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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
3 years, 4 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!

Daniel Liu - 3 years, 4 months ago

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Loved it!!!

Siddharth Brahmbhatt - 3 years, 4 months ago

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Nice discussion. Loved the way of its peculiarity.

Sharky Kesa - 3 years, 4 months ago

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You should have called the note "When 11 is greater than 100"

Nathan Ramesh - 3 years, 4 months ago

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Eh I like to keep my titles (for notes) saying what I'm about to discuss

Daniel Liu - 3 years, 4 months ago

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@Daniel Liu Can you edit this note into Fractional NUmber Bases Wiki? Thanks!

Calvin Lin Staff - 2 years, 10 months ago

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Can anyone explain me what is a|b means..???

Sudipta Biswas - 3 years, 4 months ago

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It means 'a' divides 'b'.

Aabhas Mathur - 3 years, 4 months ago

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It means that a divides b. For example, \( 2|4 \), \( 3|6 \) , \(278|278 \), etc

Siddhartha Srivastava - 3 years, 4 months ago

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Sudipta Biswas - 3 years, 4 months ago

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Like the one in this question

See the problem

Sudipta Biswas - 3 years, 4 months ago

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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.

Adrian Neacșu - 3 years, 4 months ago

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Subrata Saha - 3 years, 2 months ago

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