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Can anyone come up with a way to divide any given x by any given y without actually dividing?

For example to add any given x to any given y without adding you would just do:
$x-(-y)$

And to subtract any given x from any given y (that is, y-x) you could do:
$y+xe^{iπ}$
*edit: well since (i) is ($\sqrt{-1}$) and that is technically subtracting this one might not work perfectly but for the sake of the riddle and for the sake of example, I'm using that equation :)

How can you divide without dividing? Can anyone come up with equations that work for all $x$ and $y$ values? (For all intents and purposes we will leave out dividing by zero issues and what-not... don't worry about that...

Albert Renshaw
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    I don't really see the point of this post. Divide without dividing? Sure you can do it for simple things, like solving: $2x = 6$. You could guess rational numbers until you get it right but that's pretty silly. What about something more complicated like $2x = \pi$? How could you solve for $x$ without doing division? – Cameron Williams May 16 '13 at 02:20
  • @CameronWilliams Exactly! How would you? That's what I want to know! I think it's a fun question! – Albert Renshaw May 16 '13 at 02:21
  • $x/y=x\times y^{-1}$ – vadim123 May 16 '13 at 02:22
  • @vadim123 Brilliant! Hahaha! – Albert Renshaw May 16 '13 at 02:22
  • @vadim123, how would you compute $y^{-1}$ without division? It's a pretty self-referential approach. – Cameron Williams May 16 '13 at 02:23
  • @CameronWilliams (x^a)=(x^b)*(x^c) – Albert Renshaw May 16 '13 at 02:27
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    @CameronWilliams: Come now, Albert mentions $y+xe^{\pi i}$ as a work-around for subtraction. You could not compute this without actually doing subtraction. The question is clearly about equivalent expressions with no immediate regard to practicality. We all have to learn these things at some point :) – Eric Stucky May 16 '13 at 02:27
  • @EricStucky you make a good point. I guess I misinterpreted what OP meant. I was thinking numerical solution versus equivalent expression for some reason. – Cameron Williams May 16 '13 at 02:29
  • @CameronWilliams That was assuming you are really talking about computing, not just using algebra to divide it over. (My example was proposed for something like (2^4)=(2^6)(2^-2) so you would compute (2^6)(x)=(2^4) by guessing high and low values until you reach the proper value and therefor know what (2^-2) is without division. – Albert Renshaw May 16 '13 at 02:29
  • @CameronWilliams It's all good :) I just meant without using the symbols in the actual equation! Haha :o) – Albert Renshaw May 16 '13 at 02:30
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    Let me just add that while the question may seem a little facetious, Paul Dirac spent some time trying to join special relativity and quantum mechanics before he discovered he could take a square root without taking a square root and revolutionise physics! (Slide 13) – Nicolau Saker Neto May 16 '13 at 03:32
  • @NicolauSakerNeto I'm curious, how do you take a square-root without taking a square-root! Amazing! – Albert Renshaw May 16 '13 at 03:33
  • Unfortunately my command of physics and mathematics is far too limited to explain in detail how it works (other than pointing to what I linked and what you may find on the Dirac equation wiki page), but I thought I should at least bring it up since it seems relevant to this unusual topic. – Nicolau Saker Neto May 16 '13 at 03:45
  • You take a square root without taking a square root by $\sqrt x=exp((1/2)\log x)$. – Gerry Myerson May 16 '13 at 13:18

4 Answers4

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Look at the equation $\frac{1}{x}=a$. We use Newton's Method to approximate the solution.

Let $f(x)=\frac{1}{x}-a$. The standard Newton iteration gives $$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}=x_n -\frac{\frac{1}{x_n}-a}{-\frac{1}{x_n^2}}.$$ This simplifies to $$x_{n+1}=x_n(2-ax_n).$$

Remark: Note that only subtraction and multiplication are used. If we start with $x_0$ close enough to $\frac{1}{a}$, the method converges rapidly. It was once used to implement reciprocal in software.

André Nicolas
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  • @AlbertRenshaw: You mentioned in the comments that your original idea was to try and "guess high, guess low" until you got to the right answer. This is a formally rigorous way of implementing a strategy similar to that one. The downside is that you'll never actually reach the number, but the upside is that for integer divisions you'll be able to see pretty quickly what you're "headed towards," for example if your first guess is 0.8, this method tells you after four steps that $\frac12\approx 0.49999996$; it's not too much of a stretch to say, "Oh, that's 0.5". – Eric Stucky May 16 '13 at 02:49
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For $y \neq 0$ $$\large x\div y = \dfrac 1y\times x = y^{-1}\times x = \large y^{\left(e^{i\pi}\right)}\times x = y^{\left(i^2\right)}\times x$$

amWhy
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  • Very nice!!! For now I will accept this, I am curious what others come up with!!! (I will accept it in 7 minutes when SE lets me) – Albert Renshaw May 16 '13 at 02:23
  • Multiplying by the reciprocal could be thought of as the definition of division. This is analogous to saying that $x-y = x+(-y)$ is subtracting without subtracting, whereas an alternative would be to note that that is the definition of subtraction. – Jonas Meyer May 16 '13 at 02:30
  • @JonasMeyer I'm eager to see if other's come up with solutions that don't use that... since the only way to "really" solve mathematically for (x^-y) is to use division. But this was more of a riddle question, in which case this is a valid answer :o) – Albert Renshaw May 16 '13 at 02:31
  • @amWhy: I have seen all of the OP's comments. I'm not sure what part you want me to see. – Jonas Meyer May 16 '13 at 02:36
  • @Jonas see below the answer...he simply wants to avoid using the symbol for division. "I just meant without using the symbols in the actual equation! Haha :o) " – amWhy May 16 '13 at 02:37
  • @amWhy: Deserves another thumbs up +1 – Amzoti May 16 '13 at 05:16
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Logs turn reciprocals into minus signs: $\ln(1/y)=-\ln(y)$. Thus, $$x/y=xe^{-\ln y}.$$

(This is assuming that $y$ is positive. If $y$ is negative, then $x/y=-xe^{-\ln(-y)}$.)

Jonas Meyer
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Take the logarithm that maps multiplication/division into addition/subtraction:

$$\frac{x}{y}=e^{\log{x/y}}=e^{\log x- \log y}.$$

$x,y >0$.

Also, see my answer for multiplying natural numbers here: Advocating base 12 number system

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