04/04/2008, 07:48 PM
Catch-22
There 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.
W
A paradoxical notice
PLEASE IGNORE
THIS NOTICE
Hughes & Brecht, p. 2
Some proofs
Happiness or a ham sandwich?
Which 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.
W
Proof that either Tweedledum or Tweedledee exists
(1) TWEEDLEDUM DOES NOT EXIST
(2) TWEEDLEDEE DOES NOT EXIST
(3) AT LEAST ONE SENTENCE IN THIS BOX IS FALSE
Proof that Tweedledoo exists
(1) TWEEDLEDOO EXISTS
(2) BOTH SENTENCES IN THIS BOX ARE FALSE
Proofs that Santa Claus exists
IF THIS SENTENCE IS TRUE
THEN SANTA CLAUS EXISTS
THIS SENTENCE IS FALSE AND
SANTA CLAUS DOES NOT EXIST
Proof that there exists a unicorn
I 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:
(1) An existing unicorn exists.
(2) An existing unicorn does not exist.
Possibility (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.
Proof that you are either conceited or inconsistent
A human brain is but a finite machine, therefore there are only finitely many propositions which you believe. Let us label these propositions p 1, p 2, ..., pn , where n is the number of propositions you believe. So you believe each of the propositions p 1, p 2, ..., 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, p 1, p 2, ..., pn is false. Yet you believe each of the propositions p 1, p 2, ..., pn .
Proof that I am Dracula
(1) Everyone is afraid of Dracula.
(2) Dracula is afraid of only me.
Therefore I am Dracula.
Doesn't 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!
Smullyan (1), pp. 213-18, 224
Liars and truth-tellers
The following is often given as the "liar paradox":
Epimenides the Cretan
Eubulides, 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.
W
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.
The liar paradox
The following version is the version which wee will refer to as the liar paradox. Consider the statement in the following box:
THIS SENTENCE IS FALSE
Is that sentence true or false? If it is false then it is true, and if it is true then it is false...
The 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". Wee have a card on one side of which is written:
(1) THE SENTENCE ON THE OTHER SIDE OF THIS CARD IS TRUE
Then you turn the card over, and on the other side is written:
(2) 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.
(1) THIS SENTENCE CONTAINS FIVE WORDS
(2) THIS SENTENCE CONTAINS EIGHT WORDS
(3) EXACTLY ONE SENTENCE ON THIS CARD IS TRUE
Smullyan (1), p. 227
Hanging or beheading
Poaching 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."
Kasner & Newman, pp. 187-8
1. This Book has 597 Pages.
2. The Author of this Book is Confucius.
3. The Statements Numbered 1, 2, and 3 are all False.
Kasner & Newman, p.189
Knights and knaves
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.
Smullyan comments:
The 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.
The visiting logician
Wee 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."
Wee 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.
Now 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. "
Do wee 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."
So 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!"
Question. Is this a genuine paradox?
Smullyan (2), pp.67-8
Cellini and Bellini
... 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, wee shall assume that Cellini and Bellini were the only sign-makers of their time...
You come across the following sign:
THIS SIGN WAS MADE BY CELLINI
Who 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?
Now, 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?
The 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.
Smullyan (1), pp. 230-31
Quine's paradox
"Yields a falsehood when appended to its quotation" yields a falsehood when appended to its quotation.
W
Can you see why this is a paradox?
Infinity
Hilbert's hotel paradox
Imagine 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.
Let 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 rooms.
"Certainly, 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...
Now all odd numbered rooms become free and the infinity of new guests can easily be accommodated in them.
W
The proprietor's "just wait a minute" seems optimistic; it would surely take him an infinite time to shift the guests around.
Infinite income-tax
Now 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.
Jourdain, pp. 66-7
Schrödinger's cat
In 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 wee can say, quite happily, that the cat inside is either dead or alive. But now wee 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 wee 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 is observed.
W
Olbers' paradox - why is the night sky dark?
... the apparent paradox, stated in 1826 and now explained by postulating a finite expanding universe, that the sky is dark at night although, as there are an infinite number of stars, it should be uniformly bright.
W
The unexpected hanging
[A 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."
The 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."
"I don't see," said the prisoner.
"Let 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."
"True," said the prisoner.
