ternary expansion

**poorna** · June 18th 2009, 08:17 AM

I have been given an assignment of writing out the Cantor set using ternary expansion. But then this is a big problem, because I don't understand how to convert a decimal to ternary expansion. In the sense 1/3 . 3 = 10 (base 3) . 1/10 = 0.1
But how is 0.022222....

Can you give me some other examples.

**pomp** · June 18th 2009, 12:47 PM

Firstly it's important to lay down a definition to avoid ambiguity. So like in base 10 where we define $0.99999\ldots = 0.\dot{9} = 1$ so in base 3 we define $0.22222\ldots_3 = 0.\dot{2}_3 = 1_3$ . Similarly $0.0222\ldots_3 = 0.0\dot{2}_3 = 0.1_3$ .

I'm not going to give you a tutorial on how to convert any number to and from ternary, as it's actually not entirely necessary when we're only looking at the cantor set. We only need to consider negative powers of 3, which are easy to work out. So like you pointed out, $1/3 \times 3 = 1$ which means that

$(1/3)_{10} = (0.1)_3$

$(1/9)_{10} = 1/{3^2}_{10} = (0.01)_3$

$(1/27)_{10} = 1/{3^3} _{10} = (0.001)_3$ etc.

So, using this knowledge we can see that all the numbers between $(1/3)_{10}$ and $(2/3)_{10}$ are exactly those numbers in base 3 between $0.1_3$ and $0.2_3$

which are the numbers $0.1****$ , any number with a one in it's first decimal place.

So now looking at the cantor set in an iterative way, that is, defining

$C_0 = [0,1]$

$C_1 = [0,1] / (2,3) = C_0 / (2,3)$

$C_2 = C_1 / \left(\left(\frac{1}{9}, \frac{2}{9}\right) \cup \left( \frac{7}{9}, \frac{8}{9} \right) \right)$ etc..

$C = \cap _{i=0}^{\infty} C_i$

we can see that with each iteration we are removing the middle third of each interval that remains.

Now we consider exactly what it is we are removing. After the first removal, we have taken away all the numbers between one third and 2 thirds. That is, all the numbers between $0.1_3$ and $0.2_3$ . These numbers all share a common property that should be easy to see as I've already discussed this.

If you look for yourself what is happening after the next iteration you should see a similar thing happenening again. If you keep repeating this you should be able to see just exactly which elements belong in $C$ .

Hope this helps

-pomp

**Soroban** · June 18th 2009, 04:31 PM

Hello, poorna!

I'm not familiar with Cantor sets,
. . but here is a primitive conversion . . .

I have been given an assignment of writing out the Cantor set using ternary expansion.
But then this is a big problem, because I don't understand
how to convert a decimal to ternary expansion.

Suppose we want to convert $0.8$ to ternary.

We assume that the ternary decimal* is of the form: . $0.abcd{e}\!f\!g\hdots$

. So we have: . $0.8 \;=\; 0.abc{d}e\!f\!g\hdots$ . . [Left side: base 10. .Right side: base 3]
Multiply by 3: . $2.4 \;=\;a.bc{d}e\!f\!gh\hdots \quad\Rightarrow \quad \boxed{a \:=\:2}$

. So we have: . $0.4 \:=\:0.bc{d}e\!f\!gh\hdots$
Multiply by 3: . $1.2 \:=\:b.c{d}e\!f\!gh\hdots \quad\Rightarrow\quad \boxed{b \:=\:1}$

.So we have: . $0.2\:=\:0.c{d}e\!f\!gh\hdots$
Multiply by 3: . $0.6 \:=\:c.{d}e\!f\!gh\hdots \quad\Rightarrow\quad \boxed{c \:=\:0}$

.So we have: . $0.6 \:=\:0.{d}e\!f\!gh\hdots$
Multiply by 3: . $1.8 \:=\:d.e\!fgh\hdots \quad\Rightarrow\quad \boxed{d \:=\:1}$

. So we have: . $0.8 \:=\:0.e\!fgh\hdots$ . . . . . and the cycle repeats.

Therefore: . $0.8_{10} \;=\;0.\overline{2101}_3$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* .An interesting point . . .

In base-3 we really can't have a "decimal" . . . (deci- = ten)
. . So what do we call it? . . . a ternimal ?

In base-4, a quadrimal or tetrimal ?

