Let $a,b\in\mathbb{R}$. How can we compute $a\times b$ using only the following operations only (with any reals) :

  • $\frac{1}{*}$ (inverse)
  • $*+*$ (addition)
  • $*-*$ (subtraction)


  • $\begingroup$ I don't understand why you edited the title. The old title seemed better to me. $\endgroup$ – Ben Millwood Jun 4 '14 at 19:07
  • $\begingroup$ @BenMillwood Ok i changed it back $\endgroup$ – Hippalectryon Jun 4 '14 at 19:10
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    $\begingroup$ Are you allowed to use numbers other than $a$ and $b$ ? $\endgroup$ – lhf Jun 4 '14 at 19:15
  • $\begingroup$ Are you sure it isn't $*/*$ (division of any pair) instead of inverse $1/*$? $\endgroup$ – caub Jun 4 '14 at 22:22
  • $\begingroup$ @kwak Yep i'm sure $\endgroup$ – Hippalectryon Jun 4 '14 at 22:34

Build first the square function, as follows.

If $x\neq -1,0$, then $\displaystyle x^2= \frac{1}{\frac{1}{x}-\frac{1}{x+1}}-x$

Squaring $0$ and $-1$ is trivial.

Now, $\displaystyle xy=\frac{(x+y)^2-(x-y)^2}{4}$

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  • $\begingroup$ How do you get $x+1$ from $x$ and $y$ and the use of the three operations? $\endgroup$ – Dave L. Renfro Jun 5 '14 at 19:25
  • $\begingroup$ @DaveL.Renfro Ask OP what's allowed and what's not. $\endgroup$ – Gabriel Romon Jun 5 '14 at 19:34
  • $\begingroup$ It seems pretty clear that "adding $1$" is not allowed. As to what is intended, this also seems clear, with the possible exception of a universal quantifier at the beginning and restriction to finitely many uses of the operations: For each choice of two real numbers, $a$ and $b,$ is it possible to obtain their product by finitely many applications of the three operations (in any order) to these numbers? For some choices this is possible (i.e. $a=1$ and $b=42,$ or $a=\sqrt{2}$ and $b=42$) and for other choices this is not possible (Ben Millwood's answer), so the universal statement is false. $\endgroup$ – Dave L. Renfro Jun 5 '14 at 19:47
  • $\begingroup$ @Dave I think he modified the statement 15 mins ago "using only the following operations only (with any reals) ". But I got your point. $\endgroup$ – Gabriel Romon Jun 5 '14 at 19:49
  • $\begingroup$ I didn't notice the modification having been made until you mentioned it. I'm deleting my last comment because I don't think it's entirely appropriate for this site. (But I'm still thinking it . . .) $\endgroup$ – Dave L. Renfro Jun 5 '14 at 19:58

I claim it is impossible. Let $a = b = \sqrt{2}$. Then any number you can make with your operations is a rational multiple of $\sqrt{2}$, and in particular you cannot make $ab = 2$.

edit: this answer was given when it was unclear whether the use of arbitrary real constants was allowed. I assumed they were not, but the question has since been edited to indicate they are. I'm keeping this answer anyway because people seem to like it.

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    $\begingroup$ I'm pretty sure this answer is adequate. $\endgroup$ – Jorge Fernández-Hidalgo Jun 4 '14 at 19:16
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    $\begingroup$ @Bananarama: I believe it's correct also. To be slightly more general, the set of rational multiples of $\sqrt{2}$ (which includes $0)$ is closed under these operations, but this set is not closed under multiplication. I managed to get $\frac{a^2}{b}$ (and hence also $\frac{b^2}{a},$ $\frac{a}{b^2},$ $\frac{b}{a^2})$ by combining $\frac{1}{a-b} - \frac{1}{a+b},$ then taking the reciprocal, then adding the last result to itself, then adding $b,$ but I wasn't able to do anything with this. $\endgroup$ – Dave L. Renfro Jun 4 '14 at 19:38
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    $\begingroup$ Are you sure that $\frac{1}{\sqrt{2}-1}= 1 + \sqrt{2}$ is a rational multiple of $\sqrt{2}$? $\endgroup$ – Eric Towers Jun 4 '14 at 23:19
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    $\begingroup$ In fact, $0-\left( \frac{1}{1-\sqrt{2}} + \frac{1}{1+\sqrt{2}} \right) = 2$. $\endgroup$ – Eric Towers Jun 4 '14 at 23:24
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    $\begingroup$ @seaturtles: I see no restriction on the constants used. This is the reason I upvoted lhf's still unanswered inquiry. (In my own experimentation, I have only permitted myself constants from $\mathbb{N}$.) $\endgroup$ – Eric Towers Jun 5 '14 at 0:26

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