Happiness or a ham sandwich?
Proof that either Tweedledum or Tweedledee
Proof that Tweedledoo exists
Proofs that Santa Claus exists
Proof that there exists a unicorn
Proof that you are either conceited or inconsistent
Proof that I am Dracula
Epimenides the Cretan
The liar paradox
Hanging or beheading
Knights and knaves
The visiting logician
Cellini and Bellini
Hilbert's hotel paradox
The barber paradox
"Interesting" and "uninteresting"
Russell's paradox of classes
and algebraic paradoxes:
Proving that 2 = 1
Proving that 3 + 2 = 0
Proving that n = n + 1
was only one catch and that was Catch-22, which specified that concern
for one's own safety in the face of dangers that were real and immediate
was the process of a rational mind. Orr was crazy and could be grounded.
All he had to do was ask; and as soon as he did, he would no longer
be crazy and would have to fly more missions. Orr would be crazy
to fly more missions and sane if he didn't, but if he was sane he
had to fly them. If he flew them he was crazy and didn't have to;
but if he didn't want to he was sane and had to.
& Brecht, p. 2
is better, eternal happiness or a ham sandwich? It would appear
that eternal happiness is better, but this is really not so! After
all, nothing is better than eternal happiness, and a ham sandwich
is certainly better than nothing. Therefore a ham sandwich is better
than eternal happiness.
(1), p. 219
that either Tweedledum or Tweedledee exists
TWEEDLEDUM DOES NOT EXIST
TWEEDLEDEE DOES NOT EXIST
(3) AT LEAST ONE SENTENCE
IN THIS BOX IS FALSE
BOTH SENTENCES IN THIS BOX ARE FALSE
SENTENCE IS TRUE
THEN SANTA CLAUS EXISTS
SENTENCE IS FALSE AND
SANTA CLAUS DOES NOT EXIST
that there exists a unicorn
wish to prove to you that there exists a unicorn. To do this it
obviously suffices to prove the (possibly) stronger statement that
there exists an existing unicorn. (By an existing unicorn I of course
mean one that exists.) Surely if there exists an existing unicorn,
then there must exist a unicorn. So all I have to do is prove that
an existing unicorn exists. Well, there are exactly two possibilities:
An existing unicorn exists.
An existing unicorn does not exist.
(2) is clearly contradictory: How could an existing unicorn not
exist? Just as it is true that a blue unicorn is necessarily blue,
an existing unicorn must necessarily be existing.
that you are either conceited or inconsistent
human brain is but a finite machine, therefore there are only finitely
many propositions which you believe. Let us label these propositions
p1, p2, ..., pn, where n is the number
of propositions you believe. So you believe each of the propositions
p1, p2, ..., pn. Yet, unless you are conceited,
you know that you sometimes make mistakes, hence not everything
you believe is true. Therefore, if you are not conceited, you know
that at least one of the propositions, p1, p2, ...,
pn is false. Yet you believe each of the propositions p1,
p2, ..., pn.
Everyone is afraid of Dracula.
Dracula is afraid of only me.
I am Dracula.
that argument sound like just a silly joke? Well it isn't; it is
valid. Since everyone is afraid of Dracula, then Dracula is afraid
of Dracula. So Dracula is afraid of Dracula, but also is afraid
of no one but me. Therefore I must be Dracula!
(1), pp. 213-18, 224
is often given as the "liar paradox":
the Megarian sixth century B.C. Greek philosopher, and successor
to Euclid, invented the paradox of the liar. In this paradox, Epimenides,
the Cretan, says, "All Cretans are liars." If he is telling
the truth he is lying; and if he is lying, he is telling the truth.
& Brecht, p.7
This is not
in fact a paradox at all. Epimenides cannot be telling the truth,
but he may be lying: the truth may be that some Cretans,
including Epimenides, are liars, but not all.
following version is the version which we will refer to as the
liar paradox. Consider the statement in the following box:
that sentence true or false? If it is false then it is true, and
if it is true then it is false...
following version of the liar paradox was first proposed by the
English mathematician P. E. B. Jourdain in 1913. It is sometimes
referred to as "Jourdain's Card Paradox". We have a card
on one side of which is written:
THE SENTENCE ON
THE OTHER SIDE OF THIS CARD
you turn the card over, and on the other side is written:
THE SENTENCE ON THE OTHER SIDE OF THIS CARD IS FALSE
Another popular version of the liar paradox is given by the following
three sentences written on a card.
