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# 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 $$\sqrt{2}$$.

Note by Daniel Liu
2 years, 7 months ago

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I think this note was a little bit disorganized and "jumpy". I just noticed this intriguing fact, and wanted to share it to everyone in note-form. Enjoy! · 2 years, 7 months ago

Loved it!!! · 2 years, 7 months ago

Nice discussion. Loved the way of its peculiarity. · 2 years, 7 months ago

You should have called the note "When 11 is greater than 100" · 2 years, 7 months ago

Eh I like to keep my titles (for notes) saying what I'm about to discuss · 2 years, 7 months ago

@Daniel Liu Can you edit this note into Fractional NUmber Bases Wiki? Thanks! Staff · 2 years, 1 month ago

Can anyone explain me what is a|b means..??? · 2 years, 7 months ago

It means 'a' divides 'b'. · 2 years, 7 months ago

It means that a divides b. For example, $$2|4$$, $$3|6$$ , $$278|278$$, etc · 2 years, 7 months ago

Thanks.!! · 2 years, 7 months ago

Like the one in this question

See the problem · 2 years, 7 months ago

This is awesome. It would be interesting to do a table of addition and multiplication. So 1 + 1 = 100. This can also be justified by the fact that the digits 0 and 1 don't divide the square root of 2 into equal parts. · 2 years, 7 months ago