"Saturday, 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 day."
"I 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.
W
Here is a version from Raymond Smullyan, with his solution:
On 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."
Well, 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."
The 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 (Frida?), 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.
Actually, 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 him wrong!
Smullyan (2), pp. 8-9
Prediction paradox (Newcomb's problem)
Two closed boxes, Bl and B2, are on a table. Bl contains $1,000. B2 contains either nothing or $1 million. You do not know which. You have an irrevocable choice between two actions:
1. Take what is in both boxes.
2. Take only what is in B2.
At some time before the test a superior Being has made a prediction about what you will decide. It is not necessary to assume determinism. You only need be persuaded that the Being's predictions are "almost certainly" correct. If you like, you can think of the Being as God, but the paradox is just as strong if you regard the Being as a superior intelligence from another planet or a supercomputer capable of probing your brain and making highly accurate predictions about your decisions. If the Being expects you to choose both boxes, he has left B2 empty. If he expects you to take only B2, he has put $1 million in it. (If he expects you to randomize your choice by, say, flipping a coin, he has left B2 empty.) In all cases Bl contains $1,000. You understand the situation fully, the Being knows you understand, you know that he knows, and so on.
What should you do? Clearly it is not to your advantage to flip a coin, so that you must decide on your own. The paradox lies in the disturbing fact that a strong argument can be made for either decision. Both arguments cannot be right. The problem is to explain why one is wrong.
Let us look first at the argument for taking only B2. You believe the Being is an excellent predictor. If you take both boxes, the Being almost certainly will have anticipated your action and have left B2 empty. You will get only the $1,000 in Bl. On the other hand, if you take only B2, the Being, expecting that, almost certainly will have placed $1 million in it. Clearly it is to your advantage to take only B2.
Convincing? Yes, but the Being made his prediction, say a week ago and then left. Either he put the $1 million in B2, or he did not. "If the money is already there, it will stay there whatever you choose. It is not going to disappear. If it is not already there, it is not going to suddenly appear if you choose only what is in the second box." It is assumed that no "backward causality" is operating; that is, your present actions cannot influence what the Being did last week. So why not take both boxes and get everything that is there? If B2 is filled, you get $1,001,000. If it is empty, you get at least $1,000. If you are so foolish as to take only B2, you know you cannot get more than $1 million, and there is even a slight possibility of getting nothing. Clearly it is to your advantage to take both boxes!
"I have put this problem to a large number of people, both friends and students in class," writes Nozick. "To almost everyone it is perfectly clear and obvious what should be done. The difficulty is that people seem to divide almost evenly on the problem, with large numbers thinking that the opposing half is just being silly."
W
Hempel's ravens (the confirmation paradox)
While taking a group of benefactors on a tour through the new aviary they had just helped to build, a noted ornithologist commented, "And here wee have two of the finest examples of ravens that I have ever seen. Notice the lustrous black plumage for which all ravens are famous." The ornithologist continued his lecture, commenting on the corvine feeding and nesting habits as well as on the birds' legendary role as harbingers of ill fortune.
When the ornithologist had finished, a young man said, "Sir, excuse me, but did you say that 'All ravens are black'?"
"I don't know if I said exactly that, but it's true. All ravens are black."
"But, how do you know that - for certain, I mean?" asked the young man.
"Well, I've seen a few hundred ravens in my day and every one of them has been black."
"Yes, but a few hundred are not all. How many ravens are there, anyway?"
"I would guess several million. As for your question, many other scientists, and non-scientists for that matter, have observed ravens over thousands of years and so far the birds have all been black. At least, I don't know of a single instance in which someone has produced a non-black raven."
"That's true, but it's still not all - just most."
"True, but there is other evidence. For example, take all these lovely multicolored birds wee have seen today - the parrots, toucans, the peacocks -"
"They're lovely, but what do they have to do with your claim that all ravens are black?"
"Don't you see?" asked the ornithologist.
"No, I don't see. Please explain."
"Well, you accept the idea that every new instance of another black raven that is observed adds to the support of the generalization that all ravens are black?"
"Yes, of course."
