Can multi digit numbers be made into a single digit using a new base?

  • Thread starter Thread starter Digit
  • Start date Start date
  • Tags Tags
    Symbols
Digit
Messages
18
Reaction score
0
Digit has {b} ; b is the count of symbols
b is the base of the digit and is the symbol count

b symbols
1 0
2 0,1
3 0,1,2
...
The symbol count is necessary. The symbols can be any set of marks.
If you count and come to the end of the symbols you have, there are 3
choices:
1: give up
2: make another symbol
3: add another digit
If you add another digit you must have a carry mechanism and your digits make a polynomial in b.
It is easy to show that any multi digit number can be made into a single digit by choosing a new base that is old b to the digit power.

I make machines that do math. Anyone interested in this kind of stuff?
 
Mathematics news on Phys.org
Is that you, hlreed?

- Warren
 
Hi Warren.
I thought you would get me.
 
Originally posted by Digit
Hi Warren.
I thought you would get me.
Yes, I probably will.

- Warren
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top