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Trying to prove by Contradiction, but need to make sure negation is correct

  1. Sep 23, 2006 #1
    Hello everyone. Before I start this proof i have to make sure my negation is right of the statement and also my translation of it into universal statements before I take the negation.

    The directions says, prove the statements by method of contradition. The method of contradiction states:
    1. Suppose the statement to be proved is false. that is, suppose that the negation of the statement is true. (be very careful in writing the negation!)
    2. Show that this supposition leads logicaly to a contradiction.
    3. Conclude that the statement to be proved is true.

    Here is the problem:
    If a and b are rational numbers, b != 0, and r is an irrational number, then a+br is irrational.

    So here is me translating it into universal statements:
    [tex]\forall[/tex] real numbers a and b and r, if a and b are rational such that b !=0, and r is irrational, then a + br is irrational.

    Here is me taking the negatoin of it:
    [tex]\exists[/tex] rational numbers a and b, b != 0 and irrational number r such that a + br is rational.

    Does this look okay to you? The universal statement seemed odd, becuase i had so many ands in it but i wzsn't sure how else to write it.

    Thanks!
     
  2. jcsd
  3. Sep 23, 2006 #2

    TD

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    In your initial statement, why say that a,b are real and then restrict to Q? So, for all a,b in Q and r in R\Q: ...
     
  4. Sep 23, 2006 #3
    Your right i guess i didn't need to say real numbers.
    If i took out the real part, would i sitll end up with the same negation? It seems i would.

    There exists rational numbers a,b, and b != 0 and irrational number r such that a+br is rational.
     
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