"Well then, the statement 'All ravens are black' is logically equivalent to the statement 'All non-black things are non-ravens.' This being so and because whatever confirms a statement also confirms any logically equivalent statement, it's clear that any non-black non-raven supports the generalization 'All ravens are black.' Hence, all these colorful, non-black non-ravens also support the generalization."
"That's ridiculous," chided the young man. "In that case you might as well say that your blue jacket and gray pants also confirm the statement 'All ravens are black.' After all, they're also non-black non-ravens."
"That's correct," said the ornithologist. "Now you're beginning to think like a true scientist."
Who is reasoning correctly, the ornithologist or the young man?
W
Note that a blue jacket also confirms the statements "All ravens are white", "All ravens are yellow", or indeed any other colour except blue. Here is another confirmation paradox:
Nelson Goodman's "grue" paradox
An object is "grue" if it is green until, say, January 1, 2500, and blue thereafter. Is the law "All emeralds are grue" confirmed by observations of green emeralds?
W
Sets and subsets
The barber paradox
In 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?
Any 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. Wee are in trouble if wee say the barber shaves himself and wee are in trouble if wee say he does not.
W
Quine disarms the paradox thus:
What are wee to say to the argument that goes to prove this unacceptable conclusion? Happily it rests on assumptions. Wee 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 wee 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. Wee are confronted with nothing more than what logicians have been referring to for a couple of thousand years as a reductio ad absurdum . Wee 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.
Another "swindle", like Cellini and Bellini above?
"Interesting" and "uninteresting" numbers
The question arises: Are there any uninteresting numbers? Wee can prove that there are none by the following simple steps. If there are dull numbers, then wee 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. Wee 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.
W
Russell's paradox of classes
Most 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 self-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."
W
The following is a variant on Russell's paradox using adjectives instead of sets:
Grelling's paradox - autological and heterological
Divide the adjectives in English which are self-descriptive, such as "pentasyllabic", "awkwardnessful", and "recherché", and those which are not, such as "edible", "incomplete", and "bisyllabic". Now if wee admit "non-self-descriptive" as an adjective, to which class does it belong? If it seems questionable to include hyphenated words, wee can use two terms invented specially for this paradox: autological (= "self-descriptive"), and heterological (= "non-self-descriptive"). The question then becomes: "Is 'heterological' heterological?" Try it!
W
Berry's paradox - the least integer not nameable in fewer than nineteen syllables
(Devised by G G Berry of the Bodleian Library)
The 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.
W
Two of Zeno's paradoxes
Achilles and the tortoise
Zeno's second paradox of motion, of Achilles and the tortoise, is probably the best known of his four paradoxes of motion. In this problem, the fleet Greek warrior runs a race against a slow-moving tortoise. Assume Achilles runs at ten times the speed of the tortoise (1 meter per second to 0.1 meter per second). The tortoise is given a 100-meter handicap in a race that is 1,000 meters. By the time Achilles reaches the tortoise's starting point T0, the tortoise will have moved on to point T1. Soon, Achilles will reach point T1, but by then the tortoise would have moved on to T2, and so on, ad infinitum. Every time Achilles reaches a point where the tortoise has just been, the tortoise has moved on a bit. Although the distances between the two runners will diminish rapidly, Achilles can never catch up with the tortoise, or so it would seem.
W
Bertrand Russell commented:
This argument... shows that, if Achilles ever overtakes the tortoise, it must be after an infinite number of instants have elapsed since he started. This is in fact true; but the view that an infinite number of instants makes up an infinitely long time is not true, and therefore the conclusion that Achilles will never overtake the tortoise does not follow.
Russell
Paradox of the arrow
In the paradox of the arrow, Zeno asks us to consider an arrow in flight and argues that, in fact, the arrow must always be at rest. At each instant the arrow occupies a space equal to itself. Movement is impossible, because an instant by definition has no parts. If the arrow were capable of moving during an instant, wee would contradict the definition of an instant, for the arrow would be in one position during the first part of the instant and in another position in the other part of the instant. Thus, the arrow never seems to be moving but rather, as Russell notes in his essay on infinity, "in some miraculous way the change of position has to occur between the instants, that is to say, not at any time whatever." If the arrow does not move at any given instant, how then does it make its flight?
