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!
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