1. Floating Point (FLP) Representation
A Floating Point value:
f = m*r**e
Where: m – mantissa or fractional
r – base or radix, usually r = 2
e - exponent

2. Normalization Normalized value: 0.1011 Unnormalized value: 0.001011
=0.1011*2**-2
Value of normalized mantissa:
0.5<=m<1

3. FLP Format sign exponent mantissa sign: 0 + 1 -
1 -
Biased exponent: assume exponent q bits
-2**(q-1)<=e<=2**(q-1)-1 add bias:
+2**(q-1) to all sides, get:
0 <= eb <= 2**q -1
e – true exponent;
eb – biased exponent

4. Example f = -0.5078125*2**-2 Assume a 32-bit format:
sign – 1 bit, exponent – 10 bits (q=10),
mantissa – 21 bits
q-1 = 9, b = bias = 2**9 = 512,
e = -2, eb = e + b = = 510
f representation:
………0
since 0.5=0.1, =2**-7

5. Range of representation
In fixed point, the largest number representable in 32 bits:
2**31-1 approximately equal 10**9
In the previous 32-bit format, the largest number representable: (1-2**-21)*2**511
Approximately equal 10**153
The smallest: 0.5*2**-512
If a number falls above the largest, we have an overflow, if below the smallest, we have an underflow.

6. IEEE FLP Standard 754 1985 Single precision: 32 bits
Double precision: 64 bits
Single Precision.
f = +- 1.M*2**(E’-127) where:
M – fractional, E’ – biased exponent,
bias = 127
Format: sign: 1 bit, exponent – 8 bits,
fractional – 23 bits.
True exponent E = E’ – 127
0 < E’ < 255

7. Normalized single precision
1.xxxxxx
The 1 before the binary point is not stored, but assumed to exist.
Example: convert 5.25 to single precision representation.
5.25 = not normalized.
Normalized: *2**2
True exponent E = 2,
Biased exponent E’ = E = 129, thus:
…………0

8. Double precision Value represented: +- 1.M*2**(E’-1023)
Format: sign: 1 bit, exponent 11 bits,
fractional 52bits.
Bias = 1023
Maximal number represented in single precision, approximately: 10**38
In double precision: approximately 10**308

9. Precision
Increasing the exponent field, increases the range, but then, the fractional is decreased, decreasing the precision.
Suppose we want to receive a precision of n decimal digits. How many bits x we need in the fractional?
2**x = 10**n, take decimal log on both sides: xlog2 = n; x=n/log2=n/0.301
For n=7, need 7/0.301=23.3, 24 bits.
Achieved in single precision standard, since M has 23 bit and there is 1., not stored but existing.

10. Extended Precision 80 bits
Not a part of IEEE standard. Used primarily in Intel processors.
Exponent 15 bits.
Fractional 64 bits.
This is why FLP registers in Intel processors are 80 and not 64 bits.
Its precision is 19 decimal digits.

11. FLP Computation Given 2 FLP values: X=Xm*2**Xe; Y=Ym*2**Ye; Xe<Ye
X+-Y = (Xm*2**(Xe-Ye)+-Ym)*2**Ye
X*Y = Xm*Ym*2**(Xe+Ye)
X/Y = (Xm/Ym)*2**(Xe-Ye)