Balanced Ternary NotationDate: 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/ |
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