machine level representation of datacskedah.uitm.edu.my/roziah/public/csc159/notes/ch2...machine...

CSC159: Computer Organization Prepared by: Pn. Siti Fatimah Nor Binti Ab Wahab 1

MACHINE LEVEL

REPRESENTATION OF DATA

CHAPTER 2


Objectives

Understand how integers and fractional numbers are represented in binary

Explore the relationship between decimal number system and number systems of other bases (binary, octal, hexadecimal)

Understand signed numbers and unsigned numbers

Understand number representations in two’s complement

Perform arithmetic operations in every base

Understand the method of floating point number representation in IEEE format


Introduction

NUMBERING SYSTEM :

BASE

PLACE

5TH PLACE

4TH PLACE

3RD PLACE

2ND PLACE

1ST PLACE

SINGLE UNIT

● 1ST

PLACE 2ND

PLACE 3RD

PLACE

DECIMAL

105 104 103 102 101 100 ● 10-1 10-2 10-3

100,000 10,000 1,000 100 10 1 0.1 0.01 0.001

1/10 1/100 1/1000

BINARY

25 24 23 22 21 20 ● 2-1 2-2 2-3

32 16 8 4 2 1 0.5 0.25 0.125

1/2 1/4 1/8

OCTAL

85 84 83 82 81 80 ● 8-1 8-2 8-3

32,768 4,096 512 64 8 1 0.125 0.015625 1.953125

X 103

1/8 1/64 1/512

HEXA-DECIMAL

165 164 163 162 161 160 ● 16-1 16-2 16-3

1,048,576

65,536 4,096 256 16 1 0.0625 3.906 X

103

2.4414062 X 104

1/16 1/256 1/4096


Arithmetic Operations in Different Number

Bases The following table shows the equivalent values for decimal numbers in binary, octal and

hexadecimal:

DECIMAL BINARY OCTAL HEXADECIMAL

0 0000 0 0 1 0001 1 1 2 0010 2 2 3 0011 3 3 4 0100 4 4 5 0101 5 5 6 0110 6 6 7 0111 7 7 8 1000 10 8 9 1001 11 9

10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 F


Radix :

when referring to binary, octal, decimal, hexadecimal, a

single lowercase letter appended to the end of each

number to identify its type.

E.g.

hexadecimal 45 will be written as 45h

octal 76 will be written as 76o or 76q

binary 11010011 will be written as 11010011b

Number Bases


Decimal system:

system of positional notation based on powers of 10.

Binary system:

system of positional notation based powers of 2

Octal system:

system of positional notation based on powers of 8

Hexadecimal system:

system of positional notation based on powers of 16

Number Bases


Early computer design was decimal

Mark I and ENIAC

John von Neumann proposed binary data

processing (1945)

Simplified computer design

Used for both instructions and data

Natural relationship between

on/off switches and

calculation using Boolean logic

On Off

True False

Yes No

1 0

Why Binary?


A computer stores both instructions and data as individual electronic charges.

represent these entities with numbers requires a system geared to the concept of on and off or true and false

Binary is a base 2 numbering system

each digit is either a 0 (off) or a 1 (on)

Computers store all instructions and data as sequences of binary digit

e.g. 010000010100001001001000011 = “ABC”

Binary Numbers


Bit : each digit in a binary number

Byte : consists of 8 bits which is the basic unit of storage on nearly all computers

Each location in the computer’s memory holds exactly 1 byte or 8 bits.

A byte can hold a single instruction, a character or a number

A Word : which is 16 bits or 2 bytes long :

