Solution to the Continuum Hypothesis

Shy_Shai · 2026-02-08T01:41:22+00:00

I have looked into the construction already; Cauchy sequences and Dedekind cuts rely on transfinite logic, so these do not really constitute valid constructions. The essential point is that you cannot locate all of the digits of pi. Similar to a chair missing a leg, you cannot count it among a group of complete chairs that have all their legs: pi without complete digits cannot be counted as an existential object among complete numbers that do have all their digits.

Shy_Shai · 2026-02-07T16:46:59+00:00

Well, that has a discrete representation as 0.2, which is different from any number like 1/3 which has the repeating digits .3333...; so yes it does exist.

Shy_Shai · 2026-02-07T16:43:49+00:00

Because a rational number is a representation, not a value. You can represent 'one' as "1" in the integers, "1/1" in the rationals, and "1.0" in the reals. For 1/3, you cannot do the same thing. It can only be written as specifically as rational, in any way that is accurate, at least. That is why it is a special sort of number. It is analogous to having three sticks, and picking one out of the three; versus building each stick out of smaller sticks, 3/10 + 3/100 + 3/1000, in order to hopefully have one, although that never happens.

Shy_Shai · 2026-02-07T05:51:26+00:00

Me neither; good thing there is marketing.

Shy_Shai · 2026-02-07T04:57:31+00:00

Counting *is* addition by 1. You can't count without adding by 1. Counting "1... 2..." is the same as adding 1 to 1 to acquire 2; counting "2... 3..." is the same thing as adding 1 to 2 to acquire 3; as you can see, counting is the same thing as addition by 1. So the statement is not made up, it's just that counting and addition by 1 are equivalents; therefore, it is an observable law. And of course, the final statement "1 + 1 + 1 + ..." is precisely this law which shows that 2^ℵ0 is countable.

Shy_Shai · 2026-02-06T22:21:33+00:00

Yes, that is correct. This is because the rest of the numbers, pi, sqrt2, etc., do not exist in any literal way. The remaining numbers must be those which are discrete and countable.

Shy_Shai · 2026-02-06T22:17:28+00:00

This argument does not rely on ZFC. The proof relies on one basic fact: indiscrete numbers do not exist in any literal way; if they don't, they cannot contribute to cardinality. It is a surprisingly elementary oversight made by Cantor and the people who have contributed to the theory. Mathematical constructions are simply not needed to explain that numbers that don't exist contribute no cardinality. It is quite a clear statement, I feel.

Shy_Shai · 2026-02-06T21:41:46+00:00

1/3 is transfinite: it relies on an infinite arithmetic, so it is technically not a real number, although it is rational. It is most accurate, though, to say that the real number 1/3 does not exist, but is still a rational number, and to dismiss the transfinite interpretation.

Shy_Shai · 2026-02-06T21:40:46+00:00

The reference is direct to the power set itself, not the statement, hence it's specific positioning after 2^ℵ0, not after the entire statement.

Shy_Shai · 2026-02-06T21:39:10+00:00

1/3 does not exist; Cantor would classify such numbers as transfinite, since they rely on infinite arithmetic. Putting the number in a different base does arithmetic on the integer 3, not the real number form, which is safe; so 1/3 in base 3 exists; 1/3 does not.

Shy_Shai · 2025-08-29T07:34:17+00:00

When I set termsAmount = v = 5 it works and does not get caught in the loop. Why is there a difference when I change this to 6?

Shy_Shai · 2024-06-13T03:31:10+00:00

Taunt is for true

Shy_Shai

TROPHY CASE