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Does the barber shave himself?

 
 
Reply Wed 30 Jun, 2010 10:29 am
There is a male barber in a town who shaves all and only those men of the town who do not shave themselves.

Does the barber shave himself or not?

If he dosen't shave himself he does shave himself, and,
If he does not shave himself he does shave himself.

What is the solution to this apparent paradox?
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Type: Question • Score: 0 • Views: 6,353 • Replies: 16
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mister kitten
 
  1  
Reply Wed 30 Jun, 2010 06:45 pm
Someone else shaves the barber.
Owen phil
 
  1  
Reply Thu 1 Jul, 2010 04:53 am
@mister kitten,
mister kitten wrote:

Someone else shaves the barber.


If someone else shaves the barber then the barber does not shave himself.

If the barber does not shave himself then the barber does shave himself.
Because, he is one of 'those who do not shave themselves'.

If someone else shaves the barber then, someone else shaves him and the barber shaves himself....a contradiction.
Therfore,
He cannot be shaved by some other person.


If he dosen't shave himself he does shave himself, and,
If he does shave himself he does not shave himself.
0 Replies
 
dadpad
 
  1  
Reply Thu 1 Jul, 2010 05:03 am
It was the man from Ironbark who struck the Sydney town,
He wandered over street and park, he wandered up and down.
He loitered here, he loitered there, till he was like to drop,
Until at last in sheer despair he sought a barber's shop.
`'Ere! shave my beard and whiskers off, I'll be a man of mark,
I'll go and do the Sydney toff up home in Ironbark.'

The barber man was small and flash, as barbers mostly are,
He wore a strike-your-fancy sash, he smoked a huge cigar:
He was a humorist of note and keen at repartee,
He laid the odds and kept a `tote', whatever that may be,
And when he saw our friend arrive, he whispered `Here's a lark!
Just watch me catch him all alive, this man from Ironbark.'

There were some gilded youths that sat along the barber's wall,
Their eyes were dull, their heads were flat, they had no brains at all;
To them the barber passed the wink, his dexter eyelid shut,
`I'll make this bloomin' yokel think his bloomin' throat is cut.'
And as he soaped and rubbed it in he made a rude remark:
`I s'pose the flats is pretty green up there in Ironbark.'

A grunt was all reply he got; he shaved the bushman's chin,
Then made the water boiling hot and dipped the razor in.
He raised his hand, his brow grew black, he paused awhile to gloat,
Then slashed the red-hot razor-back across his victim's throat;
Upon the newly shaven skin it made a livid mark --
No doubt it fairly took him in -- the man from Ironbark.

He fetched a wild up-country yell might wake the dead to hear,
And though his throat, he knew full well, was cut from ear to ear,
He struggled gamely to his feet, and faced the murd'rous foe:
`You've done for me! you dog, I'm beat! one hit before I go!
I only wish I had a knife, you blessed murdering shark!
But you'll remember all your life, the man from Ironbark.'

He lifted up his hairy paw, with one tremendous clout
He landed on the barber's jaw, and knocked the barber out.
He set to work with tooth and nail, he made the place a wreck;
He grabbed the nearest gilded youth, and tried to break his neck.
And all the while his throat he held to save his vital spark,
And `Murder! Bloody Murder!' yelled the man from Ironbark.

A peeler man who heard the din came in to see the show;
He tried to run the bushman in, but he refused to go.
And when at last the barber spoke, and said, `'Twas all in fun --
'Twas just a little harmless joke, a trifle overdone.'
`A joke!' he cried, `By George, that's fine; a lively sort of lark;
I'd like to catch that murdering swine some night in Ironbark.'

And now while round the shearing floor the list'ning shearers gape,
He tells the story o'er and o'er, and brags of his escape.
`Them barber chaps what keeps a tote, By George, I've had enough,
One tried to cut my bloomin' throat, but thank the Lord it's tough.'
And whether he's believed or no, there's one thing to remark,
That flowing beards are all the go way up in Ironbark.


The Barber is female.
Owen phil
 
  1  
Reply Thu 1 Jul, 2010 06:23 am
@dadpad,
dadpad wrote:

The Barber is female.


Another contradiction.

There is a "male" barber in a town who shaves all and only those men of the town who do not shave themselves.
dadpad
 
  1  
Reply Thu 1 Jul, 2010 06:39 am
@Owen phil,
Quite correct phil. My apologies for not paying attention.
dadpad
 
  1  
Reply Thu 1 Jul, 2010 06:41 am
A bearded barber perhaps?
Its late I will give this more thought.
0 Replies
 
Owen phil
 
  1  
Reply Thu 1 Jul, 2010 06:47 am
@dadpad,
dadpad wrote:

Quite correct phil. My apologies for not paying attention.


