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Raffles10
November 8th, 2008, 10:13 PM
At college my engineering tutor told us a riddle or conundrum which has no answer, it's mind boggling.

Riddle:
Three people go into a restaurant and order a meal, the bill comes to £25. They each give £10, the waiter brings back £5 change. They each take back £1, and leave the remaining £2 for the waiter as a tip.
So, they gave £10 and got £1 back, which means they spent £9 each. OK £9 x 3= £27 + £2 (for the waiter) = £29. What happened to the missing pound ?

There has to ba an answer, someone tell me.:confused:

pp.
November 8th, 2008, 10:19 PM
No missing pounds.

Each paid (10-1)=9. Total sum paid is 3*9=27.
Amount billed 25. Tip 2. 25+2=27.

Npl
November 8th, 2008, 10:23 PM
A barber`s job is to shave every man in town who doesnt shave himself. Does the barber shave himself or not? :D

(Its a familar example in logic and proof-theorie)

AbredPeytr
November 8th, 2008, 10:24 PM
This is one of the reasons I hated math as a kid.

Ahhh, now I see

Zero Prime
November 8th, 2008, 10:30 PM
That's and old joke. It originally went like this.

Three men rent a hotel room. Each pays \$10 for a total of \$30 spent on the room. The next day the hotel owner tells the three men that they over paid for the room as it only costs \$25. The three men tell the owner to give them each a dollar back and he can keep two dollars.

If you do the math, each man paid \$9 a piece for the room for a total of \$27. The owner kept \$2 which brings the total to \$29.

The question is where did the other dollar go?

The riddle does have an answer if you realize that this is a play of words and a misdirection of facts.

pp.
November 8th, 2008, 10:30 PM
The barber is a woman.

underthechair
November 8th, 2008, 10:34 PM
You're adding the £2 when you should be subtracting it. The three men paid a total of £27 of which £25 was for the meal and £2 was for the waiter.

Npl
November 8th, 2008, 10:36 PM
The barber is a woman.Ok, its a town of vatican ministrants... damn they have no facial hair.

...its a town of homosexual apes?

DrMega
November 8th, 2008, 10:36 PM
The barber is a woman.

Women shave too you know.

chucky chuckaluck
November 8th, 2008, 10:41 PM
Women shave too you know.

but the restriction on the barber is to only shave men who don't shave themselves. it says nothing about women.

pp.
November 8th, 2008, 10:42 PM
Ok, its a town of vatican ministrants... damn they have no facial hair.

...its a town of homosexual apes?

Original riddle:

A barber`s job is to shave every man in town who doesnt shave himself.

(emphasis mine)

If the statement has to be meaningful it has to make an assertion about the shaving of every man in town. Since the statement can not answer the question about the barber's shave for a male barber in town and we expect the sentence to be meaningful, the sentence can not apply to the barber at all.

Hence, the barber is not a man or not in town.

I chose the variant of a female barber just because the seeming conundrum was written in english where a male barber is only implied but never stated.

hey_ram
November 8th, 2008, 10:48 PM
i think the 2 pounds must also be divided up among the 3 friends.

the total money spent by the friends would now be 9.666 pounds instead of 9. multiply that by 3 and u should get a figure that approaches 10, depending on the accuracy u want...

Npl
November 8th, 2008, 10:50 PM
Well, duh.. the point is that the riddle has no solution. its called "russels antinom" and that version is mathematically strictly defined.

the "barber paradox" is merely an attempt at an analogy in the real world and I dint type as much as I should have.

But you are probably knowing this and just teasing me ;)

-grubby
November 8th, 2008, 10:52 PM
A barber`s job is to shave every man in town who doesnt shave himself. Does the barber shave himself or not? :D

(Its a familar example in logic and proof-theorie)

The barber is a woman.

Note the bolded word

pp.
November 8th, 2008, 10:53 PM
Well, duh.. the point is that the riddle has no solution.

The one posted in this thread has a solution. I'd have to refer back to Russel's original wording in order to see whether it also has that solution. Do you happen to have a link to it?

pp.
November 8th, 2008, 11:00 PM
Note the bolded word

No Person shall be a Representative who shall not have attained to the Age of twenty five Years, and been seven Years a Citizen of the United States, and who shall not, when elected, be an Inhabitant of that State in which he shall be chosen.

