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Multiplication Using +, -, and 1/x


Date: 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
    
Associated Topics:
High School Number Theory
High School Puzzles

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