1 0 1 1 0 1 0 1 1 0 1 1 0 1 0 1

A 16-bit computer : its instructions can operate on 16-bit quantities

byte byte

word

bit

Binary Numbers


Place 101 100

Value 10 1

Evaluate 4 x 10 3 x1

Sum 40 3

1’s place 10’s place

43 = 4 x 101 + 3 x 100

Numeric Data Representation FOR

DECIMAL


Numeric Data Representation FOR

DECIMAL

Place 102 101 100

Value 100 10 1

Evaluate 5 x 100 2 x 10 7 x1

Sum 500 20 7

1’s place 10’s place

527 = 5 x 102 + 2 x 101 + 7 x 100

100’s place


Numeric Data Representation

6248 = 40410

Place 82 81 80

Value 64 8 1

Evaluate 6 x 64 2 x 8 4 x 1

Sum for

Base 10 384 16 4

64’s place 8’s place 1’s place


Numeric Data Representation 6,70416 = 26,37210

Place 163 162 161 160

Value 4,096 256 16 1

Evaluate 6 x 4,096 7 x 256 0 x 16 4 x 1

Sum for

Base 10 24,576 1,792 0 4

4,096’s place 256’s place 1’s place 16’s place


Numeric Data Representation

Place 27 26 25 24 23 22 21 20

Value 128 64 32 16 8 4 2 1

Evaluate 1 x 128 1 x 64 0 x 32 1 x16 0 x 8 1 x 4 1 x 2 0 x 1

Sum for

Base 10 128 64 0 16 0 4 2 0

1101 01102 = 21410


Base:

The number of different symbols required to

represent any given number

The larger the base, the more numerals are required

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

Base 2: 0,1

Base 8: 0,1,2, 3,4,5,6,7

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

Base or Radix


For a given number, the larger the base

the more symbols required

but the fewer digits needed

Example #1:

6516 10110 1458 110 01012

Example #2:

11C16 28410 4348 1 0001 11002

Number of Symbols vs Number of Digits


Decimal Other bases

Other bases Decimal

Binary Octal

Binary Hexadecimal

Octal Binary

Hexadecimal Binary

Numeric Conversion between Number Bases


CONVERSION :

DECIMAL OTHER BASES


Base 10 42

2 ) 42 ( 0 Least significant bit

2 ) 21 ( 1

2 ) 10 ( 0

2 ) 5 ( 1

2 ) 2 ( 0

2 ) 1 ( 1 Most significant bit

0

Base 2 101010

Remainder

Quotient

From Base 10 to Base 2


Convert 3710 to binary.

* MSB (most-significant-bit) : left most bit

LSB (least-significant-bit) : right most bit

37 ÷ 2 = 18 balance 1 (LSB) 18 ÷ 2 = 9 balance 0 9 ÷ 2 = 4 balance 1 4 ÷ 2 = 2 balance 0 2 ÷ 2 = 1 balance 0 1 ÷ 2 = 0 balance 1 (MSB)

Therefore , 3710 = 1001012



What is the value of 37.687510 in binary?

Steps :

1. Convert the integer to binary by using method shown in previous slide.

2. Convert the decimal point to binary by using the following method.

So, (0.6875)10 = (0.1011)2 ; Therefore, 37.687510 = 100101.10112

0.6875

X 2

(MSB) 1 1.3750 X 2

0 0.7500 X 2

1 1.5000 X 2

(LSB) 1 1.0000

The 1 is saved as result, then dropped and the process repeated

From Base 10 (decimal point) to Base 2


Base 10 135

8) 135 ( 7 Least significant bit

8) 16 ( 0

8) 2 ( 2 Most significant bit

0

Base 8 207

Quotient

Remainder



8 21 5

8 2 2

0

0.25

x 8

2 2.00

From Base 10 (decimal point) to Base 8 Convert 21.2510 to octal.

Therefore,

21.2510 = 25.28


Convert 21.2510 to octal. (OTHER METHOD)

21 ÷ 2 = 10 balance 1 (LSB)

10 ÷ 2 = 5 balance 0

5 ÷ 2 = 2 balance 1

2 ÷ 2 = 1 balance 0

1 ÷ 2 = 0 balance 1 (MSB)

So, 2110 = 101012

Now, 0.25

X 2

(MSB) 0 0.50

X 2

(LSB) 1 1.00

So, 0.2510 = 0.012


Therefore,

Refer to conversion of binary to hexadecimal

21.2510 = 010 101 . 0102

= 25.28


Base 10 5,735

16 ) 5,735 ( 7 Least significant bit

16 ) 358 ( 6

16 ) 22 ( 6


0

Base 16 1667

Quotient

Remainder



Base 10 8,039

16 ) 8,039 ( 7 Least significant bit

16 ) 502 ( 6

16 ) 31 ( 15


0

Base 16 1F67

Quotient

Remainder



Convert 2110 to hexadecimal.

