10101011011 + 1=

10101011100– Handy Trick: Leave all of the least significant

0’s and first 1 unchanged, and then “flip” the bits for all other digits.

• Eg: 01010100100 -> 10101011100

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Complements

• Note the 2’s complement of the 2’s complement is just the original number N– EX: let N = 01010100100– 2’s comp of N = M = 10101011100– 2’s comp of M = 01010100100 = N

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Two’s Complement Representation for Signed Numbers

• Let’s introduce a notation for negative digits:– For any digit d, define d = −d.

• Notice that in binary, where d {0,1}, we have:

• Two’s complement notation:– To encode a negative number, we implicitly

negate the leftmost (most significant) bit:• E.g., 1000 = (−1)000

= −1·23 + 0·22 + 0·21 + 0·20 = −8

101111

011010

1,1

dddd

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Negating in Two’s Complement

• Theorem: To negatea two’s complementnumber, just complement it and add 1.

• Proof (for the case of 3-bit numbers XYZ):

1)( 22 YZXYZX

1

1)1)(1(

111100

)1()(

2

2

222

2222

YZX

ZYX

YZXYZX

YZXYZXYZXYZX

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Signed Binary Numbers

• Two methods: – First method: sign-magnitude

• Use one bit to represent the sign– 0 = positive, 1 = negative

• Remaining bits are used to represent the magnitude

• Range - (2n-1 – 1) to 2n-1 - 1

where n=number of digits• Example: Let n=4: Range is –7 to 7 or • 1111 to 0111

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Signed Binary Numbers

• Second method: Two’s-complement• Use the 2’s complement of N to represent

-N• Note: MSB is 0 if positive and 1 if negative• Range - 2n-1 to 2n-1 -1

where n=number of digits

• Example: Let n=4: Range is –8 to 7Or 1000 to 0111

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Signed Numbers – 4-bit example

Decimal 2’s comp Sign-Mag 7 0111 0111 6 0110 0110 5 0101 0101 4 0100 0100 3 0011 0011 2 0010 0010 1 0001 0001 0 0000 0000 Pos 0

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Signed Numbers-4 bit example

Decimal 2’s comp Sign-Mag -8 1000 N/A -7 1001 1111 -6 1010 1110 -5 1011 1101 -4 1100 1100 -3 1101 1011 -2 1110 1010 -1 1111 1001 -0 0000 (= +0) 1000

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Notes:

• “Humans” normally use sign-magnitude representation for signed numbers– Eg: Positive numbers: +N or N– Negative numbers: -N

• Computers generally use two’s-complement representation for signed numbers– First bit still indicates positive or negative.– If the number is negative, take 2’s complement to

determine its magnitude• Or, just add up the values of bits at their positions,

remembering that the first bit is implicitly negative.

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Example

• Let N=4: two’s-complement• What is the decimal equivalent of

01012

Since msb is 0, number is positive

01012 = 4+1 = +510

• What is the decimal equivalent of11012 =

• Since MSB is one, number is negative• Must calculate its 2’s complement• 11012 = −(0010+1)= − 00112 or −310

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Very Important!!! – Unless otherwise stated, assume two’s-complement numbers for all problems, quizzes, HW’s, etc.

The first digit will not necessarily be explicitly underlined.

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TPS Quizzes 5-7

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Arithmetic Subtraction

• Borrow Method– This is the technique you learned in grade

school– For binary numbers, we have–

0 - 0 = 0

1 - 0 = 1

1 - 1 = 0

0 - 1 = 1 with a “borrow”1

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Binary Subtraction

• Note:– A – (+B) = A + (-B)– A – (-B) = A + (-(-B))= A + (+B)– In other words, we can “subtract” B from A by

“adding” –B to A.– However, -B is just the 2’s complement of B,

so to perform subtraction, we• 1. Calculate the 2’s complement of B• 2. Add A + (-B)

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Binary Subtraction - Example

• Let n=4, A=01002 (410), and

B=00102 (210)

• Let’s find A+B, A-B and B-A

0 1 0 0+ 0 0 1 0

(4)10 (2)10

0 11 0 6

A+B

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Binary Subtraction - Example

0 1 0 0- 0 0 1 0

(4)10 (2)10

10 0 1 0 2

A-B

0 1 0 0+ 1 1 1 0

(4)10 (-2)10

A+ (-B)

“Throw this bit” away since n=4

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Binary Subtraction - Example

0 0 1 0- 0 1 0 0

(2)10 (4)10

1 1 1 0 -2

B-A

0 0 1 0+ 1 1 0 0

(2)10 (-4)10

B + (-A)

1 1 1 02 = - 0 0 1 02 = -210

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“16’s Complement” method

• The 16’s complement of a 16 bit Hexadecimal number is just:

• =1000016 – N16

• Q: What is the decimal equivalent of B2CE16 ?

