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# boolean algebra - multiplication and division

Discussion in 'Homework Help' started by arwin, Jan 31, 2008.

1. ### arwin Thread Starter Guest

about arithmetic multiplication and division, how do we implement it in OR, AND gates.Although I know we can implement it using addition and subtraction respectively like for example:
2 x 3 = 2 + 2 + 2, we have 2 addition
6/2 = 6 - 2=4 - 2= 2 - 2; we have 3 subtraction.

But its too long if we have two or more variables multiplying together.

a x b x c. Is there a direct way of writing multiplication and division in boolean way.

Thanking you a lot in advance for spending time to read this.

2. ### Dave Retired Moderator

Nov 17, 2003
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172
This thread has been moved to a more appropriate forum.

We have made it possible for persons to leave commentary and feedback on the All About Circuits e-book, site and forums in an anonymous fashion, but we require registration to participate in the other board functions, such as technical discussion with other members and requesting information.

Dave

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
172
Multiplication is a valid operator in Boolean algebra, as it is in normal algebra (essentially the AND function). For complex multiplication, the method is as you have pointed out, a compounded addition. Division is not a valid operator in Boolean algebra and, as you point out, is nothing more than a compounded subtraction. Strangely, subtraction is not wholly valid as an operation - subtraction is achieved by addition of a negative number.

The reality is that modern computing technology that relies so much on the fundamentals of Boolean logic is so fast that the effect of a long winded method such as this is negligible.

Dave

4. ### Papabravo Expert

Feb 24, 2006
12,285
2,724
One way to do multiplcation is to "shift and add". It is the binary equivalent of the way we learned to do decimal multiplication back in the 3rd grade.

There are similar algorithms for division. In both cases the calculation time is proportional to the number of bits.

Neither of these methods is combinatorial. That's a fancy way of saying that you can't write a Boolean expression to represent them. They are examples of sequential circuits that involve the use of memory. Sequential circuits have inputs, outputs, and internal memory to keep track of the present state of the circuit. At each clock pulse they evaluate the inputs and the present state to compute the next state and the outputs.

One additional thing that might interest you is "Wallace Trees"

5. ### arwin New Member

Feb 3, 2008
1
0
Do you know exactly how many "shift and add" will be needed for the multiplication of a number?

6. ### Papabravo Expert

Feb 24, 2006
12,285
2,724
Yes. It is equal to the number of bits in the multiplier. The number of bits in the multiplicand is irrelevant.