# Can the product $AB$ be computed using only $+, -,$ and reciprocal operators?

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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Are $A$ and $B$ any real numbers? – Joe Johnson 126 Mar 17 '12 at 22:39
What kind of objects are $A$ and $B$? Integers? Reals? Matrices? – Arturo Magidin Mar 17 '12 at 22:39
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. – Jeff Mar 17 '12 at 22:40
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. – joriki Mar 17 '12 at 23:22
Of interest to some people: this question stirred up some deeper discussion: Reciprocal-based field axioms. – user2468 Mar 18 '12 at 4:59

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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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 $+2AB$. – joriki Mar 17 '12 at 22:54
@joriki I changed the signs. Thanks for the 1/2 trick! – user2468 Mar 17 '12 at 22:56
A little simpler: $\frac{1}{x-1}-\frac{1}{x+1}=\frac{2}{x^2-1}$. Then $\frac{(A+B)^2-1}{2}-\frac{A^2-1}{2}-\frac{B^2-1}{2}=AB+1$. – N. S. Mar 17 '12 at 22:59
To both J.D. and @N.S.: How do you get the $1$? – joriki Mar 17 '12 at 23:08
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). – Asaf Karagila Apr 13 '12 at 13:14

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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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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Alternatively, you could simply subtract 1/A from 1/B, which simplifies to B-A/AB. Therefore,

(1/A)-(1/B)=(B-A)/AB. so, AB=(B-A)/[(1/A-(1/B)], which is then your formula.

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You can't, because we are only given the reciprocal key, not a division key. Your solution multiplies (1/((b-a)/ab) by (b-a). – George V. Williams Mar 11 '13 at 0:43