Barber paradox

The barber paradox is a puzzle derived from Russell's paradox. It was used by Bertrand Russell as an illustration of the paradox, though he attributes it to an unnamed person who suggested it to him.[1] The puzzle shows that an apparently plausible scenario is logically impossible. Specifically, it describes a barber who is defined such that he both shaves himself and does not shave himself.

Paradox

The barber is the "one who shaves all those, and those only, who do not shave themselves". The question is, does the barber shave himself?[1]

Answering this question results in a contradiction. The barber cannot shave himself as he only shaves those who do not shave themselves. Thus, if he shaves himself he ceases to be the barber. Conversely, if the barber does not shave himself, then he fits into the group of people who would be shaved by the barber, and thus, as the barber, he must shave himself.

A barber is one who shaves those who do not shave themselves. So, can a barber shave himself?

If he does, he cannot be a barber, since a barber does not shave himself.
If he doesn't, he falls in the category of those who do not shave themselves, and so, cannot be a barber.

History

This paradox is often incorrectly attributed to Bertrand Russell (e.g., by Martin Gardner in Aha!). It was suggested to Gardner as an alternative form of Russell's paradox,[1] which Russell had devised to show that set theory as it was used by Georg Cantor and Gottlob Frege contained contradictions. However, Russell denied that the Barber's paradox was an instance of his own:

That contradiction [Russell's paradox] is extremely interesting. You can modify its form; some forms of modification are valid and some are not. I once had a form suggested to me which was not valid, namely the question whether the barber shaves himself or not. You can define the barber as "one who shaves all those, and those only, who do not shave themselves". The question is, does the barber shave himself? In this form the contradiction is not very difficult to solve. But in our previous form I think it is clear that you can only get around it by observing that the whole question whether a class is or is not a member of itself is nonsense, i.e. that no class either is or is not a member of itself, and that it is not even true to say that, because the whole form of words is just noise without meaning.
— Bertrand Russell, The Philosophy of Logical Atomism

This point is elaborated further under Applied versions of Russell's paradox.

In first-order logic

(\exists x)({\text{person}}(x)\wedge (\forall y)({\text{person}}(y)\implies ({\text{shaves}}(x,y)\iff \neg {\text{shaves}}(y,y))))

This sentence is unsatisfiable (a contradiction) because of the universal quantifier $(\forall )$ . The universal quantifier $y$ will include every single element in the domain, including our infamous barber $x$ . So when the value $x$ is assigned to $y$ , the sentence can be rewritten to ${\text{shaves}}(x,x)\iff \neg {\text{shaves}}(x,x)$ , which is an instance of the contradiction $a\iff \neg a$ .

References

The Philosophy of Logical Atomism, reprinted in The Collected Papers of Bertrand Russell, 1914-19, Vol 8., p. 228

External links

Proposition of the Barber's Paradox
Joyce, Helen. "Mathematical mysteries: The Barber's Paradox". Plus, May 2002.
Edsger Dijkstra's take on the problem
The Monist, Vol. 29, No. 3, JULY, 1919, THE PHILOSOPHY OF LOGICAL ATOMISM, page 354

This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.

[atomism-1] The Philosophy of Logical Atomism, reprinted in The Collected Papers of Bertrand Russell, 1914-19, Vol 8., p. 228