This document discusses floating point number representation using 8-bit notation. It explains that floating point numbers represent both the integer and fractional parts of a number using a sign bit, exponent field, and mantissa. The radix point separates the integer and fractional parts. Through examples, it demonstrates how to represent given decimal numbers like 2.75 and 1.125 in 8-bit floating point notation by determining the sign, exponent, and mantissa values. Representing fractional decimal numbers in binary requires noting the position of the radix or decimal point.

1. Why do we use this notation?
•Unlike the storage of integers, representing fractional parts
requires representing the number in binary and also the position of
the radix point.
What does radix point mean?
•A radix point is the symbol used in numerical representations to
separate the integer part of a number (to the left of the radix point)
from its fractional part (to the right of the radix point).
Floating point notation

2. 8 bit floating point representation:
————————
Sign bit
Exponent(excess
notation)
Mantissa
•If sign bit(msb) equals 1 then the number is negative and if it is equal 0 then the
number is negative.
•Suppose we have the number 01101011 and we want to represent in using this
notation:
•1st  sign bit is 0 so the number is positive.
•2nd  exponent is 110 which equals to “2” in excess 3 notation.

3. 3rd  we have to extract the mantissa and put a radix point on its left side( .1011)
4th  since the exponent is 2 “+ve” we will move two points to the right(if its –ve
2 then we move two points to the left).
5th  our mantissa becomes 10.11 which is the binary representation of “2.75”. (if
the number was 11101011 then the value would be “-2.75”.
Try representing this number 00111100
Answer 0.375

4. What if it’s the other way around?
How to represent 1.125
1st  we convert it to binary( 1.001)
2nd we copy the bit pattern into the mantissa field from left to right
starting with the leftmost “1”
1.001
————————
Mantissa
1 0 0 1

5. 3rdwe have to fill out the exponent field, so we imagine putting a radix at the
left of our mantissa (.1001) for it to get back to (1.001) we have to move to the
right one bit  this means that out exponent should be a +ve one in “excess
notation” which is 101.
4thour number 1.125 was positive so sign bit is 0.
so our number is 0 101 1001
Note:
when representing a number like 0.375 which is 0.11 in binary, when copying the
mantissa we start filling it with the left most one so the mantissa will be 1100
it will not be  0110 .

6. I hope this was helpful
Abdelrahman