A curious paradox

  • #1
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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 [Broken]

what do you think can the paradox be solved?
 
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  • #2
Hurkyl
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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!
 
  • #3
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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 dont know which is greater than the other then you should choose the "or" option.
 
  • #4
Hurkyl
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"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.
 
  • #5
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ok i think i understand your answer.
so there is no paradox.
 
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  • #6
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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 im not sure this method to tackle the question (ofcourse other than checking evey number from the smallest number (which i havent checked if it's a prime) 123456789 and then checking if it's not divisible by the numbers smaller than it.
 
  • #7
Hurkyl
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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.
 
  • #8
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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?
 
  • #9
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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
Hurkyl
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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
uart
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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
Hurkyl
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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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