No need for apologies here, the argument is designed to have no answer.
The conditions for the barber are not possible, that is, the presumed barber cannot exist.

By the way, do you know why I have inherited the last name 'Phil'?
0 Replies
 
Tryagain
 
  2  
Reply Fri 2 Jul, 2010 04:50 pm
This paradox as studied in Mathematics is a result of an inherent
problem with defining a set as:

{x|x possesses certain properties}

where x is an individual in the universe of entities under study (and
"|" means "such that").

This way of defining a set gets into trouble when we define the set S as

S = {x|x is not a member of x}

i.e. S is the set of sets that are not members of themselves (please
interpret this comment in conjunction with the notational definition
given above).

The paradox is that S can neither be a member of itself nor not be a
member of itself. More specifically,

x is in S <=> x is not in x ("<=>" is just "if and only if")

Substituting S for x (to find out whether S itself is in S), we have

S is in S <=> S is not in S (#)


Coming back to the barber, the set of (all) inhabitants of the
town who are shaved by the barber is

Q = {y | y does not shave y}

Thus, if the barber is b,

b shaves y <=> y does not shave y

To find out whether b is in Q or not, we substitute b for y and get

b shaves b <=> b does not shave b (##)

(##) is the same kind of paradox as (#).


There is a real paradox here. How to give an accurate account of the barber's version of this paradox may not be trivial. But whether an attempt to do so is successful or not, the paradox is there for its own sake.

However, it is therefore not a paradox but a linguistic sleight of hand. Perhaps that this is precisely the nature of a paradox and that this conclusion is the one to which the mathematical philosophers would have us lead?

The precise (classical)'fix' to the problem of the set of all sets that do not contain themselves is a hierarchy of set-LIKE entities arranged so that the paradoxical self-embedding objects do not exist - the addition, that is, of NEW categories not stated in the problem.

To wit: The man shaves himself BEFORE he starts work as a barber!





Quote:
"By the way, do you know why I have inherited the last name 'Phil'?"

Probably because there was already a 'Owen' registered here.

spikepipsqueak
 
  1  
Reply Fri 2 Jul, 2010 05:21 pm
@Owen phil,
Owen phil wrote:

There is a male barber in a town who shaves all and only those men of the town who do not shave themselves.

Does the barber shave himself or not?

If he dosen't shave himself he does shave himself, and,
If he does not shave himself he does shave himself.

What is the solution to this apparent paradox?



The barber is not a man of that town.
0 Replies
 
DrewDad
 
  1  
Reply Fri 2 Jul, 2010 05:34 pm
@Owen phil,
Owen phil wrote:
Does the barber shave himself or not?

Yes.
0 Replies
 
Owen phil
 
  1  
Reply Sun 4 Jul, 2010 05:08 am
@Tryagain,
Tryagain wrote:

This paradox as studied in Mathematics is a result of an inherent
problem with defining a set as:

{x|x possesses certain properties}

where x is an individual in the universe of entities under study (and
"|" means "such that").

This way of defining a set gets into trouble when we define the set S as

S = {x|x is not a member of x}

i.e. S is the set of sets that are not members of themselves (please
interpret this comment in conjunction with the notational definition
given above).


I don't agree that the definition of a set produces problems.

The set of individuals that satisfies the predicate F =df {x|Fx}.
How would you define a set determined by the predicate F ?

Tryagain wrote:

The paradox is that S can neither be a member of itself nor not be a
member of itself. More specifically,

x is in S <=> x is not in x ("<=>" is just "if and only if")

Substituting S for x (to find out whether S itself is in S), we have

S is in S <=> S is not in S (#)


If S exists then, x in S <-> ~(x in x).
If S exists then, S in S <-> ~(S in S).

That is, If S exists then (contradiction).
therefore, ~(S exists).
S does not exist, implies, there is no set that S is a member of, and there is no set that is a member of S.

There is no paradox here at all.
There is an logical error being made, namely that we have assumed that S exists and we can show that it does not exist.

The answer to Russell's question "Is the set of sets that are not members of themselves, a member of itself." is NO. It cannot be a member of any set, because it does not exist.

Tryagain wrote:

Coming back to the barber, the set of (all) inhabitants of the
town who are shaved by the barber is

Q = {y | y does not shave y}
Thus, if the barber is b,
b shaves y <=> y does not shave y

To find out whether b is in Q or not, we substitute b for y and get
b shaves b <=> b does not shave b (##)

(##) is the same kind of paradox as (#).

There is a real paradox here. How to give an accurate account of the barber's version of this paradox may not be trivial. But whether an attempt to do so is successful or not, the paradox is there for its own sake.