21 ÷ 16 = 1 balance 5 (LSB)

1 ÷ 16 = 0 balance 1 (MSB)

Therefore , 2110 = 1516



16 21 5

16 1 1

0

0.25

x 16

4 4.00

From Base 10 (decimal point) to Base 16 Convert 21.2510 to hexadecimal.

Therefore,

21.2510 = 15.416


Convert 21.2510 to hexadecimal. (OTHER METHOD)

21 ÷ 2 = 10 balance 1 (LSB)

10 ÷ 2 = 5 balance 0

5 ÷ 2 = 2 balance 1

2 ÷ 2 = 1 balance 0

1 ÷ 2 = 0 balance 1 (MSB)

So, 2110 = 101012

Now, 0.25

X 2

(MSB) 0 0.50

X 2

(LSB) 1 1.00

So, 0.2510 = 0.012


Therefore,

Refer to conversion of binary to hexadecimal

21.2510 = 10101 . 01002

= 15.416


CONVERSION :

OTHER BASES DECIMAL


Convert 1001102 to decimal.

1001102

= (1 X 25) + (0 X 24) +(0 X 23) +(1 X 22) + (1 X 21) + (0 X 20)

= 32 + 0 + 0 + 4 + 2 + 0

= 3810

Therefore, 1001102 = 3810

1 0 0 1 1 0

25 24 23 22 21 20

32 16 8 4 2 1

1*32 1*4 1*2

32 + 4 + 2 = 3810



Convert E516 to decimal.

E516

= (E X 161) + (5 X 160)

= (14 X 16) + (5 X 1)

= 224 + 5

= 22910

Therefore, E516 = 22910

E 5

161 160

16 1

14*16 5*1

224 + 5 = 22910



Convert E5.A816 to decimal.


E 5 . A 8

161 160 . 16-1 16-2

16 1 . 0.0625 0.00390625

14*16 5*1 . 10*0.0625 8*0.00390625

224 + 5 + 0.625 + 0.03125 = 229.6562510


Convert E5.A816 to decimal. (OTHER METHOD)

1. Firstly convert the number into binary,

2. From Example 15: E516 = 22910

3. Now,

Hex E 5 . A 8

Binary 1110 0101 . 1010 1000

(0.1010 1000)2

= (1 X 2-1) + (0 X 2-2) + (1 X 2-3) + (0 X 2-4) + (1 X 2-5) + (0 X 2-6) + (0 X 2-7) + (0 X 2-8)

= 1/2 + 1/23 + 1/25

= 1/2 + 1/8 + 1/32

= 0.5 + 0.125 + 0.03125

= (0.65625)10

Therefore, E5.A816 = 229.6562510



From Base 8 to Base 10 Convert 72638 to decimal.

7 2 6 3

83 82 81 80

512 64 8 1

7*512 2*64 6*8 3*1

3584 + 128 + 48 + 3 = 376310


72638 = 3,76310

(OTHER METHOD)

7

x 8

56 + 2 = 58

x 8

464 + 6 = 470

x 8

3760 + 3 = 3,763



From Base 8 (decimal point) to Base 10 Convert 46.328 to decimal.

4 6 . 3 2

81 80 . 8-1 8-2

8 1 . 0.125 0.015625

4*8 6*1 . 3*0.125 2*0.015625

32 + 6 + 0.375 + 0.03125 = 38.4062510


CONVERSION :

BINARY HEXADECIMAL

BINARY OCTAL


The nibble approach

Hex easier to read and write than binary

Why hexadecimal?

Modern computer operating systems and networks present variety

of troubleshooting data in hex format

The relationship between BINARY and HEXADECIMAL

1 digit hexadecimal is EQUIVALENT to 4 bits binary


Base 16 1 F 6 7

Base 2 0001 1111 0110 0111


Method:

Get the binary representation for the hexadecimal number.

Example:

Convert F116 to binary.

F116 = F 1

1111 00012

F116 = 1111 00012



Method:

Get the binary representation for the hexadecimal number.

Example:

Convert 2A.5C16 to binary.

2A.5C16 = 2 A . 5 C

0010 1010 . 0101 1100

2A.5C16 = 10 1010 . 0101 112

From Base 16 to Base 2 (decimal point)


Method:

Find the hexadecimal representation for the GROUP of 4 binary numbers.

Example:

Convert 11001112 to hexadecimal number.

