Multiplication Using +, -, and 1/xDate: 10/25/2000 at 22:33:27 From: Jonathan Silverman Subject: Puzzle: mathematical operations Is there a way to multiply two numbers using the addition, subtraction, and reciprocal functions only? Which means, I guess, express AB in terms of a function of A and B using +, -, 1/A and 1/B. I have a gut feeling that there is a way to do it, but after hours of juggling, I couldn't find the answer. Please try to quench my curiosity. Jonathan Date: 10/25/2000 at 22:56:01 From: Doctor Peterson Subject: Re: Puzzle: mathematical operations Hi, Jonathan. Here's a place to start: 1 1 A + B --- + --- = ----- A B AB See if that leads you to a formula for AB using only "+" and "reciprocal." - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 10/25/2000 at 23:20:26 From: Jonathan Silverman Subject: Reply Thanks for your prompt reply. I had already tried the step you suggested, to start from 1/a + 1/b = (a+b)/ab and further use + and 1/x operations, but it did not lead me to anything. Could you please help me a little more? This problem has been in my head now for a long time. Jonathan Date: 10/26/2000 at 16:52:41 From: Doctor Rick Subject: Re: Again puzzle: mathematical operations Hi, Jonathan. We're still thinking about your problem, which we misinterpreted at first. In the meantime, I have a question and some comments. The question is: Was this problem posed to you as a problem with a known solution, or are you just hoping there is a solution? I wonder also, is an iterative algorithm acceptable, as opposed to a closed-form identity for AB? You'll see why I ask in a moment. One comment regards your interpretation of the problem. The solution isn't necessarily restricted to +, -, 1/A, and 1/B. It could contain, for instance, 1/(A+B), or the reciprocal of any function of A and B (provided that function contains only +, -, and other reciprocals). When you say you were advised to "start from 1/a + 1/b = (a+b)/ab and further use + and 1/x operations," this may be a misunderstanding. The idea was not to take this expression and do more operations on it, but to solve it for ab and see if this leads anywhere. ab = (a+b) * 1/(1/a + 1/b) This involves additions and reciprocals, but unfortunately it also involves a multiplication. As of now, I don't see what you could do with it. Here is my main comment. Historically, peoples have used various algorithms for multiplication. You can see two such algorithms, among the earliest known, on this Web page: Babylonian and Egyptian mathematics MacTutor History of Mathematics Archive http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Babylonian_and_Egyptian.html The Babylonian algorithm makes use of a closed-form identity, A * B = ((A + B)/2)^2 - ((A - B)/2)^2 By this means, multiplication of integers can be reduced to taking the difference between two entries in a table of squares. Entry N of the table has the value (N/2)^2. The Egyptian algorithm is an iterative method, like our method of multiplication, but it does not require memorization of multiplication tables. The only operations involved are addition and subtraction. That makes it a solution to your problem, if iterative methods are accepted. It's probably not what you want, because it doesn't make use of reciprocals, but it's fascinating. This ancient method also has strong connections to the method used in binary computers. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 10/26/2000 at 17:42:29 From: Jonathan Silverman Subject: Re: Puzzle: mathematical operations Thanks for your comments, Dr. Rick. I think I better rephrase the question in this way: Suppose we have a calculator with just the operator buttons +, - and 1/x. Can we use this calculator to perform multiplication? I think now it will be clear why I wanted to find a closed identity for AB. But the problem doesn't say there has to be solution. I will just have to be content with an explanation that there is no solution. BUT the Babylonian and Egyptian methods are excellent. I enjoyed reading them immensely. Especially the Egyptian method can be performed iteratively using a piece of paper, pen and the calculator, assuming the person does not know how to multiply, without using square tables. But we can do multiplication the long way on this calculator. For example, to find 50*20, we add 50 twenty times using the + button only. It will be great if we can find a closed identity for AB to compute it using our faulty calculator, that is, if it exists. Thanks! Jonathan |
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