Re: #321 10^9999....999999 is (pi) and where 10^999...9998 is ; 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:

Same issue.

If I start counting at 0, I get to 1 at the same time as I get to 10^0.
I get to 2 before I get to 10^1.
I get to 3 before I get to 10^2.
I get to 4 before I get to 10^3.

and so on, but I get to 10^{999....998} before I get to 999....999.

Of course you do because 10^9999....99998 looks like this

10000.....00000 while 9999.....999999 is this
99999.....99999



and thus, there is at least one number between n and n + 1.


You lost your claim there.

How sad. You snipped my argument that showed why I made this claim.
I will repeat it here. It follows from what you admit right above.

If I start counting at 0, I get to 1 at the same time as I get to
10^0.
I get to 2 before I get to 10^1.
I get to 3 before I get to 10^2.
I get to 4 before I get to 10^3.

and so on, but I get to 10^{999....998} before I get to 999....999.

(The above you just conceded.)

Thus, there is some n such that

If I start counting at 0, I get to n before I get to 10^{n-1}, but
I get to 10^{n} before I get to n+1
(or maybe at the same time).

Do you agree with the above? If so, then what follows is an obvious
consequence.

As lwalke3 pointed out, this would mean:

I get to n before 10^{n-1},
to 10^{n-1} before 10^n,
to 10^n before or at the same time as n + 1

and thus, there is at least one number between n and n + 1.


--
"If you have a really big idea, you can get a measure of how big it is
by how much people resist the obvious. From what I've seen, I have a
REALLY, REALLY, *REALLY*, BIG DISCOVERY!!!"
--James Harris: If I'm not important, how come people ignore me?
.

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