1. Radix –Two Division
Most complex of the four arithmetic operations (addition, subtraction, multiplication and division).
Requires the most in terms of computation power compared to the other operations

2. Radix –Two Division The division can be represented as:
X = Q.D + R with R < D
Where X -> dividend
D -> divisor
R -> remainder
Q -> quotient
Assume that X, Q, R and D >0
X may occupy a double length register (such an accumulator holding partial results: X is a 32 bit register while R, D and Q are 16-bit registers.

3. Radix –Two Division
As such, we need to insure that Q < 2n-1 to prevent overflow.
If Q < 2n-1 and R < D then:
X < (2n-1+1)D
If Q < 2n-1 and R = 0 then:
X = QD < 2n-1D
If fractional division is performed, then to prevent overflow condition  X < D.
Division is obtained by a series of subtractions and shifts.

4. Sequential Division Algorithm
At each step i, the value (2 x remainder) is compared to the divisor D. If this value is the larger of the two, then the quotient bit qi is set to 1, otherwise set to 0.
ri = 2ri-1 - qiD , i = 1, 2, …m, Equ(1)
where m is the number of fractional bits.
where ri is the new remainder.
where ri-1 is the previous remainder.
where ro is set to be the dividend X
How the above equation will perform the division?

5. Sequential Division Algorithm
Let rm be the remainder in the last step then:
rm = 2rm-1 – qm.D
Substituting recursively using Equ(1) we get:
rm = 2(2rm-2 – qm-1.D) - qm.D = …..
rm = 2mr0 – (qm + 2qm-1 + … + 2m-1q1).D
Substituting r0 by the dividend or X and dividing both sides by 2m we get:
2-mrm = X – (q q … + qm2-m).D
2-mrm = X – Q.D

6. Division: Example Let X = 0.10000 (0.5), D = 0.110 (3/4)
X < D is satisfied -> no overflow
Remainder = r32-3 (1/32)
r0 = X
2 r0
Add -D
0.100
01.000
11.010
000
Set q1 = 1
r1 = 2r0 – D
2 r1
r2 = 2r1
2 r2
00.010
00.100 00
Set q2 = 0
Set q3 = 1
r3 = 2r2 – D
Quotient = 0. q1 q2 q3
Quotient = (5/8)
Note for signed division, get the absolute values, perform division and apply sign at the end