110 01112 = 0110 0111

6 7

11001112 = 6716



Method:

Find the hexadecimal representation for the GROUP of 4 binary numbers.

Use the decimal point as your STARTING point.

Example:

Convert 111110.001012 to hexadecimal number.

11 1110.0010 12 = 0011 1110 . 0010 1000

3 E . 2 8

111110.001012 = 3E.2816



The relationship between BINARY and OCTAL

1 digit octal is EQUIVALENT to 3 bits binary

Base 8 1 2 5 7

Base 2 001 010 101 111



Method:

Get the binary representation for every digit in the octal number.

Example 17:

Convert 538 to binary.

538 = 5 3

101 0112

538 = 1010112



Method:

Get the binary representation for every digit in the octal number.

Example 17:

Convert 27.648 to binary.


27.648 = 2 7 . 6 4

010 111 . 110 100

27.648 = 10 111 . 110 12


Method:

Find the octal representation for the GROUP of 3 binary numbers.

Example:

Convert 11001112 to octal number.

1 100 1112 = 001 100 111

1 4 7

11001112 = 1478



Method:

Find the octal representation for the GROUP of 3 binary numbers. Use the

decimal point as your STARTING point.

Example:

Convert 11110.001012 to octal number.

11 110.001 012 = 011 110 . 001 010

3 6 . 1 2

11110.001012 = 36.128



ADDITION AND SUBTRACTION OF

DIFFERENT BASES


Base Problem Largest Single Digit

Decimal 6

+3 9

Octal 6

+1 7

Hexadecimal 6

+9 F

Binary 1

+0 1

Addition


Addition and Subtraction of Binary Numbers

Example ADD

1 1 1 1 1 1

1 0 1 1 1 1 0 0

+ 1 1 0 0 1 1 1 1

1 1 0 0 0 1 0 1 1

Example SUB

0 2 0 1 1 1 2

1 1 1 0 0 0 0 1

- 1 0 1 0 1 1

1 0 1 1 0 1 1 0


Addition of Octal Numbers

Example ADD

1 1 1

3 3 3 3 + 6 = 9 – 8 = 1

+ 4 7 6 1 + 3 + 7 = 11 – 8 = 3

1 0 3 1 1 + 3 + 4 = 8 – 8 = 0

1 + 0 = 1


Subtraction of Octal Numbers

Example SUB

6 81 8

7 2 3 8 + 3 – 6 = 5

- 5 3 6 8 + 1 – 3 = 6

1 6 5 6 – 5 = 1


Addition of Hexadecimal Numbers

Example ADD

1

1 A 2 3 3 + 8 = 11 = B

+ 7 C 2 8 2 + 2 = 4

9 6 4 B 10 + 12 = 22 – 16 = 6

1 + 1 + 7 = 9


Subtraction of Hexadecimal Numbers

Example SUB

E 16 6 16

2 F 7 D 16 + 13 – 14 = 15 = F

- 9 E 16 + 6 – 9 = 13 = D

2 E D F E – 0 = E

2 – 0 = 2


THE ALPHANUMERIC

REPRESENTATION


The Alphanumeric Representation

The data entered as characters, number digits, and

punctuation are known as alphanumeric data.

3 alphanumeric codes are in common use.

ASCII (American Standard Code for Information

Interchange)

Unicode

EBCDIC (Extended Binary Coded Decimal Interchange

Code).


The Alphanumeric Representation

The following

tables show the

comparisons

between the

ASCII code in

binary and

hexadecimal for

the given

characters.

Character Binary Hex A 100 0001 41

B 100 0010 42

C 100 0011 43

D 100 0100 44

E 100 0101 45

F 100 0110 46

G 100 0111 47

H 100 1000 48

I 100 1001 49

J 100 1010 4A

K 100 1011 4B

Space 010 0000 20

Full stop 010 1110 2E

( 010 1000 28

+ 010 1011 2B

$ 010 0100 24


THE DECIMAL REPRESENTATION


The Decimal Representation

BCD (Binary

Coded Decimal) is

often used to

represent decimal

number in binary.

