# binary division

Discussion in 'Math' started by manikanta.mashetti, Jan 21, 2014.

1. ### manikanta.mashetti Thread Starter New Member

Jan 13, 2014
3
0
how to divide two binary numbers .if numerator is less than the denominator.

eg: 6/9 =0110/1001 .

how to do for the above eg.Is there any algorithms for this ..

2. ### atferrari AAC Fanatic!

Jan 6, 2004
2,674
785
Multiplying first, 6 times 10 and then dividing the result of your division by 10?

Last edited: Jan 21, 2014
3. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Same way we do it in decimal. Multiply the number 0110 with a straightforward count until it exceeds 1001, then take the number under that. You subtract (just like the decimal logarithm) the rest is the remainder.

Binary
.....___1r11
0110/1001
.....0110
..... 011

Decimal

..1r3
6/9
..6
..3

4. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,447
1,628
The only good thing about binary division is you don't have to guess at the number to multiply the divisor by: it's either larger then the remaining dividend or not, that's one or zero.

Code ( (Unknown Language)):
1.         0.10101
2.      ------
3. 1001/0110.0
4.       100 1
5.       -----
6.         1 10
7.      -     0
8.         ----
9.         1 100
10.       - 1 001
11.         -----
12.            110
13.        -     0
14.           ----
15.            1100
16.          - 1001
17.            ----
18.              11   and so on...
.

(Psst... Bill... you inverted the numbers.)

5. ### MrChips Moderator

Oct 2, 2009
12,657
3,461
Shift the dividend one bit to the left into a 4-bit register.
Subtract the divisor as needed,

0000 <- 0110
0000 <- 1100 Q = 0
0001 <- 1000 Q = 0
0011 <- 0000 Q = 0
0110 <- 0000 Q = 0

fractional part follows:

1100 <- 0000 Q = 1, R = (1100 - 1001) = 0011
0110 <- 0000 Q = 0
1100 <- 0000 Q = 1, etc

Result is 0000.101010

Last edited: Jan 24, 2014
6. ### WBahn Moderator

Mar 31, 2012
18,096
4,920
The first thing you need to do is decide what representation you are going to use for your result? quotient/remainder? Fixed point? Four bits? Eight bits? Where will the radix be? All of these affect the algorithm in ways small and great.