Arithmetic in Other Bases

Date: 02/25/2002 at 09:24:11
From: Linda 
Subject: Arithmetic in other bases

It is easy to perform the four basic algorithms in base 10 form.  But 
we have recently been asked in my math class to perform arithmetic in 
other bases. For example: 5430/13 (base 6). What I need to know is how 
to write out the steps for addition, subtraction, multiplication, and 
division in other bases. Please help or let me know where I can go for 
help.

Date: 02/27/2002 at 01:52:45
From: Doctor Mike
Subject: Re: Arithmetic in other bases

Hello Linda,
  
It's exactly the same, only different. If you follow the familiar
processes, but keep in mind the consequences of the digits meaning
powers of six instead of powers of ten, you will have it. Let's
use your example for starters.
  
Remember EVERYTHING is in base 6. To divide 13 into 5430 by the usual 
long division, you first say that since 13 cannot go into 5, you need 
to find out how many times 13 can go into 54. My first guess was 
4 times. But, (in base 6) 4 times 13 is 100, which is too big, so it 
must go 3 times. So, 3 is the first digit of the quotient, and what 
you write under the 54 is 3 times 13, which is 43. This really looks 
strange, because we are all VERY used to multiplication tables in 
base 10, so facts like 3*13 = 43 or 5*5 = 41 take some getting used 
to.
  
Getting back to the division, write 43 under the 54 and subtract to 
get 11. Bring down the next digit from the dividend to make it 113.  
If you remember the "false start" we made, namely, that 4*13 = 100, it 
should be pretty easy to do in your head that 5*13 is 113. That means 
that 5 is the next digit in the quotient, and 5*13 = 113 is written 
down below. Subtract 113 from 113 to get zero. Bringing down the final 
digit 0 from the dividend shows that we are done, and the division is 
exact, with quotient = 350 and no remainder.  
  
I will try to do this semi-pictorally to give you an idea.
   
              3 5 0
           ----------
      1 3 / 5 4 3 0
            4 3    
           -----
            1 1 3
            1 1 3
           -------
                0 0
  
One more example: Add 543 and 321. First add the 3 and the 1 in the 
units place to get 4. Write that in the units place in the answer.  
Next, add the 4 and the 2 in the six's place to get 10, so write 0 in 
the six's place in the answer, and "carry" the 1 to the six squared 
place. Add the 5 and the 3 in the six squared place to get 12, and 
then add on the carry to get 13, and write that all down below to get 
the full answer 1304.  See?
  
Visually, 

         1
         5 4 3
       + 3 2 1
        -------
       1 3 0 4 

I hope this helps. To be honest with you, even as a mathematician 
I do NOT have the base 6 multiplication tables memorized. To do the 
5 squared calculation, I first remembered the base 10 fact that 
5 squared is 25 in base ten, and then figured out that in base 10 we
have that 25 = 4*6 + 1 so that 25 base 10 equals 41 base 6. See?
As easy as pi, right?
   
- Doctor Mike, The Math Forum
  http://mathforum.org/dr.math/   

Date: 02/27/2002 at 08:06:49
From: Linda 
Subject: Arithmetic in other bases

Thank you very much for your help. You have helped out a lot!