Note the bolded word

Npl
November 8th, 2008, 11:03 PM
Bear in mind that the original problem is a mathematical one and there are no "escapes" like questioning what shaving means

pp.
November 8th, 2008, 11:09 PM
No disagreement about what Russell set out to demonstrate: "...because the whole form of words is just noise without meaning".

Besides being a profound mathematical statement, it also is a philosophical or linguistic one: not everything that can be said has a meaning. This applies to this post as well, of course.

MasterNetra
November 8th, 2008, 11:22 PM
nevermind.

SomeGuyDude
November 9th, 2008, 01:20 AM
Sounds like the Achilles riddle.

Achilles is in a race. The other man has a 100m head start. Achilles can run 10x as fast as the other racer. After he has gone forward 10 yards, the other man has only gone one.

The riddle: can Achilles ever pass the other man? If he's within one meter, he goes one meter but the other man has gone a tenth of a meter. If he tries to close the gap, the man is forward a hundredth of a meter. Mathematically, he will never overtake him.

It's a riddle that relies on a misleading framing of the question.

-grubby
November 9th, 2008, 01:49 AM
Note the bolded word

Heh that's a pretty good point.. :|

nerd0795
November 9th, 2008, 06:16 AM
I know, What is the meaning of life, the universe and everything. It has no answer. Nobody will ever know.

decoherence
November 9th, 2008, 06:25 AM
ab=a^2
ab-b^2=a^2-b^2
b(a-b)=(a+b)(a-b)
b=a+b
b=2b
1=2

that's what happens when you ignore the rules

Greyed
November 9th, 2008, 06:46 AM
i think the 2 pounds must also be divided up among the 3 friends.

the total money spent by the friends would now be 9.666 pounds instead of 9. multiply that by 3 and u should get a figure that approaches 10, depending on the accuracy u want...

Exactly.

Each pays in 10. 5 comes back. Making it 25. 25/3 = 8.3333333. Each takes 1 so now each has paid 9.333333 and gives 2 to the waiter. 9.33333 * 3 = 28. 2 to the waiter = 30.

As for the barber, yes, he shaves himself. His job is to shave all males who do not shave themselves. Since he is a male and shaves himself he, as a barber, doesn't have to do it again. There's nothing in the original statement that says he cannot (re)shave those who do shave themselves. The Wikipedia entry linked by Npl precludes this by defining his job as shaving only those who don't shave themselves. ;)

jomiolto
November 9th, 2008, 08:02 AM
Sounds like the Achilles riddle.

Achilles is in a race. The other man has a 100m head start. Achilles can run 10x as fast as the other racer. After he has gone forward 10 yards, the other man has only gone one.

The riddle: can Achilles ever pass the other man? If he's within one meter, he goes one meter but the other man has gone a tenth of a meter. If he tries to close the gap, the man is forward a hundredth of a meter. Mathematically, he will never overtake him.

It's a riddle that relies on a misleading framing of the question.

Actually that is only a paradox if you dismiss infinity (like the ancient Greeks did). In modern mathematics we have no trouble dealing with infinite sums and series, which can be used to, for example, calculate the exact moment when Achilles overtakes the other runner.

It's the same idea as in calculating the infinite sum 1/2 + 1/4 + 1/8 + 1/16 + ..., which gives you exactly one.

miggys
November 9th, 2008, 08:42 AM
As for the barber, yes, he shaves himself. His job is to shave all males who do not shave themselves. Since he is a male and shaves himself he, as a barber, doesn't have to do it again. There's nothing in the original statement that says he cannot (re)shave those who do shave themselves. The Wikipedia entry linked by Npl precludes this by defining his job as shaving only those who don't shave themselves. ;)

Well, I think that Npl was probably repeating this from memory (and left out the "shaving those and only those...") as an example of a "riddle with no answer."

But the other two questions posed do have "answers." And to me, that makes them more enjoyable. They are not just escher drawings with words :)

The missing dollar...It's already been said, but not too clearly. All three give 10. That's 30. They all get 1 back. So, everybody has paid 9 so 27 total. That original 30 is just a ruse. Only 27 is in play 25 went for food and 2 went for the tip (not a very healthy tip!).

