Existence of Digital Root

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Theorem

Let $n \in \N$ be a natural number.

Let $b \in \N$ such that $b \ge 2$ also be a natural number.

Let $n$ be expressed in base $b$.


Then the digital root base $b$ exists for $n$.


Proof

By definition, the digital root base $b$ for $n$ is the single digit resulting from:

adding up the digits in $n$, and expressing the result in base $b$
adding up the digits in that result, and again expressing the result in base $b$
repeating until down to one digit.


Let $n = d_1 + b d_2 + \dotsb + b^{m-1} d_m$ where, for all $i$, $0 \le d_i < b$.

Let $S \left({n}\right)$ be the digit sum of $n$.

Then:

$S \left({n}\right) = d_1 + d_2 + \dotsb + d_m$

Thus:

$S \left({n}\right) < n$

unless $d_2, d_3, \dotsb, d_m = 0$ in which case $n$ is a one digit number.

Similarly:

$S \left({S \left({n}\right)}\right) < S \left({n}\right)$

Every time the digit sum is taken, the result is at least one less than the previous digit sum.

As $n$ is finite, it will take a finite number of steps to reduce the result to a one digit number.

Hence the result.

$\blacksquare$