Falleta, p.192
The ship of Theseus
The ship wherein Theseus and the youth of Athens returned had thirty oars, and was preserved by the Athenians down even to the time of Demetrius Phalereus, for they took away the old planks as they decayed, putting in new and stronger timber in their place, insomuch that this ship became a standing example among the philosophers, for the logical question of things that grow; one side holding that the ship remained the same, and the other contending that it was not the same.
W
Protagoras's pupil
Another paradox which has its foundation - real or legendary - in antiquity concerns the sophist Protagoras, who lived and taught in the fifth century BC. It is said that Protagoras made an arrangement with one of his pupils whereby the pupil was to pay for his instruction after he had won his first case. The young man completed his course, hung up the traditional shingle, and waited for clients. None appeared. Protagoras grew impatient and decided to sue his former pupil for the amount owed him.
'For,' argued Protagoras, 'either I win this suit, or you win it. If I win, you pay me according to the judgement of the court. If you win, you pay me according to our agreement. In either case I am bound to be paid.'
'Not so,' replied the young man. 'If I win, then by the judgement of the court I need not pay you. If you win, then by our agreement I need not pay you. In either case I am bound not to have to pay you.'
Whose argument was right? Who knows?
W
Sorites paradox (paradox of the heap)
Suppose you have a heap of sand. If you take away one grain of sand, what remains is still a heap: removing a single grain cannot turn a heap into something that is not a heap. If two collections of grains of sand differ in number by just one grain, then both or neither are heaps. This apparently obvious and uncontroversial supposition appears to lead to the paradoxical conclusion that all collections of grains of sand, even one-membered collections, are heaps.
W
M C Escher W
Drawing hands
Belvedere
Waterfall
Mobius band W
Mobius band
A Mobius band has just one side and just one edge.
Penrose triangle W
Necker cube (Freemish crate) W
Ames room W
Curry's paradox W
Curry's paradox
Arithmetic and algebraic paradoxes
Proving that 2 = 1
Here 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, wee conclude that x + 1 = 1; that is, since x = 1, 2 = 1.
Quine, p.5
Assume that
a = b. (1)
Multiplying both sides by a,
a² = ab. (2)
Subtracting b² from both sides,
a² - b² = ab - b² . (3)
Factorizing both sides,
(a + b)(a - b) = b(a - b). (4)
Dividing both sides by (a - b),
a + b = b. (5)
If now wee take a = b = 1, wee conclude that 2 = 1. Or wee can subtract b from both sides and conclude that a, which can be taken as any number, must be equal to zero. Or wee 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.
Northrop, p. 85
The paradox 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 example:
Proving that 3 + 2 = 0
Assume A + B = C, and assume A = 3 and B = 2.
Multiply both sides of the equation A + B = C by (A + B).
Wee obtain A² + 2AB + B² = C(A + B)
Rearranging the terms wee have
A² + AB - AC = - AB - B² + BC
Factoring out (A + B - C), wee have
A(A + B - C) = - B(A + B - C)
Dividing both sides by (A + B - C), that is, dividing by zero, wee get A = - B, or A + B = 0, which is evidently absurd.
Kasner & Newman, p. 183
Proving that n = n + 1
(a) (n + 1)² = n² + 2n + 1
(b) (n + 1)² - (2n + 1) = n²
© Subtracting n(2n + 1) from both sides and factoring, wee have
(d) (n + 1)² - (n + 1)(2n + 1) = n² - n(2n +1)
(e) Adding ¼(2n + 1)² to both sides of (d) yields
(n + 1)² - (n + 1)(2n + 1) + ¼(2n + 1)² = n² - n(2n + 1) + ¼(2n + 1)²
This may be written:
(f) [(n + 1) - ½(2n + 1)]² = [(n - ½(2n + 1)]²
Taking square roots of both sides,
(g) n + 1 - ½(2n + 1) = n - ½(2n + 1)
and, therefore,
(h) n = n + 1
Kasner & 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 read:
±(n + 1 - ½(2n + 1)) = ±(n - ½(2n + 1))
don't be overwhelmed by this thread, please
There 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.
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Joseph Heller, Catch-22
A paradoxical notice
PLEASE IGNORE
THIS NOTICE
Hughes & Brecht, p. 2
Some proofs
Happiness or a ham sandwich?
Which 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.