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16’s Complement

• Since sign bit is one, number is negative. Must calculate the 16’s complement to find magnitude.

• =1000016 – B2CE16 = ?????

• We have

10000

- B2CE

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16’s Complement

FFF10

- B2CE

23D4

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16’s Complement

• So,

1000016 – B2CE16 = 4D3216

= 4×4,096 + 13×256 + 3×16 + 2

= 19,76210

• Thus, B2CE16 (in signed-magnitude)represents -19,76210.

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Sign Extension

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Sign Extension

• Assume a signed binary system

• Let A = 0101 (4 bits) and B = 010 (3 bits)

• What is A+B?– To add these two values we need A and B to

be of the same bit width.– Do we truncate A to 3 bits or add an

additional bit to B?

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Sign Extension

• A = 0101 and B=010

• Can’t truncate A!! Why?– A: 0101 -> 101– But 0101 <> 101 in a signed system– 0101 = +5– 101 = -3

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Sign Extension

• Must “sign extend” B,

• so B becomes 010 -> 0010

• Note: Value of B remains the same

So 0101 (5)

+0010 (2)

--------

0111 (7)

Sign bit is extended

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Sign Extension

• What about negative numbers?

• Let A=0101 and B=100

• Now B = 100 1100

Sign bit is extended 0101 (5)+1100 (-4)------- 10001 (1)

Throw away

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Why does sign extension work?• Note that:

(−1) = 1 = 11 = 111 = 1111 = 111…1– Thus, any number of leading 1’s is equivalent, so long

as the leftmost one of them is implicitly negative.

• Proof:111…1 = −(111…1) =

= −(100…0 − 11…1) = −(1) • So, the combined value of any sequence of

leading ones is always just −1 times the position value of the rightmost 1 in the sequence.

111…100…0 = (−1)·2n

n

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Number Conversions

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Decimal to Binary Conversion

Method I: Use repeated subtraction.

Subtract largest power of 2, then next largest, etc.

Powers of 2: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2n

Exponent: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 , n

210 2n292820 2721 22 23 2624 25

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Decimal to Binary Conversion

Suppose x = 156410 Subtract 1024: 1564-1024 (210) = 540 n=10 or 1 in the (210)’s position

Thus:156410 = (1 1 0 0 0 0 1 1 1 0 0)2

Subtract 512: 540-512 (29) = 28 n=9 or 1 in the (29)’s position

Subtract 16: 28-16 (24) = 12 n=4 or 1 in (24)’s position

Subtract 8: 12 – 8 (23) = 4 n=3 or 1 in (23)’s position

Subtract 4: 4 – 4 (22) = 0 n=2 or 1 in (22)’s position

28=256, 27=128, 26=64, 25=32 > 28, so we have 0 in all of these positions

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Decimal to Binary Conversion

Method II: Use repeated division by radix.

2 | 1564 782 R = 02|_____ 391 R = 02|_____ 195 R = 12|_____ 97 R = 12|_____

48 R = 12|_____ 24 R = 0

2|__24_ 12 R = 02|_____ 6 R = 02|_____

3 R = 02|_____ 1 R = 12|_____ 0 R = 1

Collect remainders in reverse order

1 1 0 0 0 0 1 1 1 0 0

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Binary to Hex Conversion

1. Divide binary number into 4-bit groups

2. Substitute hex digit for each group

1 1 0 0 0 0 1 1 1 0 00Pad with 0’sIf unsigned number

61C16

Pad with sign bitif signed number

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Hexadecimal to Binary ConversionExample

1. Convert each hex digit to equivalent binary

(1 E 9 C)16

(0001 1110 1001 1100)2

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Decimal to Hex Conversion

Method II: Use repeated division by radix.

16 | 1564 97 R = 12 = C16|_____ 6 R = 116|_____ 0 R = 6

N = 61C 16