Re: Cantor set

In article <461117bd$0$97236$892e7fe2@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Andersen <andersen_800@xxxxxxxxxxx> wrote:
Dave Seaman wrote:

One way to see this is to notice that 1/3 can be written as
(0.02222...)_3. Any number having a base-3 representation that contains
no 1's is a member of the Cantor set.

This is my problem. Because the Cantor set was indeed defined as the way
you defined it in my book. But 1/3 is (0.1)_3 right? So (0.0222...)_3 is
some other number.

No. Some numbers have more than one ternary representation, just as
some numbers have more than one decimal representation.

You know that 0.999999.... is equal to 1.00000... ? The same
phenomenon is being seen with 1/3.

Look at the definition: it says that a number that has ->a<- base-3
representation with the property is in the set. This means that IF a
number has two representations, and at least one of them contains no
1's, then it ->is<- in the set.

Where am I wrong? Is it so that (0.1)_3 = (0.02222...)_3 ? If so, then why?

Remember what the numbers means.

0.02222....

means the result of the convergent series

2/3^2 + 2/3^3 + 2/3^4 + .... + 2/3^n + ...

This is a geometric series, to wit

(2/9)*(1 + 1/3 + 1/9 + ... + 1/3^n + ...)

and the geometric series 1+ 1/3 + ... converges to

1/(1 - (1/3)) = 1/(2/3) = 3/2. Thus, this series converges to
(2/9)*(3/2) = 3/9 = 1/3.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

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