Raymond Smullyan's Recreational Math Paradox on Page 189: Can it be Solved?

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    Curious Paradox
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Discussion Overview

The discussion revolves around a paradox presented by Raymond Smullyan in a book on recreational mathematics. Participants explore whether the paradox can be resolved, examining the logical implications of certain mathematical propositions and their truth values.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that both propositions in the paradox can be true due to the nature of vacuous truth, where a false hypothesis leads to a true implication.
  • One participant suggests that the paradox arises from the ambiguous use of variables, specifically that the same variable is used to represent both the minimum and maximum in different contexts.
  • Another participant proposes an alternative approach to the paradox by suggesting that the statements should be framed with "or" instead of "and" to avoid contradictions.
  • In a separate question, participants discuss the smallest prime number containing each digit from 1 to 9, with one suggesting a method involving the properties of prime numbers and another noting that any such number must repeat digits to avoid divisibility by 9.
  • There is a discussion about the divisibility rules for 9 and 3, with one participant confirming that any number divisible by 9 is also divisible by 3.

Areas of Agreement / Disagreement

Participants express differing views on the resolution of the paradox, with some suggesting it can be resolved while others maintain that the ambiguity in the propositions leads to confusion. The discussion on the prime number question also reveals varying approaches and methods without a clear consensus.

Contextual Notes

Participants highlight the importance of phrasing in mathematical statements and the implications of using the same variable in different contexts, which may lead to misunderstandings. The discussion on the prime number question also indicates a lack of consensus on the best method to find the solution.

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in this webpage there is a book about recreational maths in it there is a paradox by raymond smullyan, it's on page 189:
http://www.g4g4.com/paul/BOOK.pdf

what do you think can the paradox be solved?
 
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Actually, propositions 1 and 2 can both be true.


Recall that if P is false, then P => Q is a true statement.

It turns out that both of these propositions are vacuously true; e.g. if you rewrite the first one in a less misleading fashion, it becomes:

If x > y and y > x, then x - y > y - x.

And the second one becomes

If x > y and y > x, then x - y = y - x


Obviously, the hypothesis of both of these statements is always false, so these statements are always true.

When you combine these statements to produce the "contradiction", you get:

If x > y and y > x, then x - y > y - x and x - y = y - x

Or

If false, then false.

Which is a true statement!


Mathematics is saved; there is no choice of x and y that satisfies the hypothesis of this statement, so the conclusion of the statement never matters!
 
Originally posted by Hurkyl
Actually, propositions 1 and 2 can both be true.


Recall that if P is false, then P => Q is a true statement.

It turns out that both of these propositions are vacuously true; e.g. if you rewrite the first one in a less misleading fashion, it becomes:

If x > y and y > x, then x - y > y - x.

And the second one becomes

If x > y and y > x, then x - y = y - x


Obviously, the hypothesis of both of these statements is always false, so these statements are always true.

When you combine these statements to produce the "contradiction", you get:

If x > y and y > x, then x - y > y - x and x - y = y - x

Or

If false, then false.

Which is a true statement!


Mathematics is saved; there is no choice of x and y that satisfies the hypothesis of this statement, so the conclusion of the statement never matters!
i think it should be if x>y or y>x because you don't know which is greater than the other then you should choose the "or" option.
 
"The excess of x over y, if x is greater than y" only gives numbers for x > y.

Similarly, "The excess of y over x if y is greater than x" only gives numbers for y > x.

The only way both of these statements can give a number (and thus for one to make equations and inequations from those numbers) is if both of the hypotheses are true; that is if both x > y and y > x.
 
ok i think i understand your answer.
so there is no paradox.
 
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here's another question from the book:
find the smallest prime number that contains each digit from 1 to 9 at least once.

a method i thought about is if 2^n-1 is a prime then n is also a prime but I am not sure this method to tackle the question (ofcourse other than checking evey number from the smallest number (which i haven't checked if it's a prime) 123456789 and then checking if it's not divisible by the numbers smaller than it.
 
First off, 9 will divide any 9-digit number that contains each of 1, 2, 3, ..., 9... so you have to have at least one repeat.

If I were to tackle this problem, I would simply examine the smallest 100 or so 10-digit integers that contain each digit. If none of them are prime, then I would go back to the drawing board and try to figure something out.
 
Originally posted by Hurkyl
First off, 9 will divide any 9-digit number that contains each of 1, 2, 3, ..., 9... so you have to have at least one repeat.

i didnt know that.
is there a reason for that?
 
could it be said that a number who is divided by 9 could be divided by 3 (i think so because 9 is a multiple of 3).
 
  • #10
The classic test for divisibility by 9 is:

x is divisible by 9 iff the sum of the (base 10) digits of x is divisible by 9.


For a quick proof, notice that

Σi di 10i = Σi di 1i (mod 9)


1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, and 45 is divisible by 9, so any number which has exactly one of each of these digits must be divisible by 9.


And yes, anything divisible by 9 is divisible by 3.
 
  • #11
Originally posted by Hurkyl
Actually, propositions 1 and 2 can both be true.


Recall that if P is false, then P => Q is a true statement.

It turns out that both of these propositions are vacuously true; e.g. if you rewrite the first one in a less misleading fashion, it becomes:

If x > y and y > x, then x - y > y - x.

And the second one becomes

If x > y and y > x, then x - y = y - x


Obviously, the hypothesis of both of these statements is always false, so these statements are always true.

When you combine these statements to produce the "contradiction", you get:

If x > y and y > x, then x - y > y - x and x - y = y - x

Or

If false, then false.

Which is a true statement!


Mathematics is saved; there is no choice of x and y that satisfies the hypothesis of this statement, so the conclusion of the statement never matters!

Hey, it's even simpler than that. All the first proposition is really saying is that if one of the numbers (x or y) is equal to twice the other number then the difference is equal to one times the minimum (of x and y) but equal to one half times the maximum. Of course there is no paradox when stated like this and it is perfectly compatible with proposition 2.

The apparent paradox comes from the authors use of the variable y as both the smallest number, min(x,y), in the first instance and also as the largest number, max(x,y) in the second instance. So y is not a constant unless a different set of two numbers is used for each case, which would make the comparision pointless anyway.
 
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  • #12
Well, the reason I explained it this way is because phrasing can sometimes be very important, so I didn't want to "cheat" by rewriting their claim.
 

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