According to Godel’s Proof (the book) some of these are hoax paradoxes (hence the scare quotes), stemming from some sloppy logic. You could have fooled me, though…

This statement is false.

Is that true or false? Everything has to be either true or false, right?

A two-step version:

The following sentence is false.

The preceeding sentence is true.

Russell’s Antinomy

Sets seem to be of two kinds: those which do not contain themselves, and those which do. A set will be called “normal” iff it does not contain itself as a member; otherwise it will be called “non-normal”.

Example of a normal set: The set of all letters in the alphabet. Surely, the “set of all letters” is not itself a letter in the alphabet, so it does not contain itself.

Example of a non-normal set: The set of all sets. The set of all sets, is a set, so it should be in the set of all sets. Therefore the set of all sets is a non-normal class.

Let $$N$$ by definition stand for the set of all normal sets. We ask whether $$N$$ is a normal set.

If $$N$$ is a normal set, it should be a member the set of all normal sets ($$N$$), but in that case, $$N$$ would contain itself which, by definition, makes it a non-normal set.

On the other hand, if $$N$$ is a non-normal set, it should contain itself. But if $$N$$ is contained within the set of all normal sets, then it must be a normal set.

(A kind of English-language version of the Russell or Richard paradox)

I give you a list of short words/phrases/sentences and your job is to divide them into two classes, one class is the class of phrases which accurately describe themselves, and the other are the phrases which don’t accurately describe themselves.

To get a simple intuition, here are a few examples of phrases from each class:

self descriptive not self descriptive
This is a sentence. The sky is blue.
A phrase A sentence
A six syllable phrase green
multisyllabic Where did you go
pentasyllabic edible
awkwardnessfull nonsense

Now you get the idea.

What group would the phrase “self descriptive” go into? How about “not self descriptive”?

Consider the English definition of numerical properties. For example, “x is a square number if and only if it is the product of a number and itself” could be the definition for the property of being a square number.

Take all such definitions, and sort them alphabetically. Then, number each one according to their position in the sorted list. Now, you have a unique number corresponding to every English definition of a numerical property. Let’s call that the index of the property.

Some indexes may have the property of which they’re the index. For example, if the definition for the properly of being a square number that we put forth above happened to have an index of 64, then the index 64 is an example of what we are talking about. Let’s define the numerical property being “Richardian” as having the property of which you’re the index. Likewise, we will define being “non-Richardian” as not having the property of which you’re the index.

As a starting example, is 64 non-Richardian? No. 64 is square, so it does have the property of which it’s the index, so 64 is Richardian.

Let’s say that 15 is the index of the definition of being even: “x is even iff it is divisible by 2”. Is 15 non-Richardian? Sure, 15 is not even, so 15 does not have the property of which it’s the index.

Now consider the index of the property of being non-Richardian. Let’s say that it is the number $$n$$. Is $$n$$ non-Richardian?

If $$n$$ is non-Richardian, then (by the definition of being non-Richardian) it does not have [the property of which it’s the index] (A). But [the property of which it’s the index] (A) is [the property being non-Richardian] (B, by construction). Substituting (B) for (A) in the first sentence, we get: If $$n$$ is non-Richardian, then it does not have the property of being non-Richardian. Shortened a bit, if $$n$$ is non-Richardian, $$n$$ is not non-Richardian.

Starting with the assumption that $$n$$ is not non-Richardian (i.e. is Richardian) leads to a similar conclusion. If $$n$$ is Richardian, then (by the definition of being Richardian), it has [the property of which it’s the index] (A). But [the property of which it’s the index] (A) is [the property being non-Richardian] (B, by construction). So, if $$n$$ is Richardian, is has the property of of being non-Richardian.

Where am I?

Who cares?

One reasonable reaction the paradoxes above could be something like, “Huh - that’s kinda of interesting, but who really cares?”. I guess my concern is the following. I’ve seen many “proofs” of this form:

Proof that $$A$$ is true:

1. Assume $$A$$ is not true.
3. $$A$$ must be true.