Decimal BCD

0 0000

1 0001

2 0010

3 0011

4 0100

5 0101

6 0110

7 0111

8 1000

9 1001


SIGNED AND UNSIGNED NUMBERS


Signed and Unsigned Numbers

Binary numbers may either signed or unsigned

Oddly, CPU performs arithmetic and comparison operations for both

type equally well, without knowing which type it’s operating on.

An unsigned numbers :

numbers with only positive values

for 8-bit storage location : store unsigned integer value between 0 - 255

for 16-bit storage location : store unsigned integer value between 0 - 65535

Unsigned number can be converted directly to binary numbers and

processed without any special care


For negative numbers, there are several ways used to represent it in

binary form, depending on the process take place :

i. Sign-and-magnitude representation

ii. 1’s complement representation

iii. 2’s complement representation

Signed and Unsigned Numbers


Sign-and-Magnitude

In daily usage, signed integers are represented by a plus or minus sign

and a value.

In the computer, the uses of 0’s and 1’s take place.

0 : plus ( positive)

1 : minus (negative)

The leftmost bit in a binary number is considered the sign bit.

The remaining (n-1) bits are used for magnitude.


Example 1 :

+ 0010 0101

(+37)

0000 0000 0000 0001

(+1)

0111 1111 1111 1111

(+32767)

- 1010 0101

(-37)

1000 0000 0000 0001

(-1)

1111 1111 1111 1111

(-32767)

Sign-and-Magnitude


Example 2 :

What is the sign-and-magnitude representation of the decimal

numbers –31 and +31 if the basic unit is a byte ?

3110 = 111112

Unit is a byte = 8 bits

Sign-and-Magnitude

+31 = 0 0 0 1 1 1 1 1

-31 = 1 0 0 1 1 1 1 1


Example 3 :

What is the sign-and-magnitude representation of the decimal

numbers –31 and +31 if the basic unit is a word?

3110 = 111112

Unit is a word= 16 bits

Sign-and-Magnitude

+31 = 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1

-31 = 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1


Example 4 :

What is the decimal equivalent value of the sign-and -magnitude

binary sequence 1011 1001 ?

Sign-and-Magnitude

1 0 1 1 1 0 0 1

Sign is negative 11 10012 = 5710

1011 10012 = - 5710


Addition of 2 numbers in sign-and-magnitude :

using the usual conventions of binary arithmetic

if both have same sign : magnitude are added and the same sign

copied

if the sign different : number that has smaller magnitude is

subtracted from the larger one. The sign is copied from the larger

magnitude.

Sign-and-Magnitude


Example 5 :

What is the decimal value of the sum of the binary numbers 00100110 and 00011110

if they are represented in sign-and-magnitude ? Assume that the basic unit is the byte.

Same signs : magnitude are added

Sign-and-Magnitude

1 1 1 1 1 Sign is positive 100 01002 = 6810

0100 01002= +6810

0 0 1 0 0 1 1 0

+ 0 0 0 1 1 1 1 0

0 1 0 0 0 1 0 0


Example 6 :

What is the decimal value of the sum of the binary numbers 10110011 and 00010110

if they are represented in sign-and-magnitude ? Assume that the basic unit is the byte.

Different signs : Larger magnitude - smaller magnitude

Larger magnitude : 1011 0011

Smaller magnitude : 0001 0110

Sign-and-Magnitude

0 2 0 1 2

Sign is negative 111012 = 2910

1001 11012 = - 2910

1 0 1 1 0 0 1 1

- 0 0 0 1 0 1 1 0

1 0 0 1 1 1 0 1


In base 2, the largest digit is 1.

The 1’s complement is performed simply by changing:

every 0 1 and

every 1 0.

Also known as inversion

1’S Complement Convention


Most popular among computer manufacturers since it does not present any of the

problems of the sign-and-magnitude or 1’s complement.

Positive numbers : using similar procedure as sign-and-magnitude

Given n bits, the range of numbers that can be represented in 2’s complement is

(–(2n )) to (2n-1 –1)

Notice that the range of negative numbers is one larger than the range of the positive

values


0111 1111

(127)

0000 0000

(0)

1111 1111

(-1)

1000 0000

(-128)

( 2n-1 –1) - (2n)


To represent a negative number in this convention, follow the 3 steps process

below:

Step 1 : Express the absolute value of the number in binary

Step 2 : Change all 0s to 1s and vise versa 1’s complement

Step 3 : Add 1 to the binary number of step 2

Example 1 :

What is the 2’s complement representation of -23 ?

