I have a key-ring calculator where the number symbols are intact but the operation symbols have worn off. So to rediscover which operation key is which, I enter 3, operation key, 2, Enter. If 1 appears, it was -; if 5, +; if 6, x; etc. Nothing brilliant about that, but this post reminded me of it. I'm sure this notion could be generalized into a nontrivial logical puzzle. Steve Gray Kerry Mitchell wrote:
Hi Eric,
It seems to me that since you used 1! to represent 1 (apparently to get one more symbol), you can do things with parentheses, 0!, 1^2 (either with a caret or a superscript), 0^2, 1^[2^(3^4)], etc., to get an arbitrarily large number of symbols.
Kerry
On Nov 22, 2007 10:30 AM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Hello Math-fun,
those equalities are doubly true : 2+3=5 5+1!=6 because they have respectively 5 and 6 symbols, altogether.