# Does the shifting square root method work for non-integer bases?

Under "methods of computing square roots", Wikipedia states that the digit-by-digit calculation method, of which the shifting $n^{th}$ root algorithm is a generalization, works for all bases, but the page for the shifting $n^{th}$ root algorithm lists the base $B$ with other variables as an integer.

Does the digit-by-digit method work for non-integer bases like golden ratio base, base-$\sqrt 2$ or base-$\pi$?

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why don't you try it out? :) –  VelvetThunder Jul 27 '12 at 2:00
Note that when people say 'for all bases' they almost certainly implicitly mean 'integer bases' because of the difficulties involved in cleanly defining non-integer bases. –  Steven Stadnicki Jul 27 '12 at 4:34
@Quixotic, I'd love to, but it's a bit tricky. I was hoping there was a theoretical reason that the method would work or not work for some or all non-integer bases. I want to know if the algorithm can be generalized to non-integer bases. –  hatch22 Jul 27 '12 at 15:20
@Steven - duly noted. But I still think the question is worth investigating. Other methods of finding square roots can work for arbitrary real or even complex bases. Is this method restricted to integer bases because of something in the nature of how it works, or will it also work for arbitrary real bases? –  hatch22 Jul 27 '12 at 15:20
@StevenStadnicki We can generalize the representation of a number with a summation: $\sum a_n b^n$ where $a_n$ is the digit in position $n$ in the range $[0, b)$ and where $b$ is the base. Then, for $12_\pi$, $a_1$ is $1$, $a_0$ is $2$, and $b$ is $\pi$. This value is then $\pi^2 + 2\pi$. –  Cole Johnson May 25 '14 at 21:53