The number $A = 1 + \sqrt{2}$ solve the equation $A^2 = 1 + 2 \sqrt{2} + 2 = 1+2A$. We also have that $$0 < \dots < A^{-2} < A^{-1} < 1 < A < A^2 < \dots $$ The powers are in increasing order. Any real number $x \in \mathbb{R}$ is close to a sum of powers of $A$, $x \approx \sum a_i A^i$ with $a_i = 0, 1$ or $2$.

The $\beta$-expanson or the $q$-expansion requires $0 \leq a_i \leq \lfloor 1+\sqrt{2}\rfloor=2$ in this case.
There's also an article on Golden ratio base, but that's $B = \frac{1+\sqrt{5}}{2}$ and $B^2 = B+1$.

Let's try $x = 3$. We could have $x = 1 + 1 + 1 = 3$. That's not acceptable in our base expansion, since we are writing $x = 3 \times 1$. So instead let's write $2 = A + \frac{1}{A}$. Then we have a complete answer $3 = A^{-1} + 1 + A$.

What happens with $x= \sqrt{5}$? I know it's between $1$ and $2$. We have $2 < \sqrt{5} < 1 + (1+\sqrt{2})$. Can we get ten decimal places? Or even 100?


closed as off-topic by rhsquared, Peregrine Rook, PerpetualJ, athin, Excited Raichu Jan 1 at 2:26

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  • $\begingroup$ let me know if this question is ill-posed. these are always drafts. $\endgroup$ – john mangual Jan 1 at 0:09
  • $\begingroup$ You should ask this on math.stackexchange.com instead. $\endgroup$ – Display name Jan 1 at 4:29
  • $\begingroup$ Note that $A^{-1}=-1+\sqrt{2}$, thus all powers of $A$--both positive and negative--and therefore all base-$A$ numbers are expressible in the form $p+q\sqrt{2}$ with $p$ and $q$ integers. Since $\sqrt{5}$ is not of that form, it will not have a terminating expansion; but, like other real numbers, it does have a unique nonterminating expansion, beginning $0.2010201001\ldots$ $\endgroup$ – 2012rcampion Jan 1 at 5:55
  • $\begingroup$ Also, you get about $\log_{10}(1+\sqrt{2})\approx 0.38$ decimal digits of precision for each additional base-$A$ digit, so to get 100 decimal places you need around 260 base-$A$ digits. $\endgroup$ – 2012rcampion Jan 1 at 6:11