Re: #304 decimal representation of Infinite Integers such as 9876.....54321; new textbook: Mathematical Physics (Reals & Counting Numbers/AP-adics Primer) for age 6 years onward

a_plutonium <a_plutonium@xxxxxxxxxxx> writes:

Jesse F. Hughes wrote:


Pretty good, but we can replace the infinity symbol with the world's
largest integer 999....99999

We can?

Okay, but then if we apply the same reasoning to *that* number, we
find that 999....99999 is defined as

9x10^0 + 9x10^1 + 9x10^2 + ... + 9x10^{999....9999 - 2}
+ 9x10^{999....9999 - 1} + 9x10^999....9999

So, 999....9999 > 9 x 10^999....9999. Does that seem right to you?


Okay, in my haste I forgot to consider a fine detail that the notation
requires
the infinity digit place value to hold the number 9999.....999999 -1

Which is 9999.....999998 So that the number 9876....54321 really looks
like this


9 x 10^999....9998 + 8 x 10^9999....99997 + .....

Thanks for catching that minor detail mistake.

Yes, minor indeed. But let me just check something. According to
what you've written, we see that

9999...9999 > 10^{9999...9998}.

Is this correct?

On the other hand:

1 = 10^{1-1}
2 < 10^{2-1}
3 < 10^{3-1}
....
n < 10^{n-1}

at least as far as I have checked. Indeed, 10^{n-1} seems to be
increasing faster that n. But according to what you've said, there is
at least one number (999....9999) satisfying n > 10^{n-1}.

You agree so far?

Do you also agree that this means there is some number k so that
k <= 10^{k-1} and yet k + 1 > 10^k? That is, there's some k which is
the "crossover point" for these two functions?

If so, can you help me figure out what that k is? I know it's bigger
than 100 and less than 999...999, but I can't seem to narrow it down
more than that.

Many thanks.

--
Jesse F. Hughes
"To all Leaders of the World, buy or rent the movie 'The Day
After'[...] I assure you will have a new perspective on WMDs."
-- practical advice from online petitions
.

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