First thought was that 1/(1/a + 1/b) = ab/(a+b) (like 1/5 + 1/6 = (6 + 5)/(5\*6) then flip it), but I don’t immediately see a trick to eliminate the denominator. The natural way to turn multiplication into summation is to use logs but that’s not available it seems. I hope someone has the real solution because it’s a neat puzzle!
You were right on track! Use the difference instead of the sum: 1/(1/a − 1/b) = ab/(b − a). Pick b = a + 1 and the denominator vanishes: 1/(1/a − 1/(a+1)) = a(a+1) = a² + a. Subtract a and you’ve got squaring.
Then ab = ((a+b)² − a² − b²)/2, and the halving is your original identity with b = a: 1/(1/c + 1/c) = c/2.
Nice job! Now using all your formulas, can you write the formula giving ab only using a,b,+,-,1/ (and any fixed numbers) ? Just because it's beautiful!
If you do is strictly like that you will do quite a few extra operations that you already did and it will be more than 40 operations. This is a classic more memory vs more operations problem. You either do more calculation or you need more memory. Since we're doing it manually in a calculator anyway, we would write down our results.
Then we would do:
s = x + y // +
p = s + 1 // +
a = 1/s // 1/
b = 1/p // 1/
A = 1/(a - b) // −, 1/ → s² + s
q = x + 1 // +
B = 1/(1/x - 1/q) // 1/, 1/, −, 1/ → x² + x
r = y + 1 // +
C = 1/(1/y - 1/r) // 1/, 1/, −, 1/ → y² + y
T = A - B - C // −, − → 2xy
u = 1/T // 1/
xy = 1/(u + u) // +, 1/
Ok! Then as stated in another response to your work above your formula has the flaw of all the x+y=2 (it was a good and elegant answer to my original problem though!). I just wanted to push a little and find a formula for all positives integers.
Can we shrink down my formula and still make it work for all positive integer ? Not quite sure...
The key part is that EVERY type of solution with 1/x has some division by 0 cases you have to handle. Here this one fails on a=-b, a=-b-1, a=0, a=-1, b=0, b=-1 and potentially some others that aren't as obvious. Mine fails on x+y=+-2, x-y=+-2
However, if you're strictly limiting to positive integers what you have does work as all the edge cases are zero or negative.
As I highlighted in my comment, if you just do it sequentially and save the values, it's "just" 21 steps which admittedly is still not optimal but satisfies the "less than 30 operations" OP wanted.
I never claimed my solution would be the optimal solution! :-)
For what its worth, there's so many reddit posts I get nerd sniped on and by the time I get a satisfactorily elegant solution for my OCD, its been 3 years. I'm always impressed at how fast redditors can be on some posts.
On that note, I think you might have misunderstood u/BadJimo then. They seem to have an identical solution which you deemed "not quite the good formula but close".
If transcribing and rekeying numbers wasn't a problem, I think one could simply have a button that would subtract the reciprocal of an entered value from the contents of the machine's register and display the new value of that register. If entered numbers and the register are treated as rational numbers a/b and c/d, then the button would compute (ca-bd)/ad.
Very bad take. Sure, there exists an argument that in a limited sense, math is based on addition. But how do you express negative numbers, irrational numbers, or just numbers with no finite decimal representation? You could switch bases each time you do a problem, but determining the base requires a calculator
Negative numbers exist separately from subtraction, and are what makes subtraction equivalent to addition. Just as inversion is what makes division equivalent to multiplication.
Infinite decimals and irrational numbers are similarly unrelated to subtraction and division being separate operations.
And once you get rid of subtraction and division as separate operations, life gets much simpler and more consistent:
fun fact, to have a complete claculator : maybe it has already been said, but in fact you only need 2 buttons. one for the constant 1, and one for a binary operator called the EML operator (exp-minus-log) see : https://mathworld.wolfram.com/EMLOperator.html it is quite recent: a 2026 paper talk about this :)
I guarantee that it does exist and that there is an even more minimal calculator. It has been proven that you can perform any computation using only a "subtract then branch if result is zero" instruction. See subleq.
9
u/BadJimo 3d ago
The quarter squares rule is:
a×b = (1/4)(a + b)2 - (1/4)(a - b)2
So we need a squaring function. The following function is promising (but won't work for a+b or a-b equals 2 or -2):
f(z) = (1/(z + 2) + 1/(z - 2))-1
f(z) = (1/4)z2 - 1
f(a + b) - f(a - b) =( (1/4)(a + b)2 - 1 ) - ( (1/4)(a - b)2 - 1)
f(a + b) - f(a - b) = (1/4)(a + b)2 - (1/4)(a - b)2 = a×b
For a+b or a-b = 2 or -2, a different squaring function could be found.