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Smullyan (1), p. 219
Proof that either Tweedledum or Tweedledee exists
(1) TWEEDLEDUM DOES NOT EXIST
(2) TWEEDLEDEE DOES NOT EXIST
(3) AT LEAST ONE SENTENCE IN THIS BOX IS FALSE
Proof that Tweedledoo exists
(1) TWEEDLEDOO EXISTS
(2) BOTH SENTENCES IN THIS BOX ARE FALSE
Proofs that Santa Claus exists
IF THIS SENTENCE IS TRUE
THEN SANTA CLAUS EXISTS
THIS SENTENCE IS FALSE AND
SANTA CLAUS DOES NOT EXIST
Proof that there exists a unicorn
I 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:
(1) An existing unicorn exists.
(2) An existing unicorn does not exist.
Possibility (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.
Proof that you are either conceited or inconsistent
A human brain is but a finite machine, therefore there are only finitely many propositions which you believe. Let us label these propositions p 1, p 2, ..., pn , where n is the number of propositions you believe. So you believe each of the propositions p 1, p 2, ..., 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, p 1, p 2, ..., pn is false. Yet you believe each of the propositions p 1, p 2, ..., pn .
Proof that I am Dracula
(1) Everyone is afraid of Dracula.
(2) Dracula is afraid of only me.
Therefore I am Dracula.
Doesn't 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!
Smullyan (1), pp. 213-18, 224
Liars and truth-tellers
The following is often given as the "liar paradox":
Epimenides the Cretan
Eubulides, 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.
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Hughes & 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.
The liar paradox
The following version is the version which wee will refer to as the liar paradox. Consider the statement in the following box:
THIS SENTENCE IS FALSE
Is that sentence true or false? If it is false then it is true, and if it is true then it is false...
The 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". Wee have a card on one side of which is written:
(1) THE SENTENCE ON THE OTHER SIDE OF THIS CARD IS TRUE
Then you turn the card over, and on the other side is written:
(2) 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.
(1) THIS SENTENCE CONTAINS FIVE WORDS
(2) THIS SENTENCE CONTAINS EIGHT WORDS
(3) EXACTLY ONE SENTENCE ON THIS CARD IS TRUE
Smullyan (1), p. 227
Hanging or beheading
Poaching 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."
Kasner & Newman, pp. 187-8
1. This Book has 597 Pages.
2. The Author of this Book is Confucius.
3. The Statements Numbered 1, 2, and 3 are all False.
Kasner & Newman, p.189
Knights and knaves
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.
Smullyan comments:
The 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.
The visiting logician
Wee 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."
Wee 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.
Now 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. "
Do wee 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."
So 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!"
Question. Is this a genuine paradox?
Smullyan (2), pp.67-8
Cellini and Bellini
... 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, wee shall assume that Cellini and Bellini were the only sign-makers of their time...
You come across the following sign:
THIS SIGN WAS MADE BY CELLINI
Who 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?
Now, 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?
The 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.
Smullyan (1), pp. 230-31
Quine's paradox
"Yields a falsehood when appended to its quotation" yields a falsehood when appended to its quotation.
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Quine, p. 9
Can you see why this is a paradox?
Infinity
Hilbert's hotel paradox
Imagine 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.
Let 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 rooms.
"Certainly, 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...
Now all odd numbered rooms become free and the infinity of new guests can easily be accommodated in them.
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Gamow, p. 17
The proprietor's "just wait a minute" seems optimistic; it would surely take him an infinite time to shift the guests around.
Infinite income-tax
Now 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.
Jourdain, pp. 66-7
Schrödinger's cat
In 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 wee can say, quite happily, that the cat inside is either dead or alive. But now wee 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 wee 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 is observed.
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Gribbin, pp. 2-3
Olbers' paradox - why is the night sky dark?
... the apparent paradox, stated in 1826 and now explained by postulating a finite expanding universe, that the sky is dark at night although, as there are an infinite number of stars, it should be uniformly bright.
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The Chambers Dictonary 10th edition, 2006
The unexpected hanging
[A 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."
The 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."
"I don't see," said the prisoner.
"Let 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."
"True," said the prisoner.
"Saturday, 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 day."
"I 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.
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Gardner (3), pp. 11-13
Here is a version from Raymond Smullyan, with his solution:
On 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."
