# Every positive real number x has a g-adic representation

Firstly we consider only the case $x\in \left[0,1\left[$. For such a number x we use the Gaussian function $\left[X\right]$ i $\left[x\right]=\mathrm{max}\left\{n\in ℤ|n\le x\right\}\in ℤ$. Note that$0\le x-\left[x\right]<1$.
to define recursively the sequence $\left({r}_{n}\left(x\right)\right)$ by

${r}_{1}\left(x\right)≔x\cdot g\wedge {r}_{n+1}\left(x\right)≔\left({r}_{n}\left(x\right)-\left[{r}_{n}\left(x\right)\right]\right)\cdot g$

and subsequently set  ${a}_{n}\left(x\right)≔\left[{r}_{n}\left(x\right)\right]\text{.}$ It is easily seen that $0\le {r}_{n}\left(x\right), so that $0\le \left[{r}_{n}\left(x\right)\right] is valid as well. Therefore we have  ${a}_{n}\left(x\right)\in {D}_{g}$ .

When proving the following please note that the number $x\cdot g-{a}_{1}\left(x\right)={r}_{1}\left(x\right)-\left[{r}_{1}\left(x\right)\right]$  is always a member of $\left[0,1\left[$.

1. Proof  by induction:

►   ${r}_{1}\left(x\cdot g-{a}_{1}\left(x\right)\right)=\left(x\cdot g-\left[{r}_{1}\left(x\right)\right]\right)\cdot g={r}_{2}\left(x\right)$

2. The proof  is immediate from 1.

3. For all  $n\in {ℕ}^{\ast }$ the inequality

Proof  by induction:

►   $|x-\frac{{a}_{1}\left(x\right)}{g}|=|x-\frac{\left[x\cdot g\right]}{g}|=\frac{1}{g}|x\cdot g-\left[x\cdot g\right]|\le \frac{1}{g}$

►   For the inductive step we employ the induction hypothesis with the number $x\cdot g-{a}_{1}\left(x\right)$ !

3. proves the equality  $x=\sum _{i=1}^{\infty }\frac{{a}_{i}\left(x\right)}{{g}^{i}}$. So we only need to overcome the initial restriction for x.
Now let $x\ge 0$ by any positive real. With  $n≔\mathrm{min}\left\{k\in ℕ|x<{g}^{k}\right\}$  (note: $\left({g}^{n}\right)$ is unbounded) we have $\frac{x}{{g}^{n}}\in \left[0,1\left[$ and from that we can set

${x}_{i}≔{a}_{i}\left(\frac{x}{{g}^{n}}\right)$  for $i>0$  and in addition  ${x}_{0}≔0$,

and get:  $\frac{x}{{g}^{n}}=\sum _{i=0}^{\infty }\frac{{x}_{i}}{{g}^{i}}$  and finally have:

$x=\sum _{i=0}^{\infty }\frac{{x}_{i}}{{g}^{i-n}}=\sum _{i=-n}^{\infty }\frac{{x}_{i+n}}{{g}^{i}}=\sum _{i=-n}^{0}\frac{{x}_{i+n}}{{g}^{i}}+\sum _{i=1}^{\infty }\frac{{x}_{i+n}}{{g}^{i}}={x}_{0}{g}^{n}+\dots +{x}_{n}{g}^{0}+\sum _{i=1}^{\infty }\frac{{x}_{i+n}}{{g}^{i}}$