This one's harder to explain. Essentially we are breaking up the race into smaller and smaller time intervals. Small enough in fact that it is a convergent sequence to the time in which Achilles passes his opponent. That is, looking at the problem in these increments Achilles will never pass because we've put a bound on the longest time that we will look: that time which Achilles does pass. Of course, it's much easier just to do: 10v*t=100+v*t ;)

rjmdomingo2003
November 9th, 2008, 09:04 AM
ab=a^2
ab-b^2=a^2-b^2
b(a-b)=(a+b)(a-b)
b=a+b
b=2b
1=2

that's what happens when you ignore the rules
I believe that's not correct since any no. divided by zero is indeterminate...
(a-b)=0
isn't it?

LordIshamael
November 9th, 2008, 01:17 PM
firstly, the \$30 question - as said before, each man pays \$30, of which \$25 goes to the bill, \$2 to the tip and \$3 they get back = \$30
so, no missing money :D
my grandfather explained this one to me years ago lol

the barber paradox, yes, its russell's paradox, correct me if im wrong but i think it was originally catalogs or something
anyway, it can have no answer, atleast not in the correct form, because if he shaves himself, then he has broken the rules as he shaved a man who shaved himself (and the original problem states only), but if he doesnt, then he has broken the rules as he did not shave a man who hasnt shaved himself (and the original problem says every)
i remember reading somewhere that some old king had given this punshiment to the barber as in do this or you die, for sleeping with the king's daughter - so ya, the barber is a man, and old enough to be shaving, not a boy or anything

the achilles problem is one of zeno's paradoxes, it was solved by archimedes
the idea is that achilles has to cover distances that keep getting smaller, so he gets there in time periods that keep getting smaller eg. if he took 10s to cover the first 100m while the other ran 10m, he'll take 1s to cover the next 10m while the guy runs 1m, then 0.1s to cover...
and archimedes found a way to calculate sum of incr. smaller infinite terms
that gives us the answer as to when achilles will overtake the other guy

decoherence
November 9th, 2008, 04:02 PM
I believe that's not correct since any no. divided by zero is indeterminate...
(a-b)=0
isn't it?

Bingo!

jimi_hendrix
November 9th, 2008, 04:48 PM
I know, What is the meaning of life, the universe and everything. It has no answer. Nobody will ever know.

DrMega
November 9th, 2008, 06:21 PM
but the restriction on the barber is to only shave men who don't shave themselves. it says nothing about women.

But that is sexist. In Britain that would be against the law, not to mention morally wrong. If the barber is allowed to turn women away, then what are women to do if they don't know how to shave themselves?

andras artois
November 9th, 2008, 07:28 PM
30-25=5-3=2=the tip

Also each of them don't pay exactly 9 pounds. If the bill was divided equally among them they would have to have put it 8.33 pounds in. Therefore they did not spend 9 pounds each. So the 9x3 is a lie, just like the cake.

Sounds like the Achilles riddle.

Achilles is in a race. The other man has a 100m head start. Achilles can run 10x as fast as the other racer. After he has gone forward 10 yards, the other man has only gone one.

The riddle: can Achilles ever pass the other man? If he's within one meter, he goes one meter but the other man has gone a tenth of a meter. If he tries to close the gap, the man is forward a hundredth of a meter. Mathematically, he will never overtake him.

It's a riddle that relies on a misleading framing of the question.

It's only mathematically impossible if you don't take into account the earth's round and they're running round it.

Also yards don't fit equally into metre's.
My personal favourite:

Three people go into a restaurant and order a meal, the bill comes to £25. They each give £10, the waiter brings back £5 change. They each take back £1, and leave the remaining £2 for the waiter as a tip.
So, they gave £10 and got £1 back, which means they spent £9 each. OK £9 x 3= £27 + £2 (for the waiter) = £29. What was the waiters name?

pp.
November 9th, 2008, 08:37 PM
30-25=5-3=2=the tip

Also each of them don't pay exactly 9 pounds. If the bill was divided equally among them they would have to have put it 8.33 pounds in. Therefore they did not spend 9 pounds each. So the 9x3 is a lie, just like the cake.

Er - WHAT?

schauerlich
November 9th, 2008, 08:37 PM
Sounds like the Achilles riddle.

Achilles is in a race. The other man has a 100m head start. Achilles can run 10x as fast as the other racer. After he has gone forward 10 yards, the other man has only gone one.

