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Balanced Ternary Notation

Date: 04/06/2002 at 15:50:27
From: Thomas
Subject: Weird base three

The place values in a base 3 number system are powers of 3, namely 1, 
3, 9, 27, 81 and so on. Suppose the digits in a base 3 number system 
are 1, 0 and -1, where -1 means negative one. The base 10 number 35 is 
written as 110-1  in this base 3 system. (In our worksheet the one has 
the line over it.)

  a) Write this base 3 notation for the base 10 numbers 1 through 35.

  b) Make addition and multiplication tables for this base 3 system. 
     (Use the single digits -1, 0, 1, only)

  c) Show how to multiply 100-110 by 1-10-11 in this base three number 
     system. Use the multiplication and addition tables from (a) and 
     (b). Use the standard multiplication algorithm.

  d) Prove your answer is correct by changing all numerals to ba.


Date: 04/06/2002 at 22:23:16
From: Doctor Peterson
Subject: Re: Weird base three

Hi, Thomas.

This is called "balanced ternary notation," and is very interesting to 
play with; in particular, it makes negative numbers incredibly easy to 
handle. I suggest you do just that: don't expect to find the answers 
directly, but experiment until you get a feel for how it works. Start 
with the first question, but do it backward: write out a bunch of 
numbers in balanced ternary, and work out what they mean. For example,

     0 = 0
     1 = 1
    -1 = -1
   1 0 = 1*3 + 0 = 3
   1 1 = 1*3 + 1 = 4
   1-1 = 1*3 + -1 = 2

Once you've found the numbers from 1 through 35 (and in fact, -35 
through 35), put them in order and see if you can recognize the 
pattern. Then, figure out WHY it works that way. You should be able to 
find a simple way to convert to this notation; I once found two rather 
different ways to do it. You may find yet another.

By now you will have a feel for how the notation operates, and you can 
easily write addition and multiplication tables. Actually, you could 
almost have done that first, because the tables are pretty small; but 
it's best to wait until you know what you are working with.

Actually doing a multiplication problem will take a little more work; 
multiplication in any base follows exactly the same algorithm, except 
for the details of how you do "carries." I suggest you first try some 
simple additions, and only when you can do that correctly move on to 
simple multiplication, and finally take the big challenge.

This is a wonderful project, discovering something you've never seen 
before as if you were the mathematician who discovered it. Have fun 
with it; don't feel you have to find the answers somewhere outside 
yourself. I'll be happy to guide you along if you get stuck; just show 
me what you have found so I can see where you're having trouble.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 04/08/2002 at 11:41:19
From: Thomas   
Subject: Weird base three

Hello Dr. Peterson,

First of all thank you for your prompt answer to my math problem. I 
followed your advice and made a table using base 3 for the numbers -5 
to 35. I noticed a pattern. The column of the 1s follows 1, -1 0.  The 
column of the 3s: 1,1,1, -1,-1,-1, 0, 0,0. The column of the 9s has 
nine 1s, nine -1s, and nine 0s. Consequently the next column, which is 
the column of the 27s, will have 27 ones, 27 -1s, and 27 0s.

I made the addition and multiplication tables, and this is what I got:

Addition:
 +   -1   0   1
-1 -1 1  -1   0
 0   -1   0   1
 1    0   1   1-1

Multiplication:
 x   -1   0   1
-1    1   0   -1
 0    0   0   0
 1   -1   0   1

Doing the multiplication proved to be very difficult, and there is 
where I have a big problem:

                                 1   0   0  -1   1   0
                               x     1  -1   0  -1   1
                              ------------------------
                                 1   0   0  -1   1   0
                             -1  0   0   1  -1   0
                          0   0  0   0   0   0
                      -1  0   0  1  -1   0
                    1  0  0  -1  1   0
                  -------------------------------------
                    1 -1  0  -2  3  -1   1  -2   1   0


                    1 -1  0 -11  10 -1   1  -11  1   0



1(3*9)+ -1(3*8)+0(3*7)+-1 1(3*6)+1 0(3*5)+ -1(3*4)+1(3*3)+ -1 
1(3*2)+1(3*1)+0(3*0) = 
19683 - 6561 - 729  + 0 - 81 + 27 - 9 + 3 = 12,333

When I convert the numbers in base three to base ten the answer I get 
is different by 9:

1 0 0 -1 1 0 = 1(3*5)+0(3*4)+0(3*3)+-1(3*2)+1(3*1)+0(3*0) = 237

1 -1 0 -1 1  =  1(3*4) + -1(3*3) + 0(3*2) + -1(3*1) + 1(3*0) = 52

237 x 52 = 12, 324

Please tell me where I am wrong.  Thanks a million!
Thomas


Date: 04/08/2002 at 12:20:08
From: Doctor Peterson
Subject: Re: Weird base three

Hi, Thomas.

I'm going to use T for -1 since that looks like 1 with a bar above it, 
and only takes one character space so it won't keep confusing me. 
Also, it looks as if you used * to indicate an exponent; we use that 
for multiplication (as in many computer programming languages), and we 
use ^ for exponents.

You've got the list of numbers correct, though I would list the digit 
pattern as T, 0, 1 rather than 0, 1, T because each time the digit 
to the left changes, ths digit you are observing will reset back to T, 
the start of the cycle.

Have you found a simple way to convert a decimal number to ternary, 
without looking it up in the table? It will be much like the method 
used for ordinary bases.

Your tables look fine, too.

But I think you should have followed my advice to learn to add before 
you try a big multiplication. Everything there is right, as I 
expected, until you try to add. Your answer is not written in the 
right notation: you can't have two symbols for one digit! Then when 
you tried to express the answer in base ten, you took -1 1 as if it 
meant -1 times 1.

Let's back up and try an addition.

      4 = 0  1  1
    + 1 = 0  0  1
    ---   -------
          0  1  2 <-- decimal sum of each column
          0  1 1T <-- ternary sum of each column
          0 1+1 T <-- carry: move the extra digit to the next column
          0 1T  T <-- add; but we have to carry again!
          1  T  T <-- final answer, 5

Do you see what I did? In order to have only a single digit in each 
"ternary place," I had to rewrite the two-digit sums so that the 
Threes digit is added to the column on the left. Doing this as you 
normally do addition, column by column, it works this way:

         1
      0  1  1
    + 0  0  1
    ---------
            T

      1  1
      0  1  1
    + 0  0  1
    ---------
         T  T

      1  1
      0  1  1
    + 0  0  1
    ---------
      1  T  T

This is exactly what you do in base ten; the only difference is that 
you have to use the right addition table.

Now do the same thing to add the partial products you got, and you 
should get the correct answer.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Number Theory
High School Number Theory

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