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If we take the square root of 012345, we get 111.108055514. The square root of 012345678901234567 approximates to 111111110.611. We're taking the square roots of numbers following a pattern, the numbers 0-9 repeating with an even number of digits before any decimal (I'm counting it starting with 0). Note that we get very close to 111.111 repeating, or a variation with decimal moved.

Now let's omit '8' from the pattern. square root(123456790123456) = 11111111.1111. That's a far closer approximation. I observe, ((10)^nth/9)^2, where n is positive will be represented by the digits 0-9 repeating omitting 8.

This pattern will only appear significant in a number base 10 system, without a number base 10 1234567890123 doesn't appear to be a number with a pattern. What is the significance of '8' in 100/81? in 1000/81?

Phillip Crawford
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  • The same pattern does appear in other bases: in hexadecimal, the square root of 123456789ABCDE01234 (without the F) is very close to the hexadecimal 1111111111.1111, for example. – Greg Martin Sep 09 '19 at 22:39

1 Answers1

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Consider what happens when you square $1111\cdots1.$ Near the beginning (where the most-significant digits are) you are adding the string of ones to itself shifted by every place: \begin{align*} 111111111111\cdots &\\ + 11111111111\cdots &\\ + 1111111111\cdots &\\ + 111111111\cdots &\\ + 11111111\cdots &\\ \vdots&\\ = 123456790123 \cdots & \end{align*} Notice that after the 10th digit, the sum carries; the carry causes the 9 in the previous digit of the sum to carry, so that instead of $$789(10)(11)(12)\cdots$$ you get $$790123\cdots.$$ After another ten digits the sum starts carrying twice instead of once, so that the pattern repeats, etc.

user7530
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