# Powers Of Two In Lex Order Subscribe! My latest posts can be found here:
Previous blog posts:
 I'm still waiting to hear back about how my friend would like to be credited, so this page will change when I get that information. This page has been Tagged As Maths
Recently I was sent a lovely little twitchet by a friend, so I thought I'd share it for people to think about. No answers provided ...

## Firstly, let's do something odd ...

Take the powers of 2 up to $2^9$.

• 1, 2, 4, 8, 16, 32, 64, 128, 256, 512.

Now list them in lexicographic order (personally, I find this a bit tricky!)

• 1, 128, 16, 2, 256, 32, 4, 512, 64, 8.

Now put a decimal point after the first digit:

• 1.0, 1.28, 1.6, 2.0, 2.56, 3.2, 4.0, 5.12, 6.4, 8.0

So that gives us a sequence of numbers in the range from 1 to 8.

Now put that list to one side and ...

## ... Let's do something completely different ...

 n Power Ans 0 $10^{0.0}$ 1.000 1 $10^{0.1}$ 1.259 2 $10^{0.2}$ 1.585 3 $10^{0.3}$ 1.995 4 $10^{0.4}$ 2.512 5 $10^{0.5}$ 3.162 6 $10^{0.6}$ 3.981 7 $10^{0.7}$ 5.012 8 $10^{0.8}$ 6.310 9 $10^{0.9}$ 7.943

Here are some powers of 10. For each number $n$ from 0 to 9 we are computing 10 to the power of $n/10$. This seems to be unconnected with our previous exercise, I know, but having done this we can have a look at the numbers we get.

There's no surprise that the answers are in the range from 1 to 8, and they are not especially "nice" numbers, but that is what you get when you raise a number to a fractional power (whatever that means).

But looking closely at the last column, with the eye of faith, you may notice something interesting.

## Combining the results ...

Let's augment the table of powers by adding a last column with the results of the deeply odd process we started with.

 n Power Ans Oddity 0 $10^{0.0}$ 1.000 1.00 1 $10^{0.1}$ 1.259 1.28 2 $10^{0.2}$ 1.585 1.60 3 $10^{0.3}$ 1.995 2.00 4 $10^{0.4}$ 2.512 2.56 5 $10^{0.5}$ 3.162 3.20 6 $10^{0.6}$ 3.981 4.00 7 $10^{0.7}$ 5.012 5.12 8 $10^{0.8}$ 6.310 6.40 9 $10^{0.9}$ 7.943 8.00

The agreement really is remarkably good (says he, remarking on it, and thus making it a self-fulfilling statement). But it really is remarkably good, and the question one is left with is:

 How does that work?

I know how to do the calculations, I know how to prove that this is close, but what I'm lacking is a clear, "intuitive" explanation for people who are not comfortable with detailed calculation.

Explanations invited.

 <<<< Prev <<<< Emerging E Expanded : >>>> Next >>>> Seventy Versus One Hundred ... You can follow me on Mathstodon.

 Of course, you can alsofollow me on twitter:  ## Send us a comment ...

 You can send us a message here. It doesn't get published, it just sends us an email, and is an easy way to ask any questions, or make any comments, without having to send a separate email. So just fill in the boxes and then

 Your name : Email : Message :

# Contents Suggest a change ( <-- What does this mean?) / Send me email