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Can the product of $A, B$ be computed using only $+, -,$ and reciprocal operators using a calculator? You can use calculator's memory function (multiply and divide are broken though).

Additional: I should have mentioned earlier, in addition to the 3 operators, the numberpad of the calculator can be used so yes 1 can be used.

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    $\begingroup$ Are $A$ and $B$ any real numbers? $\endgroup$
    – J126
    Mar 17, 2012 at 22:39
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    $\begingroup$ What kind of objects are $A$ and $B$? Integers? Reals? Matrices? $\endgroup$ Mar 17, 2012 at 22:39
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    $\begingroup$ Sure. $A$ times $B$ equals $A$ plus $A$ plus $A$ plus ... etc. (a total of $B$ times). You will get very bored if $B$ is very large. $\endgroup$
    – Jeff
    Mar 17, 2012 at 22:40
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    $\begingroup$ You accepted an answer that uses a constant $1$. If that was intentional, I think you should clarify the question to reflect that not only $A$ and $B$ but also constants can be entered. $\endgroup$
    – joriki
    Mar 17, 2012 at 23:22
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    $\begingroup$ Of interest to some people: this question stirred up some deeper discussion: Reciprocal-based field axioms. $\endgroup$
    – user2468
    Mar 18, 2012 at 4:59

4 Answers 4

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Edit: previous answer was wrong. Posted new answer. Hopefully right this time

  1. We can compute and store $A^2$ using $$ \frac{1}{A} - \frac{1}{A+1} = \frac{1}{A^2 + A} $$ We can extract $A^2$ using only $+, -, ^{-1}.$ Similarly we can compute and store $B^2.$

  2. Then

$$\frac{1}{A+B-1} - \frac{1}{A+B} = \frac{1}{(A+B)(A+B-1)} = \frac{1}{A^2 + B^2 + 2AB - A - B} $$

where we can extract $2AB,$ again, using only $+, -, ^{-1}$ and the values for $A^2, B^2$ we computed in step $1$ above.

Thanks to joriki, now to get $AB$ from $2AB$, add $\frac{1}{2AB} + \frac{1}{2AB},$ and take the reciprocal.

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    $\begingroup$ You can extract $2AB$, not $AB$, but you can add $1/(2AB)$ to itself to get $1/(AB)$. By the way you've got two sign errors; the first equation has the difference the wrong way around, and in the end result it should be $+2AB$. $\endgroup$
    – joriki
    Mar 17, 2012 at 22:54
  • $\begingroup$ @joriki I changed the signs. Thanks for the 1/2 trick! $\endgroup$
    – user2468
    Mar 17, 2012 at 22:56
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    $\begingroup$ A little simpler: $\frac{1}{x-1}-\frac{1}{x+1}=\frac{2}{x^2-1}$. Then $\frac{(A+B)^2-1}{2}-\frac{A^2-1}{2}-\frac{B^2-1}{2}=AB+1$. $\endgroup$
    – N. S.
    Mar 17, 2012 at 22:59
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    $\begingroup$ To both J.D. and @N.S.: How do you get the $1$? $\endgroup$
    – joriki
    Mar 17, 2012 at 23:08
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    $\begingroup$ I would also mention the case of $A=0$ or $B=0$ (and in the case both equal zero, none of the above is defined). $\endgroup$
    – Asaf Karagila
    Apr 13, 2012 at 13:14
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J.D. and N. S. have shown how to do it if a constant $1$ is allowed. Here's a proof that it can't be done if only $A$ and $B$ can be entered, no constants.

We can show by structural induction that all expressions we can generate change sign if both $A$ and $B$ change sign.

Base case: The two atomic expressions $A$ and $B$ change sign when both $A$ and $B$ change sign.

Induction step: $x+y$ changes sign when both $x$ and $y$ change sign, $x-y$ changes sign when both $x$ and $y$ change sign, and $x^{-1}$ changes sign when $x$ changes sign.

Since $AB$ doesn't change sign when both $A$ and $B$ change sign, it follows that it can't be generated from $A$ and $B$ using only these operations.

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  • $\begingroup$ That's only true if you can't find out what A or B is or if they are non-integers. Otherwise $\sum^{A} \frac{1}{A} = 1$. $\endgroup$
    – Fax
    Dec 6, 2019 at 10:57
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    $\begingroup$ @Fax: That's true. On the other hand, if you can perform $A$ additions dependent on $A$, with $A$ an integer, then you don't need a complicated detour through $1$; then you can just compute $AB=\sum^AB$. My answer only shows that the product can't be computed through some fixed set of operations. $\endgroup$
    – joriki
    Dec 6, 2019 at 11:14
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    $\begingroup$ Good point. Also, now that I think about it, I could loosen the restriction in my comment from integers to rationals by $A = \frac{x}{y} \implies (\sum^y A) = x$. $\endgroup$
    – Fax
    Dec 6, 2019 at 11:33
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user2468's solution ends up requiring quite a lot of operations -- three reciprocals to get $A^2$, then the same to get $B^2$, then another three to get $A^2+B^2+2AB-A-B$. Then to get rid of that pesky factor of 2 you need another three reciprocals, for a total of 12 reciprocal operations (and 31 binary operations).

It's possible to do better by employing some tricks:

  • Instead of $(A+B)^2 - A^2 - B^2$, which involves three terms, use $(A+B)^2 - (A-B)^2$, which involves only two terms.
  • Instead of needing to divide by a constant near the end of the process, which involves a lot more reciprocals, identify where in the formula you actually introduced the constant, and scale your other constants to match.

Using those tricks, I was able to reduce this down to 6 reciprocal operations and 11 binary operations:

$$ \frac{1}{\frac{1}{A+B-2} - \frac{1}{A+B+2}}-\frac{1}{\frac{1}{A-B-2} - \frac{1}{A-B+2}} = AB $$

I suspect this is probably the most beautiful solution to this challenge.

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We have $$\frac{A^2}4=\frac1{\dfrac1{A}-\dfrac1{A+4}}-A\tag1$$ and $$AB=\frac14\,(A+B)^2-\frac14\,(A-B)^2\tag2.$$

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