B Squaring finite decimals of 2/3 and 1/3 - growing patterns

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The discussion explores the patterns observed when squaring finite decimal approximations of 2/3 and 1/3, revealing a series of growing segments of identical digits. Squaring these approximations, such as 0.666 and 0.333, produces results that exhibit a consistent structure, with the digits separating into distinct segments. This phenomenon is linked to the properties of base 10 and the nature of recurring decimals, particularly how they interact during multiplication. The mathematical expressions provided illustrate how these patterns emerge through specific calculations involving the digits and their placements. Overall, the regularity in the results stems from the unique characteristics of fractions like 1/3 and 2/3 in decimal form.
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The number of decimal repeats feeds segments of the square with same digits
I was doing some probability calculations that include squaring a number between 0 and 1.
When I approximate 2/3 using 0.6 or 0.66 or 0.666 etc. I get an interesting series of growing same digit segments...

0.6^2=0.36
0.66^2=0.4356
0.666^2=0.443556
0.6666^2=0.44435556
0.66666^2=0.4444355556
0.666666^2=0.444443555556

And (2/3)^2=0.4444444444444...

Similar thing squaring 1/3 approximated as 0.3 or 0.33 or 0.333 etc.

What is this called?
Is it an artifact of base 10?
Sometimes a long division yields a repeating remainder so a similar string of repeats forms, but this is multiplication that produces two growing strings that preserve the digits that separate the strings.
 
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You can investigate this by observing that the n-th item in your sequence is
\begin{align*}\left(\frac23 \times (1 - 10^{-n})\right)^2
&=\left(\frac23\right)^2 \times \left(1 - 2 \times 10^{-n} + 10^{-2n}\right)
\\&=0.\dot4 - 0.0...(n\mathrm{\ zeros})...0\dot8 + 0.0...(2n\mathrm{\ zeros})...0\dot4
\end{align*}
That gives a decimal with zero before the decimal point and 2n digits after the decimal point, with those digits given by:
$$0.4...(n\mathrm{\ fours})...40...(n\mathrm{\ zeros})...0 - 0.0...(n\mathrm{\ zeros})...04...(n\mathrm{\ fours})...4$$
eg the 3rd number is:
\begin{align*}
&0.444000-\\
&0.000444\\
=&0.443556\end{align*}
You can see how the regularity arises from the regularity of that subtraction.

Similarly, with 1/3 we have that the n-th item is
$$0.1...(n\mathrm{\ ones})...10...(n\mathrm{\ zeros})...0 - 0.0...(n\mathrm{\ zeros})...01...(n\mathrm{\ ones})...1$$
eg the 3rd number is:
\begin{align*}
&0.111000-\\
&0.000111\\
=&0.110889\end{align*}

The neat patterns arise for 1/3 and 2/3 because d/9 has base 10 representation of d-recurring. Other bases will have similar patterns for different numerals.
 
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