Reciprocal of 81 being the sequence of all natural numbers?

Question

According to this document: http://www.answering-christianity.com/fakir60/81.htm describing the theory of scientist Peter Plichta, the reciprocal of 81 is: the sequence of all natural numbers (0.01234567890...).

Myself not quite being a mathematician, but still trying to reproduce everything I come across, typed it in the calculator: reciproc(81). And there it was:

0.0123456790...

Uuhm, 8 where art thou?

The missing 8 is accounted for in the document as it goes on describing it is merely an illusion because the reciprocal calculation should be written differently and it should be read as Gauchy product (<< does this make any sense?):

1 : 81 = 1 / 9 x 1 / 9 = 0,111....x 0,111.... = 0,0(1)(1+1)(1+1+1)(1+1+1+1)....etc

= 0,123456789(10)(11)(12)(13)....( Cauchy product)

Can somebody explain in layman's terms what's going on in this last step? Am I correct and can this be referred to as base 9 counting?

(p.s.: my rep doesn't let me correctly tag this question)

Answer 1

because when you do $0.00000008 +0.000000009 + 0.0000000010 + 0.00000000011$ carry rules apply and it becomes $0.0000000901...$

\begin{align} 0&.01 &+\\ 0&.002 &+\\ 0&.0003 &+\\ 0&.00004 &+\\ 0&.000005 &+\\ 0&.0000006 &+\\ 0&.00000007 &+\\ 0&.000000008 &+\\ 0&.0000000009 &+\\ 0&.00000000010 &+\\ 0&.000000000011 &+\\ 0&.0000000000012 &=\\ 0&.0123456790122 \end{align}

Edit: $\frac1{81}$ is a rational number, thus it has a periodic representation in base 10. It can be easily verified that $$\frac1{81} = \frac{12345679}{999999999} = 0.\overline{012345679}$$ so $8$ never appears in the decimal expansion of $\frac1{81}$

Answer 2

This is really just a comment, but I didn't want you to miss it:

Stay away from that site! It will rot your brain!

Just about every sentence is false. There aren't many crank sites you can say that about, but this is one of them.