Re: #326 Counting an infinite set versus a finite set ; new textbook: Mathematical Physics (Reals & Counting Numbers/AP-adics Primer) for age 6 years onward



a_plutonium <a_plutonium@xxxxxxxxxxx> writes:


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.)

I conceded not because of your argument, although it hastened it, for
I conceded because I have the illogic that 999....99999 +1 =
10000....00000
which is (pi) the South Pole and yet that cannot be consistent with
10^9999.....99999 which is also the South Pole since it has one more
digit place
value than 9999.....999999 itself.

So it is not your argument that force me to concede although it helped
to hasten.

Whatever.


[...]


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.


No, I buy none of that.

Okay, then you're just too broke to reason with.

You claim that every AP-adic aside from zero has a predecessor, yes?
And that, starting at zero and adding one repeatedly, we will
eventually reach any AP-adic?

Let P be the property "We count to n before we count to 10^{n-1}." We
know that P is true at n=1, 2, 3. We know it is false at n =
999...999.

Now, I claim there is some n such that P(n) is true, but P(n+1) is
false. You claim not. Let's suppose you're right. Then it follows
that, if P(n) is true, so is P(n+1).

Start counting at 1 and we get:

1: P(1) is true.
2: P(2) is true because P(1) is true.
3: P(3) is true because P(2) is true.
4: P(4) is true because P(3) is true.

and so on. Each time we reach a new number n, P(n) must be true,
since it was true at the predecessor. According to you, eventually,
we reach 999...9997 by counting and thus we see:

999...99997: P(999...99997) is true because P(999...99996) is true.
999...99998: P(999...99998) is true because P(999...99997) is true.
999...99999: P(999...99999) is true because P(999...99998) is true.

Oops!

Thus, what I said must be correct. There is a number n so that P(n)
is true and P(n+1) is false.

I've wasted enough time on this obvious point to continue, I reckon.
--
Jesse F. Hughes
"Certainly he who can digest a second or third fluxion need
not, methinks, be squeamish about any point in divinity."
George Berkeley, 1734
.



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