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ITEC 1011 Introduction to Information Technologies

4. Floating Point Numbers

Chapt. 5

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Exponential Notation

The representations differ

in that the decimal place –

the “point !! “floats to

the left or right "#ith the

appropriate ad$ustment in

the exponent%.

 p. &''

( The follo#ing are e)ui*alentrepresentations of &+',4

123,400.0 x 10-2

 12,340.0 x 10-1

  1,234.0 x 100

  123.4 x 101

  12.34 x 102

  1.234 x 103

  0.1234 x 104

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Parts of a Floating Point Number 

-0.9876 x 10-3

 p. &',

-ign of 

mantissa

ocation of 

decimal point/antissa

Exponent

-ign of 

exponent

0ase

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1EEE 254 -tandard

( /ost common standard for representing floating point numbers

( -ingle precision3 ,' bits+ consisting of...( -ign bit "& bit%

( Exponent " bits%( /antissa "', bits%

( ouble precision3 64 bits+ consisting of7( -ign bit "& bit%

( Exponent "&& bits%( /antissa "5' bits%

 p. &,,

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-ingle Precision Format

,' bits

/antissa "', bits%

Exponent " bits%

-ign of mantissa "& bit%

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 Normali8ation

( The mantissa is normalized ( 9as an implied decimal place on left

( 9as an implied “& on left of the decimal

 place

( E.g.+

( /antissa→

( :epresents7

  10100000000000000000000

1.1012 = 1.625

10

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Excess Notation

( To include ;*e and –*e exponents+ “excessnotation is used

( -ingle precision3 excess &'2

( ouble precision3 excess &

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Example

( -ingle precision

0 10000010 11000000000000000000000

&.&&'

&,< – &'2 = ,

< = positi*e mantissa

;&.&&' x ', = &&&

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9exadecimal

( 1t is con*enient and common to representthe original floating point number in

hexadecimal

( The preceding example7

0 10000010 11000000000000000000000

4 1 6 0 0 0 0 0

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Con*erting from Floating Point

( E.g.+ >hat decimal *alue is represented b?the follo#ing ,'!bit floating point number@

C17B000016

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( -tep &( Express in binar? and find -+ E+ and /

C17B000016=

1 10000010 111101100000000000000002

- E /

& = negati*e

< = positi*e

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( -tep '( Find “real exponent+ n

(   n = E – &'2

= &

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( -tep ,( Put -+ /+ and n together to form binar? result

( "onAt forget the implied “&. on the left of the

mantissa.%

-1.11110112x 2n =

 

-1.11110112x 23 = 

-1111.10112

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( -tep 4( Express result in decimal

-1111.10112

!&5 '!& =

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Con*erting to Floating Point

( E.g.+ Express ,6.56'5&

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( -tep &( Express original *alue in binar?

36.562510 =

100100.10012

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( -tep '(  Normali8e

100100.10012=

1.0010010012 x 25

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( -tep ,( etermine -+ E+ and /

+1.0010010012 x 25

- = < "because the *alue is positi*e%

/-   n E = n ; &'2

= 5 ; &'2

= &,'

= 100001002

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( -tep 4( Put -+ E+ and / together to form ,'!bit binar?

result

0 10000100 001001001000000000000002

- E /

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( -tep 5( Express in hexadecimal

0 10000100 001001001000000000000002 =

0100 0010 0001 0010 0100 0000 0000 00002 =

  4 2 1 2 4 0 0 016

Bns#er3 4'&'4

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Than ?ou