Hypermathematics

Hypermathematics is a variant of mathematics which concatenates numbers' digits as a base function, instead of adding them.Googol is a tiny dot, Hypermathematics 2009/02/12. retrieved 2022/03/10 Concatenation means gluing two numbers together, which means numbers produced by hypermathematical operations are much larger than those produced by usual operations. So in hypermathematics, 2 + 2 is equal to 22, and 43 + 27 is equal to 4,327.
==Multiplication==
It should be noted that in hypermathematics, multiplication isn't commutative nor associative, in contrast to normal mathematics. A chain of multiplications should be solved from the left to the right. Some examples are given below:
\(3 \cdot 3 \cdot 3 = (3+3+3)\cdot 3 = 333+333+333 = 333333333\)
\(4 \cdot 3 \cdot 3 = 444\cdot 3 = 444+444+444 = 444444444\)
\(3 \cdot 4 \cdot 3 = 3333\cdot 3 = 3333+3333+3333 = \underbrace{333...333}_{12}\)
\(3 \cdot 3 \cdot 4 = (3+3+3)\cdot 4 = 333 \cdot 4 = 333+333+333+333 = \underbrace{333...333}_{12}\)
\(4 \cdot 4 \cdot 4 = 4444 \cdot 4 = \underbrace{444...444}_{16}\)
\(5 \cdot 5 \cdot 5 = 55555 \cdot 5 = \underbrace{555...555}_{25}\)
For any positive integers a, b, and c, abc = acb because
abc = a(usual multiplication of b and c) = a(usual multiplication of c and b) = acb
==Exponentiation==
Exponentiation in hypermathematics should be solved from the top to the bottom, like in normal mathematics.
\(3^3\) = \(3 \cdot 3 \cdot 3\) = \(333 \cdot 3\) = \(\underbrace{333...333}_{9} = (E9-1)/3\)
(to avoid confusion of (normal / hyper)-mathematics, hyper-E notation is used.)
\(3^4\) = \(\underbrace{333...333}_{9} \cdot 3 = \underbrace{333...333}_{27} = (E27-1)/3 \)
\(3^5\) = \(\underbrace{333...333}_{27} \cdot 3 = \underbrace{333...333}_{81} = (E81-1)/3 \)
\(3^n\) = \((E(E[3](n-1))-1)/3 \)
So exponentiation in hypermathematics exhibits growth rate.
\(3^{3^2}\) = \(3^{3\cdot3}\) = \(3^{333} = (E(E[3]332)-1)/3 \approx E2.2\#3 \)
\(m^n\) = \(m\cdot(E[m](n-1))\)
== Wrong calculation ==
It was written in this article as
\(3 \cdot 3 \cdot 3 = (3+3+3)\cdot 3 = 333 \cdot 3 = \underbrace{3+3+3...3+3+3}_{333} = \underbrace{333...333}_{333}\)
\(4 \cdot 3 \cdot 3 = 444\cdot 3 = \underbrace{3+3+3...3+3+3}_{444} = \underbrace{333...333}_{444}\)
\(3^4\) = \(\underbrace{333...333}_{\underbrace{333...333}_{333}}\)
\(3^5\) = \(\underbrace{333...333}_{\underbrace{333...333}_{\underbrace{333...333}_{333}}}\)
and so on. It conflicts the calculation of the first source. As the multiplication of hypermathematics is non-commutative, a*b should be solved as a+a+...+a (b a's), as in the source. In this calculation, \(4 \cdot 3 \cdot 3 = 444\cdot 3\) is solved in this order, but \(444\cdot 3\) is solved as a*b = b+b+...+b (a b's). This mistake of calclation has been overlooked in Googology Wiki until Seruranblue pointed out in Japanese version of this articleSeruranblue's comment in Japanese version of this article on 2022/03/08 and P進大好きbot reported in the talk page.P進大好きbot's comment on 2022/03/08
==See also==
*coogol
*bigoogol
== Sources ==
ja:ハイパー数学
Category:Notations
Category:Hypermathematics