The riddle: can Achilles ever pass the other man? If he's within one meter, he goes one meter but the other man has gone a tenth of a meter. If he tries to close the gap, the man is forward a hundredth of a meter. Mathematically, he will never overtake him.

It's a riddle that relies on a misleading framing of the question.

That assumes that both racers are slowing down exponentially. If achilles is 10x faster, couldn't he maintain his speed? Also, limit((1/a)^n,n,inf) where a is some real number approaches zero. a in this case is 10.)

CJ Master
November 9th, 2008, 08:44 PM
just like the cake.
The cake is a lie!

...And you guys are too good with math. *runs off with headache*

SomeGuyDude
November 9th, 2008, 08:50 PM
Actually that is only a paradox if you dismiss infinity (like the ancient Greeks did). In modern mathematics we have no trouble dealing with infinite sums and series, which can be used to, for example, calculate the exact moment when Achilles overtakes the other runner.

It's the same idea as in calculating the infinite sum 1/2 + 1/4 + 1/8 + 1/16 + ..., which gives you exactly one.

Or you could just reframe the question and instead of doing it in pure relationships, look at things in absolute terms and realize that in however long it took the first guy to run 11.2m Achilles would have run 112 and thus be 0.8m ahead (for example).

The point is that the riddle becomes difficult when you get tricked by how the question is worded instead of backing up and looking at just what happened.

miggys
November 9th, 2008, 09:07 PM
Here's my favorite 0=1 bits:

0=(1-1)+(1-1)+(1-1)+...
0=1+(-1+1)+(-1+1)+(-1+1)+...
0=1+0+0+0+..
0=1

LordIshamael
November 10th, 2008, 02:24 AM
30-25=5-3=2=the tip

Also each of them don't pay exactly 9 pounds. If the bill was divided equally among them they would have to have put it 8.33 pounds in. Therefore they did not spend 9 pounds each. So the 9x3 is a lie, just like the cake.

ah, but here you assume that the bill is \$25
its not, the entire point is that the bill was originally mistaken to be \$30, so each paid 10
the waiter brought 5 back, so they each SHOULD have paid 8.33, but they didnt coz the waiter pocketed 2 and only brought 3 back
so they ended up paying 9 each, =27 including 25 for the bill and 2 that the waiter pocketed

Three people go into a restaurant and order a meal, the bill comes to £25. They each give £10, the waiter brings back £5 change. They each take back £1, and leave the remaining £2 for the waiter as a tip.
So, they gave £10 and got £1 back, which means they spent £9 each. OK £9 x 3= £27 + £2 (for the waiter) = £29. What was the waiters name?

nice one!im going to use this the next time the math nerd in my class starts freaking everybody out

Phenax
November 10th, 2008, 03:01 AM

schauerlich
November 10th, 2008, 04:59 AM
Here's my favorite 0=1 bits:

0=(1-1)+(1-1)+(1-1)+...
0=1+(-1+1)+(-1+1)+(-1+1)+...
0=1+0+0+0+..
0=1

there's a -1 at the end that isn't accounted for.

miggys
November 10th, 2008, 05:30 AM
It's an infinite series. There is no "end."

SomeGuyDude
November 10th, 2008, 06:40 AM
It's an infinite series. There is no "end."

Yes, but it's an infinite series of (1-1) added together. You need an even number of 1's for the series to hold. If you break up a (1-1) in order to move your parenths around then you need to have a -1 at the end to balance off your +1 at the beginning. It's an infinite series, not an irrational fraction. The series in this case is "infinite" in that no matter where you terminate it, the series is still true. It's not like an asymptotic series like "(1 + 1/2 + 1/4 + 1/8...) < 2" where the termination has an affect on its value. You can't graph out this one.

Think of it like this:

0 = (1-1) + (1-1) + (1-1)... <- true

0 = 1 - 1 + 1 - 1 + 1 - 1... <- NOT true

You took concept B as your infinite series, forgetting that the definition of the series necessitates all the 1's to come in pairs. The way the "trick" works is like that old algebra joke:

a = b

a^2 = a*b

a^2 - b^2 = a*b - b^2

(a+b)(a-b) = b(a-b)

a+b = b

2b = b

1 = 2

Here, the thing seems to work on paper, only if you forget that dividing both sides by (a-b) is, as per our definition, dividing by zero. Same with your infinite series. You cannot add a SINGLE one to the series, it has to be a (1-1) pair.

