# Write $\sqrt{5}$ as sums of powers of $1 + \sqrt{2}$? [closed]

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?

• let me know if this question is ill-posed. these are always drafts. – john mangual Jan 1 '19 at 0:09
• You should ask this on math.stackexchange.com instead. – Display name Jan 1 '19 at 4:29
• 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 $0.2010201001\ldots$ – 2012rcampion Jan 1 '19 at 5:55
• 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. – 2012rcampion Jan 1 '19 at 6:11