THIS SENTENCE CONTAINS FIVE WORDS
THIS SENTENCE CONTAINS EIGHT WORDS
(3) EXACTLY ONE SENTENCE
ON THIS CARD IS TRUE
(1), p. 227
on the hunting preserves of a powerful prince was punishable by
death, but the prince further decreed that anyone caught poaching
was to be given the privilege of deciding whether he should be hanged
or beheaded. The culprit was permitted to make a statement - if
it were false, he was to be hanged; if it were true, he was to be
beheaded. One logical rogue availed himself of this dubious prerogative
- to be hanged if he didn't and to be beheaded if he did - by stating:
"I shall be hanged." Here was a dilemma not anticipated.
For, as the poacher put it, "If you now hang me, you break
the laws made by the prince, for my statement is true, and I ought
to be beheaded, but if you behead me, you are also breaking the
laws, for then what I said was false and I should therefore be hanged."
& Newman, pp. 187-8
Book has 597 Pages.
The Author of this Book is Confucius.
3. The Statements Numbered
1, 2, and 3 are all False.
& Newman, p.189
There is a wide
variety of puzzles about an island in which certain inhabitants
called "knights" always tell the truth, and others called
"knaves" always lie. It is assumed that every inhabitant
of the island is either a knight or a knave...
Suppose A says,
"Either I am a knave or else two plus two equals five."
What would you conclude.?
If A is a knight,
then two plus two equals five, which is not true. If A is a knave,
then he is speaking the truth, which is not possible.
only valid conclusion is that the author of this problem is not
a knight. The fact is that neither a knight nor a knave could make
such a statement.
are back on the Island of Knights and Knaves, where the following
three propositions hold: (1) knights make only true statements;
(2) knaves make only false ones; (3) every inhabitant is either
a knght or a knave. These three propositions will be collectively
referred to as the "rules of the island."
recall that no inhabitant can claim that he is not a knight, since
no knight would make the false statement that he isn't a knight
and no knave would make the true statement that he isn't a knight.
suppose a logician visits the island and meets a native who makes
the following statement to him: "You will never know that
I am a knight."
we get a paradox? Let us see. The logician starts reasoning as follows:
"Suppose he is a knave. Then his statement is false, which
means that at some time I will know that he is a knight,
but I can't know that he is a knight unless he really is
one. So, if he is a knave, it follows that he must be a knight,
which is a contradiction. Therefore he can't be a knave; he must
be a knight."
far so good - there is as yet no contradiction. But then he continues
reasoning: "Now I know that he is a knight, although he said
that I never would. Hence his statement was false, which means that
he must be a knave. Paradox!"
Is this a genuine paradox?
whenever Cellini made a sign, he inscribed a false statement on
it, and whenever Bellini made a sign, he inscribed a true statement
on it. Also, we shall assume that Cellini and Bellini were the only
sign-makers of their time...
come across the following sign:
SIGN WAS MADE BY CELLINI
made the sign? If Cellini made it, then he wrote a true sentence
on it - which is impossible. If Bellini made it, then the sentence
on it is false - which is again impossible. So who made it?
you can't get out of this one by saying that the sentence on the
sign is not well-grounded! It certainly is well-grounded; it states
the historical fact that the sign was made by Cellini; if it was
made by Cellini then the sign is true, and if it wasn't, the sign
is false. So what is the solution?
solution, of course, is that I gave you contradictory information.
If you actually came across the above sign, then it would mean either
that Cellini sometimes wrote true inscriptions on signs (contrary
to what I told you) or that at least one other sign-maker sometimes
wrote false statements on signs (again, contrary to what I told
you). So this is not really a paradox, but a swindle.
(1), pp. 230-31
G G Berry of the Bodleian Library)
number of syllables in the English names of finite integers tend
to increase as the integers grow larger, and must gradually increase
indefinitely, since only a finite number of names can be made with
a given finite number of syllables. Hence the names of some integers
must consist of at least nineteen syllables, and among them there
must be a least. Hence "the least integer not nameable in fewer
than nineteen syllables" must denote a definite integer; in
fact, it denotes 111,777. But "the least integer not nameable
in fewer than nineteen syllables" is itself a name consisting
of eighteen syllables; hence the least integer not nameable in fewer
than nineteen syllables can be named in eighteen syllables.
& Whitehead, p. 61
a hotel with a finite number of rooms, and assume that all the rooms
are occupied. A new guest arrives and asks for a room. "Sorry"
- says the proprietor - "but all the rooms are occupied."