**alunw** · June 18th 2009, 05:45 PM

poorna used the usual expression for base 3 fractions "ternary". Not many number bases are important enough to have names. The usual names are
2: binary
3: ternary
8: octal
10: decimal
12: duodecimal
16: hexadecimal
60: sexagesimal

I don't think there is enough consitency in these names to guess what names for other bases should be: they seem to mix up Latin and Greek. The book "Collapse of Chaos" uses "octimal" for base 8, but this is probably a joke.

**poorna** · June 19th 2009, 05:44 AM

@ pomp
oh thanks a lot! i'm working on it. so i'm just taking a wild guess. are we left with all elements which do not contain 1 in them. I'm yet to check by actual computation.

thanks a lot! i understood

But by what you said ,
1/3 = 0.33333333333..
multiplying by 3 by once, we have 0.9999999... so a here is 0
then again multiplying 2.99999999999.... so b is 2
similarly c, d, e .... are 2
So 1/3 = .0222222...
but going by this way, we donot explicitly get 1/3 is 0.1 right?

**pomp** · June 19th 2009, 10:02 AM

Yeh nice work man, that should be the result you arrive at

**alunw** · June 19th 2009, 01:29 PM

Originally Posted by poorna

@ pomp
oh thanks a lot! i'm working on it. so i'm just taking a wild guess. are we left with all elements which do not contain 1 in them. I'm yet to check by actual computation.

Yes, as Pomp said you really don't need to do any complicated base conversions. The classic Cantor set with all middle thirds deleted contains all the ternary fractions without a 1 in them. But since numbers that end with 1 followed by infinitely many zeros can also be expressed by replacing the 1000... with 0222. those numbers, and also the set is usually defined by removing open rather than closed intervals (so that you do not delete 1/3 or 2/3 on the first iteration) you probably should be including all those numbers.

I found a nice article about the set here Cantor Set (PRIME) (though probably you already found it for yourself)

**HallsofIvy** · June 20th 2009, 04:42 AM

Originally Posted by Soroban

Hello, poorna!

I'm not familiar with Cantor sets,
. . but here is a primitive conversion . . .

Suppose we want to convert $0.8$ to ternary.

We assume that the ternary decimal* is of the form: . $0.abcd{e}\!f\!g\hdots$

. So we have: . $0.8 \;=\; 0.abc{d}e\!f\!g\hdots$ . . [Left side: base 10. .Right side: base 3]
Multiply by 3: . $2.4 \;=\;a.bc{d}e\!f\!gh\hdots \quad\Rightarrow \quad \boxed{a \:=\:2}$

. So we have: . $0.4 \:=\:0.bc{d}e\!f\!gh\hdots$
Multiply by 3: . $1.2 \:=\:b.c{d}e\!f\!gh\hdots \quad\Rightarrow\quad \boxed{b \:=\:1}$

.So we have: . $0.2\:=\:0.c{d}e\!f\!gh\hdots$
Multiply by 3: . $0.6 \:=\:c.{d}e\!f\!gh\hdots \quad\Rightarrow\quad \boxed{c \:=\:0}$

.So we have: . $0.6 \:=\:0.{d}e\!f\!gh\hdots$
Multiply by 3: . $1.8 \:=\:d.e\!fgh\hdots \quad\Rightarrow\quad \boxed{d \:=\:1}$

. So we have: . $0.8 \:=\:0.e\!fgh\hdots$ . . . . . and the cycle repeats.

Therefore: . $0.8_{10} \;=\;0.\overline{2101}_3$

Which proves, by the way, that $0.8_{10}$ is not in the Cantor ternary set. Do you see why, Poorna?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* .An interesting point . . .

In base-3 we really can't have a "decimal" . . . (deci- = ten)
. . So what do we call it? . . . a ternimal ?

In base-4, a quadrimal or tetrimal ?

**poorna** · June 20th 2009, 07:34 AM

@ alunw

Yeah, we donot remove 1/3 and 2/3 in the first iteration.
But as I was working out, I sort of concluded that we donot remove them ever.
In the sense, all the boundary elements, if I may call them so, are a part of the final Cantor set. Am I right?

**ptolemy** · July 25th 2009, 06:48 PM

@pomp

Can you please throw some light on the notation used in the
iterations for describing the construction of the cantor set?

Thanks,

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