There is no paradox here either. Nor is there a need for set theory to resolve the problem.

The description of the barber is contradictory and therefore this described barber does not exist.

The answer to Russell's question: Does the barber shave himself is NO.
there is no thing that non-existent described things can do.

We can prove the barber does not exist, within first order predicate logic.
~(some y)(all x)(yRx <-> ~(xRx)), is a theorem.

That is to say: the y such that (all x)(yRx <-> ~(xRx)), does not exist.

There is a y such that (all x)(y shaves x <-> ~(x shaves x)), is false.
There is a set S such that (all x)(x in S <-> ~(x in x)), is false.

~(some R)(some y)(all x)(xRy <-> ~(xRx)), is a theorem.

Tryagain wrote:

The precise (classical)'fix' to the problem of the set of all sets that do not contain themselves is a hierarchy of set-LIKE entities arranged so that the paradoxical self-embedding objects do not exist - the addition, that is, of NEW categories not stated in the problem.


Wrong again.
Russell's theory of types does not resolve the psuedo-paradox, but rather, type theory 'avoids' the problem by denying the predicate (x in x).

The resolution to the apparent paradoxes....

~(some y)(all x)(yRx <-> ~(xRx)), is a theorem.

Proof:
(all x)(yRx <-> ~(xRx)) -> (yRy <-> ~(yRy)), when x=y.
But, (yRy <-> ~(yRy)), is a contradiction for all y.
That is, ~(all x)(yRy <-> ~(yRy)) for all y.
ie. ~(some y)(all x)(yRy <-> ~(yRy)), is tautologous.

Quote:
"By the way, do you know why I have inherited the last name 'Phil'?"

Probably because there was already a 'Owen' registered here.

Thanks. That makes sense.


[/quote]
Owen phil
 
  1  
Reply Sun 4 Jul, 2010 09:04 am
@Owen phil,
I noticed a typo in the proof..

~(some y)(all x)(yRx <-> ~(xRx)), is a theorem.

Proof:
(all x)(yRx <-> ~(xRx)) -> (yRy <-> ~(yRy)), when x=y.
But, (yRy <-> ~(yRy)), is a contradiction for all y.
That is, ~(all x)(yRx <-> ~(xRx)) for all y.
ie. ~(some y)(all x)(yRx <-> ~(xRx)), is tautologous.
0 Replies
 
Tryagain
 
  1  
Reply Wed 7 Jul, 2010 03:38 pm
Bore da Owen. It is one of the most important tools of (classical) mathematics to setup a system that is "paradoxical" - inconsistent - in order to demonstrate that a set of premises cannot ALL be true. It is perhaps unfortunate that this methodology does not reliably discriminate between methodological errors, false premises, and equitable but mutually inconsistent hypotheses and this may be the psychology underlying the
rejection of this methodology by some mathematicians.

Of course, one might adopt the stance of an ultra-Whorfian and suppose
that the structure of math is in fact determined by some coincidental feature of the languages of all cultures that have explored number theory - but there are enough of these and sufficiently varied that this feature would seem to have the status of an "accidental universal" - it would require (in light of the evidence) something of an intellectual leap to suppose that there are languages with the proviso that their speakers are LOGICALLY INCAPABLE of verifying the proofs of the various "anti-recursion" theorems.

Actually, of course, what is going on is that the human mind is finite
and we can only do APPROXIMATIONS of mathematics. Those who hold to the doctrine of infinite language are forced to postulate that we can equally aspire only to approximations of language. Small wonder we get confused occasionally.

I do not say you are wrong; I myself thought I was wrong once...On closer inspection, I was wrong about that.


For what it is worth: I thought the answer by Spike was by far the best given the constrictions of the question.
0 Replies
 
Jean1233475453
 
  1  
Reply Tue 2 Nov, 2010 07:03 pm
@Owen phil,
His is male but he is not yet a man. Perhaps he is a young boy.
Owen phil
 
  1  
Reply Thu 4 Nov, 2010 07:30 am
@Jean1233475453,
Jean1233475453 wrote:

His is male but he is not yet a man. Perhaps he is a young boy.


Hi Jean,
I suppose the conditions for this barber could be made clearer.

Perhaps, There is a bearded man in a town who shaves all and only those men who do not shave themselves.

The point of the exercise is to realise that the barber so described cannot exist.
Any description which has a contradictory predication, does not exist.

For example (the x such that Px & ~Px) cannot exist.
0 Replies
 
Artificial
 
  1  
Reply Mon 11 Jul, 2011 07:58 pm
It's a paradox. There is no solution. If there was a solution it would not be a paradox. No matter what you say or do, there's no way you can resolve the issue that this barber will, and will not shave himself.
0 Replies
 
 

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