Now let us imagine a hotel with an infinite number of rooms, and
all the rooms are occupied. To this hotel, too, comes a new guest
and asks for a room. "But of course!" - exclaims the proprietor,
and he moves the person previously occupying room N1 into room N2,
the person from room N2 into room N3, the person from room N3 into
room N4, and so on... And the new customer receives room N1, which
becomes free as a result of these transpositions.
us imagine now a hotel with an infinite number of rooms, all taken
up, and an infinite number of new guests who come in, and ask for
gentlemen," says the proprietor, "just wait a minute."
He moves the occupant of N1 into N2, the occupant of N2 into N4,
the occupant of N3 into N6, and so on, and so on...
all odd numbered rooms become free and the infinity of new guests
can easily be accommodated in them.
"just wait a minute" seems optimistic; it would surely
take him an infinite time to shift the guests around.
if a man had an unlimited income, it is an immediate inference that,
however low income-tax may be, he would have to pay annually to
the Exchequer of his nation a sum equal in value to his whole income.
Further, if his income was derived from a capital invested at a
finite rate of interest (as is usual), the annual payments of income-tax
would each be equal in value to the man's whole capital. If, then,
the man with an unlimited income chose to be discontented, he would
be sure of a sympathetic audience among philosophers and business
acquaintances; but discontent could not last long, for the thought
of the difficulties he would put in the way of the Chancellor of
the Exchequer, who would find the drawing up of his budget most
puzzling, would be amusing. Again, the discovery that, after paying
an infinite income-tax, the income would be quite undiminished,
would obviously afford satisfaction, though perhaps the satisfaction
might be mixed with a slight uneasiness as to any action the Commissioners
of Income Tax might take in view of this fact.
the world of quantum mechanics, the laws of physics that are familiar
from the everyday world no longer work. Instead, events are governed
by probabilities. A radioactive atom, for example, might decay,
emitting an electron, or it might not. It is possible to set up
an experiment in such a way that there is a precise fifty-fifty
chance that one of the atoms in a lump of radioactive material will
decay in a certain time and that a detector will register the decay
if it does happen. Schrödinger, as upset as Einstein about the implications
of quantum theory, tried to show the absurdity of these implications
by imagining such an experiment set up in a closed room, or box,
which also contains a live cat and a phial of poison, so arranged
that if the radioactive decay does occur then the poison container
is broken and the cat dies. In the everyday world, there is a fifty-fifty
chance that the cat will be killed, and without looking inside the
box we can say, quite happily, that the cat inside is either dead
or alive. But now we encounter the strangeness of the quantum world.
According to the theory, neither of the two possibilities
open to the radioactive material, and therefore to the cat, has
any reality unless it is observed. The atomic decay has neither
happened nor not happened, the cat has neither been killed nor not
killed, until we look inside the box. Theorists who accept the pure
version of quantum mechanics say that the cat exists in some indeterminate
state, neither dead nor alive, until an observer looks into the
box to see how things are getting on. Nothing is real unless it
man condemned to be hanged] was sentenced on Saturday. "The
hanging will take place at noon," said the judge to the prisoner,
"on one of the seven days of next week. But you will not know
which day it is until you are so informed on the morning of the
day of the hanging."
judge was known to be a man who always kept his word. The prisoner,
accompanied by his lawyer, went back to his cell. As soon as the
two men were alone, the lawyer broke into a grin. "Don't you
see?" he exclaimed. "The judge's sentence cannot possibly
be carried out."
don't see," said the prisoner.
me explain They obviously can't hang you next Saturday. Saturday
is the last day of the week. On Friday afternoon you would still
be alive and you would know with absolute certainty that the hanging
would be on Saturday. You would know this before you were
told so on Saturday morning. That would violate the judge's decree."
said the prisoner.
then is positively ruled out," continued the lawyer. "This
leaves Friday as the last day they can hang you. But they can't
hang you on Friday because by Thursday only two days would remain:
Friday and Saturday. Since Saturday is not a possible day, the hanging
would have to be on Friday. Your knowledge of that fact would violate
the judge's decree again. So Friday is out. This leaves Thursday
as the last possible day. But Thursday is out because if you're
alive Wednesday afternoon, you'll know that Thursday is to be the
get it," said the prisoner, who was beginning to feel much
better. "In exactly the same way I can rule out Wednesday,
Tuesday and Monday. That leaves only tomorrow. But they can't hang
me tomorrow because I know it today!"