Well, 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."
The 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 (Frida?), 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.
Actually, 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 him wrong!
Smullyan (2), pp. 8-9
Prediction paradox (Newcomb's problem)
Two closed boxes, Bl and B2, are on a table. Bl contains $1,000. B2 contains either nothing or $1 million. You do not know which. You have an irrevocable choice between two actions:
1. Take what is in both boxes.
2. Take only what is in B2.
At some time before the test a superior Being has made a prediction about what you will decide. It is not necessary to assume determinism. You only need be persuaded that the Being's predictions are "almost certainly" correct. If you like, you can think of the Being as God, but the paradox is just as strong if you regard the Being as a superior intelligence from another planet or a supercomputer capable of probing your brain and making highly accurate predictions about your decisions. If the Being expects you to choose both boxes, he has left B2 empty. If he expects you to take only B2, he has put $1 million in it. (If he expects you to randomize your choice by, say, flipping a coin, he has left B2 empty.) In all cases Bl contains $1,000. You understand the situation fully, the Being knows you understand, you know that he knows, and so on.
What should you do? Clearly it is not to your advantage to flip a coin, so that you must decide on your own. The paradox lies in the disturbing fact that a strong argument can be made for either decision. Both arguments cannot be right. The problem is to explain why one is wrong.
Let us look first at the argument for taking only B2. You believe the Being is an excellent predictor. If you take both boxes, the Being almost certainly will have anticipated your action and have left B2 empty. You will get only the $1,000 in Bl. On the other hand, if you take only B2, the Being, expecting that, almost certainly will have placed $1 million in it. Clearly it is to your advantage to take only B2.
Convincing? Yes, but the Being made his prediction, say a week ago and then left. Either he put the $1 million in B2, or he did not. "If the money is already there, it will stay there whatever you choose. It is not going to disappear. If it is not already there, it is not going to suddenly appear if you choose only what is in the second box." It is assumed that no "backward causality" is operating; that is, your present actions cannot influence what the Being did last week. So why not take both boxes and get everything that is there? If B2 is filled, you get $1,001,000. If it is empty, you get at least $1,000. If you are so foolish as to take only B2, you know you cannot get more than $1 million, and there is even a slight possibility of getting nothing. Clearly it is to your advantage to take both boxes!
"I have put this problem to a large number of people, both friends and students in class," writes Nozick. "To almost everyone it is perfectly clear and obvious what should be done. The difficulty is that people seem to divide almost evenly on the problem, with large numbers thinking that the opposing half is just being silly."
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Gardner (3), pp. 582-3
Hempel's ravens (the confirmation paradox)
While taking a group of benefactors on a tour through the new aviary they had just helped to build, a noted ornithologist commented, "And here wee have two of the finest examples of ravens that I have ever seen. Notice the lustrous black plumage for which all ravens are famous." The ornithologist continued his lecture, commenting on the corvine feeding and nesting habits as well as on the birds' legendary role as harbingers of ill fortune.
When the ornithologist had finished, a young man said, "Sir, excuse me, but did you say that 'All ravens are black'?"
"I don't know if I said exactly that, but it's true. All ravens are black."
"But, how do you know that - for certain, I mean?" asked the young man.
"Well, I've seen a few hundred ravens in my day and every one of them has been black."
"Yes, but a few hundred are not all. How many ravens are there, anyway?"
"I would guess several million. As for your question, many other scientists, and non-scientists for that matter, have observed ravens over thousands of years and so far the birds have all been black. At least, I don't know of a single instance in which someone has produced a non-black raven."
"That's true, but it's still not all - just most."
"True, but there is other evidence. For example, take all these lovely multicolored birds wee have seen today - the parrots, toucans, the peacocks -"
"They're lovely, but what do they have to do with your claim that all ravens are black?"
"Don't you see?" asked the ornithologist.
"No, I don't see. Please explain."
"Well, you accept the idea that every new instance of another black raven that is observed adds to the support of the generalization that all ravens are black?"
"Yes, of course."
"Well then, the statement 'All ravens are black' is logically equivalent to the statement 'All non-black things are non-ravens.' This being so and because whatever confirms a statement also confirms any logically equivalent statement, it's clear that any non-black non-raven supports the generalization 'All ravens are black.' Hence, all these colorful, non-black non-ravens also support the generalization."
