When Adding and Multiplying are the Same

David A. Wheeler

This is another little essay is an exercise in mathematical recreations; I hope you find it amusing!

When is addition the same as multiplication? In other words, when is the following true?

  x + y = x * y

A bit of thinking will help you realize that {x,y} = {2,2} and {0,0} are two solutions to the problem, because:

 2 + 2 = 2 * 2 = 4
 0 + 0 = 0 * 0 = 0

But are there other solutions? To find out, we need to solve the equation using a little algebra:

  x + y = xy
  0 = xy - x - y
  0 = (x-1)(y-1) - 1     (Confused? See the Postscript below)
  1 = (x-1)(y-1)
  1/(x-1) = y-1
  y = 1/(x-1) + 1

So, there are an infinite number of real number solutions; to find a given y, just compute y = 1/(x-1)+1 (as long as x is not equal to 1; there's no real number solution in that case). For example, the pair {1.5,3} works, because:

  1.5 + 3 = 1.5 * 3 = 4.5

Now, let's narrow the question - are there any other real number solutions where x=y, other than 0 and 2? It turns out the answer is no, and here's why, starting with the equation above and changing it since now x=y:

  0 = (x-1)(y-1) - 1        (see above)
  0 = (x-1)^2 -1            (because x=y now)
  1 = (x-1)^2
  +1 or -1 = x-1
  x = 0 or 2

Now, instead of limiting ourselves to x=y, let's allow any value of x and y, but only if they are both integers. Given that variation, are there any other integer pair solutions? The short answer is no - if you limit yourself to integers, 0 and 2 are all that's possible. Here's why. Since integers are a subset of the real numbers, the equation given above applies:

  y = 1/(x-1)+1
So, for y to be an integer, 1/(x-1) has to be an integer. For a fraction to produce an integer when there's a one on top, its denominator has to have an absolute value less than one, so:
  |x-1| <= 1
This is only true for x = {0,1,2}. x cannot be 1, because that would require y to be infinite. So 0 and 2 are the only integer solutions, too.

Conclusion

Bottom line: there are an infinite number of real number pairs where adding and multiplying the pair will give the same answer. But if you require the pair to have the same value, or to both be integers, there are only two answers: {0,0} and {2,2}.

Postscript: Huh? What's that?

Sadly, some people have sent me messages, asking me to explain why:
  xy - x - y
is the same as:
  (x-1)(y-1) - 1

Tsk, tsk! Complain to your algebra teachers, they skipped some important material.

It turns out this isn't hard to show; just start with "(x-1)(y-1)" and multiply it out. These are two trivial expressions; you can just multiply them out the same way you'd multiply multiple-digit numbers:

        x - 1
    *   y - 1
   ===============
       -x + 1
    xy -y
   ===============
   xy -x -y + 1

So (x-1)(y-1) is almost the same as xy-x-y, except that it has an extra "+1". No problem, just subtract one, and you have it. That means that:

  xy - x - y  =  (x-1)(y-1) - 1

Now, to be honest, I instantly recognized "xy-x-y" as easily rewritten as "(x-1)(y-1)-1"; I didn't have to "figure it out". But if you haven't seen that before, hopefully this will convince you.

Other Stuff

If you enjoyed this article, you might enjoy my articles on the Four fours problem or weird bases.

Feel free to see my home page at www.dwheeler.com.

David A. Wheeler, 10-September-2002

This is Copyright (C) 2002-2005 David A. Wheeler.