1. Binary Real Numbers

2. Introduction Computers must be able to represent real numbers (numbers w/ fractions) Two different ways:  Fixed-point  Floating-point NOTE: Everything in binary uses powers of two

3. Decimal Review Digits to the right of the decimal point correspond to negative powers of 10 10 2 10 1 10 0.10 -1 10 -2 100101.0.10.01

4. Binary Fractions 2 -1 2 -2 2 -3 2 -4 2 -5 2 -6 0.50.250.1250.06250.03125 0.0015625 1/21/41/81/161/321/64

5. Fixed Point Notation 1. Multiply each 1 by the corresponding power of 2 2. Add up the resulting powers of 2 Example: 11.01 2 = 2 + 1 + ¼ = 3.25 10 00111.010 2 = 4 + 2 + 1 + ¼ = 7.25 10

6. Floating-Point Notation Floating-point notation is essentially the computer’s way of storing a number that has been normalized 3 different parts of any number:  Mantissa: normalized number  Exponent: power to which the base is raised  Sign: of both mantissa and exponent Decimal Example: 12.5 = 0.125 x 10 2  normalized!

7. Normalization Steps 1. Beginning with a fixed point number 2. Normalize the number such that the radix point (decimal point) is all the way to the left (produces the mantissa) 3. Multiply the resulting number by the base raised to an exponent

8. Floating-Point Example What is 12.5 in floating-point representation? 1. Convert 12.5 to binary fixed point 12.5 10 = 1100.1 2 2. Normalize the number by moving the radix point, producing the mantissa 1100.1 2 = 0.11001 * 2 4 3. Fill in the bits for each of the three parts of any real number: 1. Sign (2 bits) 2. Mantissa (# bits varies) 3. Exponent (# bits varies) 4. NOTE: 2’s complement may be applied to the mantissa or the exponent if either are negative

9. Placing the Bits Assume you have the following:  1 bit for the mantissa sign  8 bits for the mantissa  1 bit for the exponent sign  6 bits for the exponent S M M M M M M M M M S E E E E E E E Example: 0.11001 * 2 4  0 00011001 0 000100

10. Another Example Convert -12.5 10 to binary: 1. Convert 12.5 to fixed point  01100.1 2. Normalize  0.11001 * 2 4 3. Convert exponent base to binary: 4  0100 4. 2s complement the mantissa by flipping bits and adding 1:011001  100111 5. Final number  1 00111 0100

11. Upper & Lower Bounds Assume you have the following:  1 bit for the mantissa sign  8 bits for the mantissa  1 bit for the exponent sign  6 bits for the exponent What is the upper bound for the floating-point number? What is the lower bound for the floating-point number? What happens if we convert a floating-point number to an integer?

12. Integers vs. Floating-point integers:  smaller range than floating-point  all numbers within the range are 100% accurate floating-point  large range of numbers  not all numbers within the range can be represented accurately Example: 2.9999999999999 repeating

13. Possible Errors truncation error  round off errors using floating-point numbers because not all real numbers can be represented accurately overflow error  attempting to represent a number that is greater than the upper bound for the given number of bits underflow error  attempting to represent a number that is less than the lower bound for the given number of bits