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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)

?

Hippalectryon
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2 Answers2

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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.

Ben Millwood
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    I'm pretty sure this answer is adequate. – Asinomás Jun 04 '14 at 19:16
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    @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. – Dave L. Renfro Jun 04 '14 at 19:38
  • (Several comments have been deleted above) – Ben Millwood Jun 04 '14 at 19:54
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    Are you sure that $\frac{1}{\sqrt{2}-1}= 1 + \sqrt{2}$ is a rational multiple of $\sqrt{2}$? – Eric Towers Jun 04 '14 at 23:19
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    In fact, $0-\left( \frac{1}{1-\sqrt{2}} + \frac{1}{1+\sqrt{2}} \right) = 2$. – Eric Towers Jun 04 '14 at 23:24
  • @EricTowers How did you get $1$ by adding, subtracting and reciprocating with $\sqrt{2}$? – anon Jun 05 '14 at 00:06
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    @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 $\mathbb{N}$.) – Eric Towers Jun 05 '14 at 00:26
  • @Eric Towers: I just saw your comments and I don't understand their relevance to what I wrote. Specifically, let $\cal{S} = {r\sqrt{2}:; r \in {\mathbb Q}}.$ Then my claim is that none of the three specified operations applied to elements in $\cal S$ produce, when defined, an element not in $\cal{S}.$ (Note I was incorrect in saying this set is closed under these operations, since the reciprocal of zero is undefined.) None of $1 + \sqrt{2},$ $1 - \sqrt{2},$ or $\sqrt{2} - 1$ belong to the set $\cal{S},$ so I don't see their relevance. – Dave L. Renfro Jun 05 '14 at 13:53
  • The question has now been edited so that this answer is wrong. But I still like it, so I'm leaving it there. – Ben Millwood Jun 10 '14 at 19:15
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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, $\displaystyle xy=\frac{(x+y)^2-(x-y)^2}{4}$

Gabriel Romon
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  • How do you get $x+1$ from $x$ and $y$ and the use of the three operations? – Dave L. Renfro Jun 05 '14 at 19:25
  • @DaveL.Renfro Ask OP what's allowed and what's not. – Gabriel Romon Jun 05 '14 at 19:34
  • 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. – Dave L. Renfro Jun 05 '14 at 19:47
  • @Dave I think he modified the statement 15 mins ago "using only the following operations only (with any reals) ". But I got your point. – Gabriel Romon Jun 05 '14 at 19:49
  • 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 . . .) – Dave L. Renfro Jun 05 '14 at 19:58
  • How do you divide by $4$? – Ben Millwood Jun 10 '14 at 19:23
  • I suppose you can make $4xy$, then make $1/4xy$, then add it to itself four times to get $1/xy$, then reinvert it. – Ben Millwood Jun 10 '14 at 19:25
  • @Ben I think OP authorized the use of arbitrary numbers. It's as if he had a calculator where the $\times$ button is broken. – Gabriel Romon Jun 10 '14 at 19:28
  • @G.T.R: yes, but they didn't authorise the use of division :) – Ben Millwood Jun 12 '14 at 18:51
  • @Ben $\frac{a}{4}=\frac{1}{\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}}$ – Gabriel Romon Jun 12 '14 at 18:53