He is convinced, by what appears to be unimpeachable logic, that
he cannot be hanged without contradicting the conditions specified
in his sentence. Then on Thursday morning, to his great surprise,
the hangman arrives. Clearly he did not expect him. What is more
surprising, the judge's decree is now seen to be perfectly correctly.
The sentence can be carried out exactly as stated.
(3), pp. 11-13.
Here is a version
from Raymond Smullyan, with his solution:
a Monday morning, a professor says to his class, "I will give
you a surprise examination someday this week. It may be today, tomorrow,
Wednesday, Thursday, or Friday at the latest. On the morning of
the examination, when you come to class, you will not know that
this is the day of the examination."
a logic student reasoned as follows: "Obviously I can't get
the exam on the last day, Friday, because if I haven't gotten the
exam by the end of Thursday's class, then on Friday morning I'll
know that this is the day, and the exam won't be a surprise. This
rules out Friday, so I now know that Thursday is the last possible
day. And, if I don't get the exam by the end of Wednesday, then
I'll know on Thursday morning that this must be the day (because
I have already ruled out Friday), hence it won't be a surprise.
So Thursday is also ruled out."
student then ruled out Wednesday by the same argument, then Tuesday,
and finally Monday, the day on which the professor was speaking.
He concluded: "Therefore I cannot get the exam at all; the
professor cannot possibly fulfil his statement." Just then,
the professor said: "Now I will give you your exam." The
student was most surprised!
Let me put myself in the student's place. I claim that I could get
a surprise examination on any day, even on Friday! Here is
my reasoning: Suppose Friday morning comes and I haven't got the
exam yet. What would I then believe? Assuming I believed the professor
in the first place (and this assumption is necessary for the problem),
could I consistently continue to believe the professor on Friday
morning if I hadn't gotten the exam yet? I don't see how I could.
I could certainly believe that I would get the exam today (Friday),
but I couldn't believe that I'd get a surprise exam today.
Therefore, how could I trust the professor's accuracy? Having doubts
about the professor, I wouldn't know what to believe. Anything could
happen as far as I'm concerned, and so it might well be that I could
be surprised by getting the exam on Friday.
the professor said two things: (1) You will get an exam someday
this week; (2) You won't know on the morning of the exam that this
is the day. I believe it is important that these two statements
should be separated. It could be that the professor was right
in the first statement and wrong in the second. On Friday morning,
I couldn't consistently believe that the professor was right about
both statements, but I could consistently believe his first statement.
However, if I do, then his second statement is wrong (since I will
then believe that I will get the exam today.) On the other
hand, if I doubt the professor's first statement, then I won't know
whether or not I'll get the exam today, which means that the professor's
second statement is fulfilled (assuming he keeps his word and gives
me the exam). So the surprising thing is that the professor's second
statement is true or false depending respectively on whether I do
not or do believe his first statement. Thus the one and only way
the professor can be right is if I have doubts about him; if I doubt
him, that makes him right, whereas if I fully trust him, that makes
(2), pp. 8-9
Sets and subsets
a certain village there is a man, so the paradox runs, who is a
barber; this barber shaves all and only those men in the village
who do not shave themselves. Query: Does the barber shave himself?
man in this village is shaved by the barber if and only if he is
not shaved by himself. Therefore in particular the barber shaves
himself if and only if he does not. We are in trouble if we say
the barber shaves himself and we are in trouble if we say he does
the paradox thus:
are we to say to the argument that goes to prove this unacceptable
conclusion? Happily it rests on assumptions. We are asked to swallow
a story about a village and a man in it who shaves all and only
those men in the village who do not shave themselves. This is the
source of our trouble; grant this and we end up saying, absurdly,
that the barber shaves himself only if he does not. The proper conclusion
to draw is just that there is no such barber. We are confronted
with nothing more than what logicians have been referring to for
a couple of thousand years as a reductio ad absurdum. We
disprove the barber by assuming him and deducing the absurdity that
he shaves himself if and only if he does not. The paradox is simply
a proof that no village can contain a man who shaves all and only
those men in it who do not shave themselves.
like Cellini and Bellini above.
question arises: Are there any uninteresting numbers? We can prove
that there are none by the following simple steps. If there are
dull numbers, then we can divide all numbers into two sets - interesting
and dull. In the set of dull numbers there will be only one number
that is the smallest. Since it is the smallest uninteresting number
it becomes, ipso facto, an interesting number. We must therefore
remove it from the dull set and place it in the other. But now there
will be another smallest uninteresting number. Repeating this process
will make any dull number interesting."
