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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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Are $A$ and $B$ any real numbers? –  Joe Johnson 126 Mar 17 '12 at 22:39
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What kind of objects are $A$ and $B$? Integers? Reals? Matrices? –  Arturo Magidin Mar 17 '12 at 22:39
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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
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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
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Of interest to some people: this question stirred up some deeper discussion: Reciprocal-based field axioms. –  user2468 Mar 18 '12 at 4:59
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3 Answers 3

up vote 16 down vote accepted

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
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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
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To both J.D. and @N.S.: How do you get the $1$? –  joriki Mar 17 '12 at 23:08
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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
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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
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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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