Decimal to base 1

Question

Explain why base 1 radial expansions are impossible.

I think If b is a natural number greater than 1, to write the abbreviated base b radial expansion the number of symbols required is b.

For example, the digits 0, 1, and 2 are sufficient to write the abbreviated base 3.

But then, to write base 1 radial expansions is impossible because very such expansion would evaluate to 0. But I'm not sure. Can someone please explain to me why base 1 expansions are impossible, is there a theorem behind it? Thanks

Answer 1

Some people would consider the tally system to be "base 1." You only have one digit to use, so don't choose $0$, choose $1$. Then, for instance,

$$5 = 1\times 1^4 +1\times 1^3 + 1\times 1^2 + 1\times 1^1 +1\times 1^0 =11111_1, $$

just like any other radix.

Answer 2

Base $n$ radial expansions are possible, for $n \in \Bbb{Z_{\geq 2}}$ precisely because

• The sequence $~n^1, n^2, n^3, \cdots ~$ grows unbounded.

• If $M \in \Bbb{Z^+}$, and $~n^k \leq M < n^{k+1} ~: ~k \in \Bbb{Z^+}, ~$ then you can choose an element $t$ from $\{1,2,\cdots,n-1\}$ such that $~~0 \leq (M - tn^k) < n^k.~~$ So, the number $M$ can be whittled down to $(M - tn^k)$ which will result in the leftmost digit in the base $n$ expansion of $M$ being $t$.

For $n = 1$, the 2nd consideration above becomes moot, because $~1^1, 1^2, 1^3, \cdots~$ does not grow unbounded.