Base $\sqrt{2}$ and other peculiar bases

You're probably familiar with the different integer bases: for example, $45$ is $231_4$, or $231\text{ base }4$. Why? Because $45=2\times 4^2+3\times 4^1+1\times 4^0$. This should be old news to you. If you need a refresher on bases and base conversion, read through this note: Number Base Conversion

But do bases have to be integers? For example, is there any meaning to something like base "$\sqrt{2}$"?

Well, let's try it. Any number $(\overline{d_nd_{n-1}\cdots d_1d_0})_{\sqrt{2}}=d_0\times \sqrt{2}^0+d_1\times \sqrt{2}^1+\cdots +d_n\times \sqrt{2}^n$

So what would $110_{\sqrt{2}}$ be in base $10$?

It would be $0\times \sqrt{2}^0+1\times \sqrt{2}^1+1\times \sqrt{2}^2=2+\sqrt{2}$. Interesting.

But how would we convert $3$ to base $\sqrt{2}$? Well, $3=2+1=\sqrt{2}^2+\sqrt{2}^0=101_{\sqrt{2}}$

Everything more or less is going as expected right now. We have that $110_{\sqrt{2}} > 101_{\sqrt{2}}$, and $2+\sqrt{2} > 3$.

But things in $\sqrt{2}$ aren't always normal. Let's compare the numbers $11_{\sqrt{2}}$ with $100_{\sqrt{2}}$:

$11_{\sqrt{2}}=\sqrt{2}+1$ $100_{\sqrt{2}}=2$

Wait, what? We should have $100_{\sqrt{2}} > 11_{\sqrt{2}}$, because that's just common sense. But clearly $2 < \sqrt{2}+1$. Where did we go wrong?

Rest assured, nothing went wrong. Base $\sqrt{2}$ is just a peculiar base that has these sort of strange properties.

A question is then brought into mind: For what bases does intuition fail us, that a $k$ digit numeral is larger than a $k+1$ digit numeral? Think about it yourself before reading below.

To solve this problem, we compare the largest $k$ digit numeral with the smallest $k+1$ digit numeral in base $B$. First off, we know that $B < 2$ because when $B=2$, the normal intuitive results hold (it's just binary). Also, $B > 1$ or else there cannot be any digit except $0$. Thus, the smallest $k+1$ digit numeral is $(1\underbrace{00\cdots 00}_{k\text{ zeroes}})_B=B^k$

and the largest $k$ digit numeral is clearly $(\underbrace{11\cdots 11}_{k\text{ ones}})_B=\sum_{i=0}^{k-1}B^{i}=\dfrac{B^k-1}{B-1}$

Thus we want $\dfrac{B^k-1}{B-1} > B^k$

This means $B^k-1 > B^{k+1}-B^k$

or $B^{k+1}-2B^k+1 < 0$

When $k=1$, which means $1_B > 10_B$, then $B < 1$ which is not possible.

When $k=2$, which means $11_B > 100_B$, then it turns out $1 < B < \varphi$ where $\varphi=\dfrac{1+\sqrt{5}}{2}$ is the golden ratio. This brings up a good point: $11_{\varphi}=100_{\varphi}$ because of the identity $\varphi^2=\varphi + 1$.

When $k=3$, which means $111_B > 1000_B$, then $1 < B \lesssim 1.839$ where the upper limit is the only real constant that satisfies the identity $x^3=x^2+x+1$.

As $k$ increases, the upper bound tends to $2$. Thus, for sufficiently large $k$, all bases under $2$ and greater than $1$ exhibit the strange property of a $k$ digit numeral being larger than a $k+1$ digit numeral.

I hope you found this little tidbit of non-integral bases and their strange behavior pretty interesting. Thanks for reading!

As an exercise, see if you can:

somehow relate base $\sqrt{2}$ to base $2$

prove or disprove that every integer and real number of the form $a+b\sqrt{2}$ can be uniquely represented in base $\sqrt{2}$.