(1), p. 131
sets, it would seem, are not members of themselves - for example,
the set of walruses is not a walrus, the set containing only Joan
of Arc is not Joan of Arc (a set is not a person) - and so on. In
this respect, most sets are rather "run-of-the-mill".
However, some "self-swallowing" sets do contain
themselves as members, such as the set of all sets, or the set of
all things except Joan of Arc, and so on. Clearly, every set is
either run-of-the-mill or self-swallowing, and no set can be both.
Now nothing prevents us from inventing R: the set of all run-of-the-mill
sets. At first, R might seem rather a run-of-the-mill invention
- but that opinion must be revised when you ask yourself, "Is
R itself a run-of-the-mill set or a swallowing set?" You will
find that the answer is: "R is neither run-of-the-mill nor
self-swallowing, for either choice leads to paradox."
2 = 1
is the version offered by Augustus De Morgan: Let x = 1.
Then x² = x. So x² - 1 = x -1. Dividing
both sides by x -1, we conclude that x + 1 = 1; that
is, since x = 1, 2 = 1.
= b. (1)
both sides by a,
= ab. (2)
b² from both sides,
- b² = ab - b² . (3)
+ b)(a - b) = b(a - b). (4)
both sides by (a - b),
+ b = b. (5)
now we take a = b = 1, we conclude that 2 = 1. Or
we can subtract b from both sides and conclude that a,
which can be taken as any number, must be equal to zero.
Or we can substitute b for a and conclude that any
number is double itself. Our result can thus be interpreted in a
number of ways, all equally ridiculous.
arises from a disguised breach of the arithmetical prohibition on
division by zero, occurring at (5): since a = b, dividing
both sides by (a - b) is dividing by zero, which renders
the equation meaningless. As Northrop goes on to show, the same
trick can be used to prove, e.g., that any two unequal numbers are
equal, or that all positive whole numbers are equal.
Here is another
A + B = C, and assume A = 3 and B
both sides of the equation A + B = C by (A
obtain A² + 2AB + B² = C(A +
the terms we have
+ AB - AC = - AB - B² + BC
out (A + B - C), we have
+ B - C) = - B(A + B - C)
both sides by (A + B - C), that is, dividing by zero,
we get A = - B, or A + B = 0, which
is evidently absurd.
& Newman, p. 183
+ 1)² = n² + 2n + 1
+ 1)² - (2n + 1) = n²
n(2n + 1) from both sides and factoring, we have
+ 1)² - (n + 1)(2n + 1) = n² - n(2n
¼(2n + 1)² to both sides of (d) yields
+ 1)² - (n + 1)(2n + 1) + ¼(2n + 1)² = n²
- n(2n + 1) + ¼(2n + 1)²
may be written:
+ 1) - ½(2n + 1)]² = [(n - ½(2n + 1)]²
square roots of both sides,
+ 1 - ½(2n + 1) = n - ½(2n + 1)
= n + 1
& Newman, p. 184
The trick here
is to ignore the fact that there are two square roots for any positive
number, one positive and one negative: the square roots of 4 are
2 and -2, which can be written as ±2. So (g) should properly
1 - ½(2n + 1)) = ±(n - ½(2n + 1))
Gamow, One, Two, Three ... Infinity, 1967
Martin Gardner, Mathematical Puzzles and Diversions, 1959
Martin Gardner, Further Mathematical Diversions, 1969
Gribbin, In Search of Schrödinger's Cat, 1984
Douglas R Hofstadter, Godel, Escher Bach: An Eternal Golden Braid,
Brecht: Patrick Hughes and George Brecht, Vicious Circles and
Infinity: An Anthology of Paradoxes, 1975
E B Jourdain, The Philosophy of Mr B*rtr*nd R*ss*ll, 1918.
Newman: Edward Kasner and James Newman, Mathematics and the Imagination,
Morris, The Ivan Morris Puzzle Book, 1972
Eugene P Northrop, Riddles in Mathematics, revised edition,
Quine: W V Quine,
The Ways of Paradox, and Other Essays, revised edition, 1976
Whitehead: Bertrand Russell and A N Whitehead, Principia Mathematica,
Raymond Smullyan, What Is the Name of This Book?, 1978
Raymond Smullyan, Forever Undecided: A Puzzle Guide to Gödel,
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