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CRN 12107: HW 2

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p=1+p1p2p3pn.
 
p=1+p1p2p3pn.
 
Then p leaves a remainder of 1 when divided by any of the pn's and thus p must be a prime not on the list of all primes.
 
Then p leaves a remainder of 1 when divided by any of the pn's and thus p must be a prime not on the list of all primes.
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 +
[[Image:Alice3.gif]]

Latest revision as of 22:28, 10 September 2013

Problem 6. Is the statement !xR:(x2=x+7) true or false? Prove your conjecture.

Problem 7. Let A,B and C be arbitrary sets. Recall that AB={x | xA  xB}. We define AB:=(AB)(BA). Prove or disprove:

  1. AB=BA.
  2. (AB)C=A(BC).

Problem 8. Let A and B be arbitrary sets. Prove or disprove the following power set relations:

  1. P(AB)P(A)P(B).
  2. P(A)P(B)P(AB).

Problem 9. Given two real numbers a<b, the open interval (a,b) is defined to be the set {xR | (a<x)(x<b)}.

For nN, let An be the open interval (1212n,12+13n). Find nNAn and nNAn. Confirm your conjectures by proofs.


Problem 10. Critique the following proof. Is the proof correct or flawed? Explain!

Recall that a positive integer p is prime if it is divisible by exactly two positive integers, namely 1 and p. The five smallest primes are 2,3,5,7,11.

Theorem. There are infinitely many primes.

Proof: Suppose there are only finitely many primes, say the list of all primes is {p1,p2,p3,,pn} for some positive integer n. Set p=1+p1p2p3pn. Then p leaves a remainder of 1 when divided by any of the pn's and thus p must be a prime not on the list of all primes.

Alice3.gif

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