"That's ridiculous," chided the young man. "In that case you might as well say that your blue jacket and gray pants also confirm the statement 'All ravens are black.' After all, they're also non-black non-ravens."
"That's correct," said the ornithologist. "Now you're beginning to think like a true scientist."
Who is reasoning correctly, the ornithologist or the young man?
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Falleta, pp. 126-73
Note that a blue jacket also confirms the statements "All ravens are white", "All ravens are yellow", or indeed any other colour except blue. Here is another confirmation paradox:
Nelson Goodman's "grue" paradox
An object is "grue" if it is green until, say, January 1, 2500, and blue thereafter. Is the law "All emeralds are grue" confirmed by observations of green emeralds?
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Gardner (3) , p. 544
Sets and subsets
The barber paradox
In 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?
Any 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. Wee are in trouble if wee say the barber shaves himself and wee are in trouble if wee say he does not.
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Quine, p.4
Quine disarms the paradox thus:
What are wee to say to the argument that goes to prove this unacceptable conclusion? Happily it rests on assumptions. Wee 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 wee 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. Wee are confronted with nothing more than what logicians have been referring to for a couple of thousand years as a reductio ad absurdum . Wee 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.
Another "swindle", like Cellini and Bellini above?
"Interesting" and "uninteresting" numbers
The question arises: Are there any uninteresting numbers? Wee can prove that there are none by the following simple steps. If there are dull numbers, then wee 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. Wee 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.
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Gardner (1), p. 131
Russell's paradox of classes
Most 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 self-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."
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Hofstadter, p. 20
The following is a variant on Russell's paradox using adjectives instead of sets:
Grelling's paradox - autological and heterological
Divide the adjectives in English which are self-descriptive, such as "pentasyllabic", "awkwardnessful", and "recherché", and those which are not, such as "edible", "incomplete", and "bisyllabic". Now if wee admit "non-self-descriptive" as an adjective, to which class does it belong? If it seems questionable to include hyphenated words, wee can use two terms invented specially for this paradox: autological (= "self-descriptive"), and heterological (= "non-self-descriptive"). The question then becomes: "Is 'heterological' heterological?" Try it!
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Hofstadter, pp. 20-21
Berry's paradox - the least integer not nameable in fewer than nineteen syllables
(Devised by G G Berry of the Bodleian Library)
The 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.
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Russell & Whitehead, p. 61
Two of Zeno's paradoxes
Achilles and the tortoise
Zeno's second paradox of motion, of Achilles and the tortoise, is probably the best known of his four paradoxes of motion. In this problem, the fleet Greek warrior runs a race against a slow-moving tortoise. Assume Achilles runs at ten times the speed of the tortoise (1 meter per second to 0.1 meter per second). The tortoise is given a 100-meter handicap in a race that is 1,000 meters. By the time Achilles reaches the tortoise's starting point T0, the tortoise will have moved on to point T1. Soon, Achilles will reach point T1, but by then the tortoise would have moved on to T2, and so on, ad infinitum. Every time Achilles reaches a point where the tortoise has just been, the tortoise has moved on a bit. Although the distances between the two runners will diminish rapidly, Achilles can never catch up with the tortoise, or so it would seem.
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Falletta, pp. 190-91
Bertrand Russell commented:
This argument... shows that, if Achilles ever overtakes the tortoise, it must be after an infinite number of instants have elapsed since he started. This is in fact true; but the view that an infinite number of instants makes up an infinitely long time is not true, and therefore the conclusion that Achilles will never overtake the tortoise does not follow.
Russell
Paradox of the arrow
In the paradox of the arrow, Zeno asks us to consider an arrow in flight and argues that, in fact, the arrow must always be at rest. At each instant the arrow occupies a space equal to itself. Movement is impossible, because an instant by definition has no parts. If the arrow were capable of moving during an instant, wee would contradict the definition of an instant, for the arrow would be in one position during the first part of the instant and in another position in the other part of the instant. Thus, the arrow never seems to be moving but rather, as Russell notes in his essay on infinity, "in some miraculous way the change of position has to occur between the instants, that is to say, not at any time whatever." If the arrow does not move at any given instant, how then does it make its flight?