Capt. Mac
November 10th, 2008, 06:57 AM
A barber`s job is to shave every man in town who doesnt shave himself. Does the barber shave himself or not? :D
The barber does shave himself, but it's not his job. Simple question simple solution.

Here's a simple solution to the first 'impossible riddle:'

3 people pay \$10; that's \$30 total. They each take \$1 back; the total is now at \$27. Hotel owner/waiter/whoever takes a \$2 tip; the total is \$25. The remaining \$25 is taken by the establishment (used to pay the bill). There is no missing dollar.

Now here's where some people get confused:
They each gave \$10, and got \$1 back, which means which means they spent \$9 apiece (including the \$2 tip). That's \$27. Add the \$3 they did not pay and you get the full \$30. I repeat, there is no missing dollar.

lisati
November 10th, 2008, 07:05 AM
ab=a^2
ab-b^2=a^2-b^2
b(a-b)=(a+b)(a-b)
b=a+b
b=2b
1=2

that's what happens when you ignore the rules

There's a variant of this one which starts with a=b+c and similarly violates the rules but I can't remember it at the moment.....

Trail
November 10th, 2008, 09:44 AM
1^1 == 1^0
1 == 0

sisco311
November 10th, 2008, 10:18 AM
Yes, but it's an infinite series of (1-1) added together. You need an even number of 1's for the series to hold. If you break up a (1-1) in order to move your parenths around then you need to have a -1 at the end to balance off your +1 at the beginning. It's an infinite series, not an irrational fraction. The series in this case is "infinite" in that no matter where you terminate it, the series is still true. It's not like an asymptotic series like "(1 + 1/2 + 1/4 + 1/8...) < 2" where the termination has an affect on its value. You can't graph out this one.

Think of it like this:

0 = (1-1) + (1-1) + (1-1)... <- true

0 = 1 - 1 + 1 - 1 + 1 - 1... <- NOT true

You took concept B as your infinite series, forgetting that the definition of the series necessitates all the 1's to come in pairs. The way the "trick" works is like that old algebra joke:

a = b

a^2 = a*b

a^2 - b^2 = a*b - b^2

(a+b)(a-b) = b(a-b)

a+b = b

2b = b

1 = 2

Here, the thing seems to work on paper, only if you forget that dividing both sides by (a-b) is, as per our definition, dividing by zero. Same with your infinite series. You cannot add a SINGLE one to the series, it has to be a (1-1) pair.

S = 1 - 1 + 1 - 1 + ...
1 - S = 1 - ( 1 - 1 + 1 - 1 + ...) = 1 - 1 + 1 - 1 + ... = S
1 - S = S
S = 1/2

1 - 1 + 1 - 1 + ... = 1/2
:lolflag:

miggys
November 10th, 2008, 04:41 PM
Sorry SomeGuyDude, but you are wrong. It's simply an infinite series of repeat +1,-1. In the two cases, I simply put parens to group in ways that make it seem like there are two different answers. I'm not changing the series in either case. I'm not "adding a 1" and, again, there is no end. I probably should have written it out without the parens first to eliminate some of this confusion (what you called "concept B")

S = 1 - 1 + 1 - 1 + ...
1 - S = 1 - ( 1 - 1 + 1 - 1 + ...) = 1 - 1 + 1 - 1 + ... = S
1 - S = S
S = 1/2

1 - 1 + 1 - 1 + ... = 1/2
:lolflag:

I believe that those who say there is an answer (who know what they are talking about) say that it's 1/2. However, here's something that I wonder (and really I don't know!) Can you mathematically swap every other term in an infinite series and say it is the same sum? If so:

1-S=1-(1 - 1 + 1 - 1 + ...)
*swap every other term*
1-S=1-(-1 + 1 + -1 + 1 + ...)
1-S=1--(1-1+1-1+...)
1-S=1+S
0=2S

But I've never seen it done this way so that swapping is probably illegal.

pp.
November 10th, 2008, 04:50 PM
Can you mathematically swap every other term in an infinite series and say it is the same sum?

But I've never seen it done this way so that swapping is probably illegal.

(1) We know that a+b = b+a.
(2) We also know that 1+1 = 1 + (+1)
(3) ... and 1-1 = 1 + (-1).

