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Binary Operations

Date: 04/07/97 at 21:23:44
From: Matt
Subject: Binary Addition, Subtraction, Multiplication, Division

Dr. Math,

Can you explain binary addition, subtraction, multiplication, and 
division in a non-complex manner?


Date: 07/09/97 at 02:00:06
From: Doctor Mandel
Subject: Re: Binary Addition, Subtraction, Multiplication, Division

Hi Matt,


Binary addition is the simplest of the binary operations, so let's 
start there. To add two binary numbers, you only need to know three 

   0 + 0 is 0 carry 0 (pretty easy) 
   0 + 1 is 1 carry 0
   1 + 1 is 0 carry 1.  

The rest of the operation is just as in base ten.  

For example, let's add 1010 (10 in base ten) and 1011 (11 in base 

   + 1011

   0 + 1 is 1 carry 0
   1 + 1 is 0 carry 1, 
   0 + 0 + 1 (carried over) is 1 carry 0
   1 + 1 is 0 carry 1
   1 + 0 is 1 carry 0

so the final sum is 10101, 21 in base ten, which is correct.


The next easiest operation is multiplication. We'll set up the problem 
like one in base ten, with the larger number above the smaller, and 
for each digit on the bottom that you go to the left, you move that 
answer one digit to the left. 

For multiplication you just have to remember that 1 times a number is 
that number and 0 times a number is 0.  Once you have all of the 
little multiplications done, you add up the products.  

Let's try multiplying 101 (5) and 10 (2): 

    x 10

From the bottom right, 0 times 101 is 0 (because that is the product
of any number and zero) and 1 times 101 is 101 moved over 1 column to
the left (because the 1 was moved over one column to the left),
giving 0 plus 1010, or 1010, which is in base 10 is 10, and correct.


Next, subtraction. This is just like regular subtraction, with 
borrowing.  Let's try 1101 (13 in base 10) minus 110 (6 in base 10):

   - 110

From the right, 1 minus 0 is 1. Next, 0 minus 1 is impossible, so you 
borrow from the next number to the left, making the full top number 
10(10)1; then 10 minus 1 is 1. Next 0 minus 1 is impossible again, so 
we go through the same thing, making the number 0(10)01, and 10 minus 
1 is again 1. Thus the final result is 111, or 7 in base 10, which is 


This is mostly trial and error. First you take the first digit of the 
number you are dividing and see whether the number you are dividing 
into it goes in (in other words is greater than the number you are 
dividing by, because in base 2, if it goes in more than once, the 
previous step must have been incorrect). If it does, great - you can 
move on to the next digit by itself; if it doesn't go in, see whether 
the number goes into the first two digits. Keep going on like this 
until it does go in.  

Let's divide 10110 (22) by 111 (7). First, see if 111 goes into the 
first digit, 1. It doesn't, so try the first two.  111 doesn't go into 
10 or 101, but it does go into 1011 once, so the first digit of the 
answer is 1.  

Now you subtract 111 from the 1011, leaving you with 100, and start 
over saying "does 111 go into 100?"  Since it doesn't, you add the 
next digit of the number you are dividing: bring down the 0 from the 
end of 10110 to make the number you divide 1000, which 111 does go 
into. Subtracting 111 from 1000, we get 1 with 1 left over. 111 does 
not go into 1 so that is our remainder, and the final answer is 11 
remainder 1, which is 3 remainder 1, which is correct.  

You could also continue this process indefinitely after the decimal 
point, because 1/7 is a repeating decimal so you will always have a 

I hope these directions have been helpful.

-Doctor Mandel,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Number Theory

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