Falleta, p.192
The ship of Theseus
The ship wherein Theseus and the youth of Athens returned had thirty oars, and was preserved by the Athenians down even to the time of Demetrius Phalereus, for they took away the old planks as they decayed, putting in new and stronger timber in their place, insomuch that this ship became a standing example among the philosophers, for the logical question of things that grow; one side holding that the ship remained the same, and the other contending that it was not the same.
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Plutarch, Vita Thesei, 22-23
Protagoras's pupil
Another paradox which has its foundation - real or legendary - in antiquity concerns the sophist Protagoras, who lived and taught in the fifth century BC. It is said that Protagoras made an arrangement with one of his pupils whereby the pupil was to pay for his instruction after he had won his first case. The young man completed his course, hung up the traditional shingle, and waited for clients. None appeared. Protagoras grew impatient and decided to sue his former pupil for the amount owed him.
'For,' argued Protagoras, 'either I win this suit, or you win it. If I win, you pay me according to the judgement of the court. If you win, you pay me according to our agreement. In either case I am bound to be paid.'
'Not so,' replied the young man. 'If I win, then by the judgement of the court I need not pay you. If you win, then by our agreement I need not pay you. In either case I am bound not to have to pay you.'
Whose argument was right? Who knows?
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Northrop, p.188
Sorites paradox (paradox of the heap)
Suppose you have a heap of sand. If you take away one grain of sand, what remains is still a heap: removing a single grain cannot turn a heap into something that is not a heap. If two collections of grains of sand differ in number by just one grain, then both or neither are heaps. This apparently obvious and uncontroversial supposition appears to lead to the paradoxical conclusion that all collections of grains of sand, even one-membered collections, are heaps.
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Sainsbury, p.23
M C Escher W
Drawing hands
Belvedere
Waterfall
Mobius band W
Mobius band
A Mobius band has just one side and just one edge.
Penrose triangle W
Necker cube (Freemish crate) W
Ames room W
Curry's paradox W
Curry's paradox
Arithmetic and algebraic paradoxes
Proving that 2 = 1
Here 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, wee conclude that x + 1 = 1; that is, since x = 1, 2 = 1.
Quine, p.5
Assume that
a = b. (1)
Multiplying both sides by a,
a² = ab. (2)
Subtracting b² from both sides,
a² - b² = ab - b² . (3)
Factorizing both sides,
(a + b)(a - b) = b(a - b). (4)
Dividing both sides by (a - b),
a + b = b. (5)
If now wee take a = b = 1, wee conclude that 2 = 1. Or wee can subtract b from both sides and conclude that a, which can be taken as any number, must be equal to zero. Or wee 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.
Northrop, p. 85
The paradox 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 example:
Proving that 3 + 2 = 0
Assume A + B = C, and assume A = 3 and B = 2.
Multiply both sides of the equation A + B = C by (A + B).
Wee obtain A² + 2AB + B² = C(A + B)
Rearranging the terms wee have
A² + AB - AC = - AB - B² + BC
Factoring out (A + B - C), wee have
A(A + B - C) = - B(A + B - C)
Dividing both sides by (A + B - C), that is, dividing by zero, wee get A = - B, or A + B = 0, which is evidently absurd.
Kasner & Newman, p. 183
Proving that n = n + 1
(a) (n + 1)² = n² + 2n + 1
(b) (n + 1)² - (2n + 1) = n²
© Subtracting n(2n + 1) from both sides and factoring, wee have
(d) (n + 1)² - (n + 1)(2n + 1) = n² - n(2n +1)
(e) Adding ¼(2n + 1)² to both sides of (d) yields
(n + 1)² - (n + 1)(2n + 1) + ¼(2n + 1)² = n² - n(2n + 1) + ¼(2n + 1)²
This may be written:
(f) [(n + 1) - ½(2n + 1)]² = [(n - ½(2n + 1)]²
Taking square roots of both sides,
(g) n + 1 - ½(2n + 1) = n - ½(2n + 1)
and, therefore,
(h) n = n + 1
Kasner & 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 read:
±(n + 1 - ½(2n + 1)) = ±(n - ½(2n + 1))
don't be overwhelmed by this thread, please