(4) Hence we know that 1+1+1-1-1-1=(+1) + (+1) + (+1) + (-1) + (-1) + (-1)

We can legally swap. Now it's your problem to show that all those properties still apply for a series of infinite length.

sisco311
November 10th, 2008, 05:11 PM
sorry someguydude, but you are wrong. It's simply an infinite series of repeat +1,-1. In the two cases, i simply put parens to group in ways that make it seem like there are two different answers. I'm not changing the series in either case. I'm not "adding a 1" and, again, there is no end. I probably should have written it out without the parens first to eliminate some of this confusion (what you called "concept b")

i believe that those who say there is an answer (who know what they are talking about) say that it's 1/2. However, here's something that i wonder (and really i don't know!) can you mathematically swap every other term in an infinite series and say it is the same sum? If so:

1-s=1-(1 - 1 + 1 - 1 + ...)
*swap every other term*
1-s=1-(-1 + 1 + -1 + 1 + ...)
1-s=1--(1-1+1-1+...)
1-s=1+s
0=2s

But i've never seen it done this way so that swapping is probably illegal.

s = 1 - 1 + 1 - 1 + ...
| * (-1)

(-1) * s = (-1) * (1 - 1 + 1 - 1 + ...)
| +1

1 + (-1) * s = 1 + (-1) * (1 - 1 + 1 - 1 + ...)
1 - s = 1 - 1 + 1 - 1 + ...
1 - s = s
1 = 2 * s
s = 1/2

Idefix82
November 10th, 2008, 05:33 PM
Yes, but it's an infinite series of (1-1) added together. You need an even number of 1's for the series to hold. If you break up a (1-1) in order to move your parenths around then you need to have a -1 at the end to balance off your +1 at the beginning. It's an infinite series, not an irrational fraction. The series in this case is "infinite" in that no matter where you terminate it, the series is still true. It's not like an asymptotic series like "(1 + 1/2 + 1/4 + 1/8...) < 2" where the termination has an affect on its value. You can't graph out this one.

Think of it like this:

0 = (1-1) + (1-1) + (1-1)... <- true

0 = 1 - 1 + 1 - 1 + 1 - 1... <- NOT true

Wrong. First of all, it doesn't make sense to talk about even and odd infinity. An infinite number of items is neither even nor odd. It's as non-sensical as to say that love is green with wrinkles. Similarly non-sensical is it to talk about "the end" of an infinite series or about its last term.
The issue is that this series does not converge, meaning that there is no number x such that the sum of the first n terms is close to x when n is large enough. Just have look at the formal definition of convergence at the very beginning of http://en.wikipedia.org/wiki/Convergent_series and think about whether the conditions are satisfied for our series.

To sum it up, the value of the infinite sum is neither 0 nor 1, it does not have a value, just like the series 1+1+1+... has no value.

There are more subtle tricks you can play with such series by using the fact that there exist series which are convergent but not absolutely convergent (see the wikipedia article above as well as this one: http://en.wikipedia.org/wiki/Riemann_series_theorem

But again, there is no paradox there. It's just a matter of carefully defining your words before using them. I hope this clears up the matter (at least for those who can be bothered to think about it).

miggys
November 10th, 2008, 05:53 PM
@sisco

I'm familiar with what you did. I was saying that in a similar fashion we can also get the value to be 0.

However, after thinking about it further, I feel like commutation in an infinite series is not necessarily legal. But after a quick bit of googling I can not find a source either way...

But it's still my favorite series.

miggys
November 10th, 2008, 05:55 PM
(1) We know that a+b = b+a.
(2) We also know that 1+1 = 1 + (+1)
(3) ... and 1-1 = 1 + (-1).

(4) Hence we know that 1+1+1-1-1-1=(+1) + (+1) + (+1) + (-1) + (-1) + (-1)

We can legally swap. Now it's your problem to show that all those properties still apply for a series of infinite length.

What's the due date? :)

Idefix82
November 10th, 2008, 06:05 PM
What's the due date? :)

This will take you a while to prove since it's wrong :)
Your objection to swapping is an absolutely valid one and in fact you are only allowed to swap terms in an absolutely convergent series (see my previous post). This series is not only not absolutely convergent, but it is not convergent at all.

johnb820
November 10th, 2008, 07:48 PM
Here's a simple one.

"This statement is false."

Is that statement in the quotations true or false?

Choosing one or the either, the statement will cause it to contradict its own truth value.