1. Floating Point Number system corresponding to the decimal notation
1,837 * 10
significand exponent
A great number of corresponding binary standards exists.
There is one common standard:
IEEE (IEC 559) 4

2. IEEE 754-1985 Number representations: Single precision (32 bits)
sign: 1 bit
exponent: 8 bits
fraction: 23 bits
Double precision (64 bits)
exponent: 11 bits
fraction: 52 bits

3. Single Precision Format1 8 23
Sign S S E M
Exponent E:
excess 127
binary integer
Mantissa M:
normalized binary significand w/ hidden integer bit: 1.M
Excess 127; actual exponent is
e = E - 127
N = (-1)S * (1.M [bit-string])*2e

4. Example 1 S E M 1
e = E - 127
e = = -1
N = (-1)1 * (1.1 [bit-string]) *2-1
N = -1 * 0.11 [bit-string]
N = -1 * (2-1 * *1)
N = -1 * (0.5* *1) = -0.75

5. Single Precision Range
Magnitude of numbers that can be
represented is in the range:
2-126 *(1.0) to *(2-223)
which is approximately:
1.8* to *1038

6. IEEE 754-1985 Single Precision (32 bits)
Fraction part: 23 bits; 0 x < 1
Significand: 1 + fraction part. “1” is not stored; “hidden bit”. Corresponds to 7 decimal digits.
Exponent: 127 added to the exponent. Corresponds to the range to 10 39
Double Precision (64 bits)
Fraction part: 52 bits; 0 x < 1
Significand: 1 + fraction part. “1” is not stored; “hidden bit”. Corresponds to 16 decimal digits.
Exponent: 1023 added to the exponent; Corresponds to the range to

7. IEEE 754-1985 Special features:
Correct rounding of “halfway” result (to even number).
Includes special values:
NaN Not a number
 Infinity
-  - Infinity
Uses denormal number to represent numbers less than 2 -E min
Rounds to nearest by default; Three other rounding modes exist.
Sophisticated exception handling.

8. Add / Sub (s1 * 2e1) +/- (s2 * 2 e2 ) = (s1 +/- s2) * 2 e3 = s3 * 2 e3
s = 1.s, the hidden bit is used during the operation.
1: Shift summands so they have the same exponent:
e.g., if e2 < e1: shift s2 right and increment e2 until e1 = e2
2: Add/Sub significands using the sign bits for s1 and s2.
set sign bit accordingly for the result.
3: Normalize result (sign bit kept separate):
shift s3 left and decrement e3 until MSB = 1.
4: Round s3 correctly.
more than 23 / 52 bits is used internally for the addition.

9. Multiplication (s1 * 2e1) * (s2 * 2 e2 ) = s1 * s2 * 2 e1+e2
so, multiply significands and add exponents.
Problem:
Significand coded in sign & magnitude;
use unsigned multiplication and take care of sign.
Round 2n bits significand to n bits significand.
Normalize result, compute new exponent with respect to bias.

10. Division (s1 * 2e1 ) / (s2 * 2 e2 ) = (s1 / s2) * 2 e1-e2
so, divide significands and subtract exponents
Problem:
Significand coded in signed- magnitude - use unsigned division (different algoritms exists) and take care of sign
Round n + 2 (guard and round) bits significand to n bits significand
Compute new exponent with respect to bias