>PS: As far as I know, there is no such calculator actually
>manufactured.

But there's a Broken Calculator page:

https://www.cut-the-knot.org/arithmetic/Calculator.shtml

>It's just there to set up the puzzle:
>
>Can you represent multiplication as a composition of
>additions, subtractions and reciprocations?

xy = (x + y)/(1/x + 1/y),

except that it's unclear whether the reciprocal could indeed be always represented exactly in a TI.

Thankyou

sfwc
<><

Aha, now I see. With (a lot of) memory it's still more or less straightforward. Perhaps, for a simple calculator with only three operations you may want to limit the number of memory registers, or seek a minimumal number. Would be an achievement even then.

a: Put the number a onto the stack, or return ERROR if the stack is full. This simulates the user entering the number a, which (s)he presumably knows.

b: Put the number b onto the stack, or return ERROR if the stack is full.

1: Put the number 1 onto the stack, or return ERROR if the stack is full.

+: Take the top 2 numbers off the stack, or return ERROR if the stack only has 1 number on it. Put their sum onto the stack.

-: Take the top number off the stack, or return ERROR if the stack is empty. Put minus that number onto the stack.

R(k) Where k is a natural number: Take the top number off the stack, or return ERROR if the stack is empty. Put the reciprocal of that number onto the stack, or return ERROR if that number is 0. The number k is just a reference for the user of the calculator.

The stack is initally empty. The aim is to come up with a string which will finish with ab as the only number on the stack, for as small a value of n as possible. Of course, if you get ERROR from R(k), you may start again, with a new empty stack, using the new information you have gained. I have refenced what you do if you get ERROR when calculating R(k) after the heading R(k). Here is a solution for n = 3:

ab+1+1+R(1)-ab+R(2)+R(3)a1+1+R(4)aR(5)-+R(6)+b1+1+R(7)bR(8)-+R(9)+
R(1): a1+R(10)aR(11)-+R(12)a+-
R(2): a1+R(13)aR(14)-+R(15)-a+
R(3): Will never occur.
R(4): bb+-
R(5): 1-1+
R(6): Will never occur.
R(7): aa+-
R(8): 1-1+
R(9): Will never occur.
R(10): 1
R(11): 1-1+
R(12): Will never occur.
R(13): 1-
R(14): 1-1+
R(15): Will never occur.

Although that looks a little complex, what it is doing is simple enough. I guess I must have made a mistake somewhere in there, so feel free to correct it.

There is no solution for n = 2. I leave this as a challenge for the moment, but will post a proof if you want me to.

Thankyou

sfwc
<><

1/(x-1) + (y-1) = <1 + (x-1)(y-1)>/(x-1) (reciprocal and sum)
(x-1)/<1 + (x-1)(y-1)> (reciprocal)
1/(y-1) + (x-1) = <1 + (y-1)(x-1)>/(y-1) (reciprocal and sum)
(y-1)/<1 + (y-1)(x-1)> (reciprocal)
(x+y-2)/<1 + (x-1)(y-1)> = (x+y-2)/(2-x-y+xy) (sum)
(2-x-y+xy)/(x+y-2) = -1 + xy/(x+y-2) (reciprocal)
xy/(x+y-2) (add 1)
(x+y-2)/xy (reciprocal)
1/y + 1/x - 2/xy
1/x + 1/y (2 reciprocals and sum)
(1/x + 1/y) - (1/y + 1/x - 2/xy) (subtract)
2/xy
xy/2 (reciprocal)
xy/2 + xy/2 = xy (sum)

1/x - 1/(x+2) = 2/x(x+2)
1/y - 1/(y+2) = 2/y(y+2)
1/(x+y) - 1/(x+y+2) = 2/(x+y)(x+y+2)
(x+y)(x+y+2)/2 - x(x+2)/2 - y(y+2)/2 = xy.

This seems a little simpler (and it only needs a stack size of 3, which was why I posted it).

Thankyou

sfwc
<><

If u is not 1,

u² = u - 1/(1/u + 1/(1-u)).
u/2 = 1/(1/u + 1/u).

uv = ((u+v)/2)² - ((u-v)/2)².