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Representation of real number

Representation of real number. Presented by: Pawan yadav Puneet vinayak. Contents:-. Floating Point Numbers Decimal  Binary conversion Floating point representation Mantissa Exponent Normalization IEEE Floating Point Representation Floating point airhtematic

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Representation of real number

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  1. Representation of real number Presented by: Pawanyadav Puneetvinayak

  2. Contents:- • Floating Point Numbers • Decimal  Binary conversion • Floating point representation • Mantissa • Exponent • Normalization • IEEE Floating Point Representation • Floating point airhtematic • Error in floating point airthematic

  3. Floating Point Numbers • In computer science real number is also called floating point number. • In the decimal system, a decimal point (radixpoint) separates the whole numbers from the fractional part Examples: 37.25 ( whole=37, fraction = 25) 123.567 10.12345678

  4. Floating Point Numbers For example, 37.25 can be analyzed as: 101 100 10-1 10-2 Tens Units Tenths Hundredths 3 7 2 5 37.25 = 3 x 10 + 7 x 1 + 2 x 1/10 + 5 x 1/100

  5. Binary Equivalent In the binary representation of a floating point number the column values will be as follows: …26 25 24 23 22 21 20 . 2-1 2-2 2-3 2-4 … …64 32 16 8 4 2 1 . 1/2 1/4 1/8 1/16 … … 64 32 16 8 4 2 1 . .5 .25 .125 .0625…

  6. Decimal  Binary conversion Repeatedly multiply fraction by two until fraction becomes zero. 0.8125  1.625 0.625 1.25 0.25  0.5 0.5  1.0

  7. Scientific notation of floating numbers • Decimal: -123,000,000,000,000  -1.23 × 1014 0.000 000 000 000 000 123  +1.23× 10-16 • Binary: 110 1100 0000 0000  1.1011× 214 -0.0000 0000 0000 0001 1011  -1.1101 × 2-16

  8. Floating Point Number Representation • If x is a real number then its normal form representation is: x =f•Base E where f : mantissa E: exponent exponent Example: 125.3210 =0.12532•103 mantissa - 125.3210 = - 0.12532 • 103 0.054610 = 0.546 • 10 –1

  9. Normalized and Unnormalized

  10. Normalization Process

  11. Floating Point Format for Binary Numbers

  12. IEEE Floating Point Representation • more exponent bits  greater range • more significant bits  greater accuracy

  13. IEEE Floating Point Representation • The first, or leftmost, field of our floating point representation will be the sign bit: • 0 for a positive number, • 1 for a negative number.

  14. IEEE Floating Point Representation • The second field of the floating point number will be the exponent. • Since we must be able to represent both positive and negative exponents, we will use a convention which uses a value known as a bias of 127 to determine the representation of the exponent. • An exponent of 5 is therefore stored as 127 + 5 or 132; • an exponent of -5 is stored as 127 + (-5) OR 122. • The biased exponent, the value actually stored, will range from 0 through 255. This is the range of values that can be represented by 8-bit, unsigned binary numbers.

  15. IEEE Floating Point Representation • The mantissa is the set of 0’s and 1’s to the left of the radix point of the normalized (when the digit to the left of the radix point is 1) binary number. • ex:1.00101 X 23 • The mantissa is stored in a 23 bit field,

  16. Normalizing Numbers Example: 134.1510 = 0.13415 x 103 0.002110 = 0.21 x 10-2101.11B = .1011 x 23 or 1.011 x 22(hidden1)0.011B = .11 x 2-1 or 1.1 x 2-2(hidden1)AB.CDH= .ABCD x 162 0.00ACH= .AC x 16-2 • Note that the concept of a hidden 1 only applied to binary.

  17. Converting decimal floating point values to stored IEEE standard values. Example: Find the IEEE FP representation of 40.15625. Step 1. Compute the binary equivalent of the whole part and the fractional part. ( convert 40 and .15625. to their binary equivalents) 40.1562510= 101000.001012

  18. Converting decimal floating point values to stored IEEE standard values. Step 2. Normalize the number by moving the decimal point to the right of the leftmost one. 101000.00101 = 1.0100000101 x 25 Step 3. Convert the exponent to a biased exponent 127 + 5 = 132 ==> 13210 = 100001002

  19. Converting decimal floating point values to stored IEEE standard values. Step 4. Store the results from above Sign Exponent (from step 3) Mantissa ( from step 2) 0 10000100 01000001010 .. 0

  20. Convert 10.37 to single precision floating point

  21. Floating point arithmetic

  22. Floating-Point Addition Assume 4 decimal digit for mantissa

  23. floating point subtraction(using 4 digit mantissa) • Addition must be of terms of the same scale: • 0.2361106 - 0.1455104 • 0.2361106 - 0.001455106 {both106} • (0.2361 - 0.001455) 106 • 0.147861 106 • 0.234645 106 • 0.2346  106 {4 digit mantissa}

  24. Real Number Multiplication(using 4 digit mantissa) • Multiplication problem is in the mantissa • (0.2361102)  (0.1455 104) • 0.2361  0.1455 102+4 {add indices} • 0.03435255 106 = 0.3435255 105 • 0.3435 105 {4 digit mantissa} Notice that multiplication must work from the largest digit downwards since at some point the number is going to have to be truncated.

  25. Real Number division(using 4 digit mantissa) • (0.2361102) /(0.1455 104) • (0.2361/0.1455)102-4{sub indices} • 1.6226804 10-2 = 0.3435255 105 • 0.16226804 10-1 • 0.162310-1{4 digit mantissa}

  26. Errors in floating point arithmetic • Round off error Ex- 5.6999=5.7 7.238=7.24 • Truncation error 4.67444444=4.674 5.45676767=5.4567

  27. thanks

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