# How to multiply on a calculator which only allows add, subtract and reciprocal [duplicate]

Assume we have a calculator with the following flaw. The only operations can be done by it are $+$ and $-$ and $\dfrac{1}{x}$ i.e. you only can add or subtract two numbers and also calculate the reciprocal but you can't multiply or divide. Using this calculator how can you multiply two numbers?

I twiddled with lots of formulas but I got nowhere. Sorry if I can't add any further information or idea. I appreciate any solution on this....

## marked as duplicate by Claude Leibovici, Parcly Taxel, Dirk, Guy Fsone, Ethan BolkerJan 29 '18 at 14:26

• Can you do something like $5 \times 3 = 5+5+5 = 15$? – Landuros Jan 28 '18 at 8:26
• $xy=\underbrace {x+x+x+...+x+x+x}_{y \text{ times}}$ – Mohammad Zuhair Khan Jan 28 '18 at 8:33
• What if the numbers are non integer? – Mostafa Ayaz Jan 28 '18 at 8:41
• How do you calculate $\sqrt 2 \times \sqrt 3?$ – Mostafa Ayaz Jan 28 '18 at 8:41
• Should we tag this puzzle? – smci Jan 28 '18 at 20:29

I assume that you can also put additional numbers (except your two operands) into the calculator, i.e. that you can calculate $x+4$, for instance. Then, $$\frac14\,x^2=\frac1{\dfrac1x-\dfrac1{x+4}}-x\tag1$$ and $$xy=\frac14\,(x+y)^2-\frac14\,(x-y)^2\tag2.$$ Of course, (1) is valid only if $x\neq0,-4$, but there's no need to calculate $0^2$, and if $x=-4$, we use $x^2=(-x)^2$.

• In case anyone wants to see the whole thing written out: $$xy=\cfrac1{\cfrac1{x+y}-\cfrac1{x+y+4}}-y-\cfrac1{\cfrac1{x-y}-\cfrac1{x-y+4}}-y.$$ (The $\pm x$'s cancel out.) – Rahul Jan 29 '18 at 5:36

Just for curiosity because it is just a silly method to get $xy$, but Prof Vector answer made me think of that.

Let's define $H(x,y)=\dfrac 2{\frac 1x+\frac 1y}$ and $A(x,y)=\dfrac{x+y}2$ respectively harmonic and arithmetic means.

Multiplying or dividing by $2$ is not an issue since $2a=a+a$ and $\dfrac a2=\dfrac 1{\frac 1a+\frac 1a}$.

Then it is known that the arithmetic-harmonic mean $\begin{cases} x_0=x & x_{n+1}=A(x_n,y_n)\\y_0=y & y_{n+1}=H(x_n,y_n)\end{cases}$

converge quickly to the geometric mean $\sqrt{xy}$.

Finally we can use $a^2=\dfrac 1{\frac 1a-\frac 1{a+1}}-a$ to conclude.