Does Counting Exist? - An Incomplete Mathematical Argument Using Cantor's Methods
Uploader Comments (theboombody)
Video Responses
All Comments (62)
-
@JamesTR4 Yeah, I made an error in this video, that's why I made a follow up one. But the follow up one is interesting in that it shows that the set of whole numbers is uncountable only if you include whole numbers with infinite digits, like the ever-increasing sum of a diverging infinite series. I call these "beginningless numbers."
-
ok, so what if i take the other diagonal. And then construct a number with different digits from the top-right bottom-left diagonal. We still get a different number that isn't counted. Cantor's diagonal argument still holds, because you haven't removed any elements from the set of real numbers.
-
@00wassup00 What about all the beginningless numbers and endless numbers inbetween?
-
@theboombody i see.
Try this concept. Yoctation, Zeptation, Autation, Femptation, Pication, Nanation, Micration, Millation, Centation, Negative Decation, Noventation, Negative Octation, Septation, Sexation, Quintation, Quadation, Negative Exponentiation, Division, Subtraction, Zeration, Addition, Multiplication, Exponentiation, Tetration, Pentation, Hexation, Heptation, Octation, Nonation, Decation, Hectation, Kilation, Megation, Gigation, Terration, Petation, Exation, Zettation and Yattation.
RJL738 2 months ago
@RJL738 Sounds awesome.
theboombody 2 months ago
First of all, the definition of a countable set is a bijective to the naturals. Claiming that a set isn't bijective to itself isn't possible in even the most minimal axiom systems.
Second of all, Cantor's diagonal is a general one about cardinality, the reals-naturals example is a classic result, but still an example. True, it's useful to functional analysis (and frankly, to all fields of math), but it's true home is set theory.
Tikeslar 7 months ago
@Tikeslar I made a follow up to this video that clears some things up. It introduces the concept of beginningless numbers to show that the whole numbers are not countable if you classify beginningless numbers as whole numbers. The naturals probably wouldn't include these numbers.
I also made a video where I show that a two-to-one bijection exists between two particular sets where a one-to-one bijection also exists. This creates a problem for me.
theboombody 7 months ago
@theboombody
The problem with your argument is that you say "beginningless numbers" can be classified as whole numbers. By doing this you have redefined whole numbers. The important thing to note here is that while whole numbers can counted infinitely, they by definition must have a least significant term. This is not the case with the Real numbers, which can by definition have no end.
Your argument only applies to your modified version of whole numbers.
claymullis 5 months ago
@claymullis Well said. I have to make a distinction between natural numbers and whole numbers for my argument to work. For a while I actually took my own video seriously.
theboombody 5 months ago