Example 2 :

What is the decimal positive value of the 2’s complement number 11100011 ?



Example 1 :

What is the 2’s complement representation of -23 ?


Step:

23 in binary 0 0 0 1 0 1 1 1

1’s (inverse) 1 1 1 0 1 0 0 0

2’s + 1

2’s complement representation

1 1 1 0 1 0 0 1


Example 2 :

What is the decimal positive value of the 2’s complement number 11100011 ?


Step:

2’s complement representation

1 1 1 0 0 0 1 1

1’s (inverse) 0 0 0 1 1 1 0 0

2’s + 1

Decimal positive value 0 0 0 1 1 1 0 1


The IEEE Floating Point Representation

2 problems with integers;

They can’t express fractions, AND

The range number is limited to the number of bits used.

An efficient way of storing fractions floating point method

involves splitting the fraction into two parts, an exponent and a

mantissa

Computer industry agreed upon a standard for the storage of floating

point numbers

the IEEE 754 standard; uses 32 bits of memory (single precision) or

64 bits (double precision)


IEEE short real : 32 bits (Exponent system = excess 127)

IEEE short real : 64 bits (Exponent system = excess 1023)

Total number

of bits :

1

8

23

Sign Biased exponent Mantissa (fraction)

Total number

of bits :

1

11

52

Sign Biased exponent Mantissa (fraction)



Convert 153.7510 to the IEEE floating point format.

Step 1 : convert into binary

Step 2 : put into 1.xxxx X 2y format

Step 3 : get the biased exponent

Step 4 : get the sign (from the question)

Step 5 : identify significand and mantissa

Step 6 : put into the IEEE single precision format

Step 7 : convert into hexadecimal



Convert to binary 153.75 = 1001 1001.112

Put into 1.xxxx X 2y = 1.0011 0011 12 x 27

Biased exponent

Exponent = 7

Biased exponent = 127 + y = 127 + 7 = 134

= 1000 01102 (8 bits)

Sign Sign = +ve (0)

Significand

Mantissa

Significand = 1.0011 0011 1

Mantissa = 001 1001 1100 0000 0000 0000 (23bits)

IEEE format

IEEE format :

Sign Biased Exponent Mantissa (23)

0 1000 0110 001 1001 1100 0000 0000 0000

Hexadecimal

0100 0011 0001 1001 1100 0000 0000 0000

4 3 1 9 C 0 0 0

= 4319 C000 h



Convert to binary 29 / 128 = 111012 / 27 = 111012 x 2-7

Put into 1.xxxx X 2y = 1.11012 x 24 x 2-7 = 1.11012 x 2-3

Biased exponent

Exponent = -3

Biased exponent = 127 + y = 127 + (-3) = 124

= 0111 11002 (8 bits)

Sign Sign = -ve = 1

Significand

Mantissa

Significand = 1.1101

Mantissa = 110 1000 0000 0000 0000 0000 (23bits)

IEEE format

IEEE format :


1 0111 1100 110 1000 0000 0000 0000 0000

Hexadecimal

1011 1110 0110 1000 0000 0000 0000 0000

B E 6 8 0 0 0 0

= BE68 0000 h

Eg: Convert -29 / 12810 The IEEE Floating Point Representation


Convert C2F0 000016 in IEEE floating point format to the decimal number.

Step 1 : in hexadecimal

Step 2 : convert into binary

Step 3 : put into the IEEE single precision format

Step 4 : get biased exponent

Step 5 : get the sign

Step 6 : identify significand and mantissa

Step 7 :put into 1.xxxx X 2y

Step 8 : identify actual number



In hexadecimal C2F0000016

Convert to binary C 2 F 0 0 0 0 0

1100 0010 1111 0000 0000 0000 0000 0000

Put into IEEE format

IEEE format :


1 1000 0101 111 0000 0000 0000 0000 0000

Get the biased exponent Biased Exponent:: 1000 0101 = 133

Exponent : 133 – 127 = 6

Sign 1 = -ve

Mantissa

Significand

111 0000 0000 0000 0000 0000

1.111

Put into 1.xxxx X 2y

1.111 x 26

=1111 0002

= 12010

The actual number - 120


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