You can use the digits 1,2,and 3 once only and any mathematical symbols you are aware of, but no symbol is to be used more than once. The challenge is to see if you can make the smallest positive number.
Special rules: You cannot use Euler's number or pi or infinity.
Special thanks to: Rhonda Wendel for Make the most of these digits and for the problem text which was slightly altered.
Okay now I think I have finally obtained the smallest positive number which can be formed using the digits 1,2,and 3 once only and any mathematical symbols that we are aware of, and also no symbol being used more than once. Here is how I figured it out, though I do not know whether it is the smallest or not. So I might need help once again from Charlie or anyone who can help me with the calculations.
It has been observed from the previous comments that: 2^(-(31!)) is about 1/(10^(2.5*10^33)) while 3^(-(21!)) is about 1/(10^(2.4*10^19)), so the tiniest so far is 2^(-(31!)), and extended precision arithmetic tells us this is about 1.02192753424799x10^-2,475,321,084,412,797,072,581,101,485,711,599.
Now all I have to do is use a sine function because I know that sin(x) approaches 0 (zero) as x approaches 0 (zero).
Therefore the smallest positive number which can be formed using the digits 1,2,and 3 once only and any mathematical symbols that we are aware of, and also no symbol being used more than once is as given below:
sin [2^{-(31!)}].
That is, sin(x), where x = 2^(-(31!))
That's it. This is the SMALLEST. I got it.
