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Falsity of assumptions question.

  1. Aug 10, 2011 #1
    In the course of proving that [itex] \sqrt{3} [/itex] is irrational, I had another question pop up. To prove that [itex] \sqrt{3} [/itex] is irrational, I first assumed 2 things: [itex] \sqrt{3}[/itex] is rational, and the rational form of [itex] \sqrt{3} [/itex] is in it's lowest form. I then broke the proof up into cases and showed that none of these cases could occur.

    My question boils down to: did I actually show that [itex] \sqrt{3} [/itex] is irrational?

    From a purely logical standpoint, let's say that the 2 assumptions I made were named A and B. I successfully showed that A [itex] \wedge [/itex] B is false. However, this doesn't mean that BOTH A and B are false. More specifically, A could be true and B could be false, and I would still arrive at A [itex] \wedge [/itex] B being false.

    On the other hand, the second assumption that was made (the rational form of [itex] \sqrt{3} [/itex] is in it's lowest form) shouldn't (doesn't?) change the problem.

    Could someone give me solace and explain this little technicality I have? Thank you very much!
     
  2. jcsd
  3. Aug 10, 2011 #2

    Hurkyl

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    If you are merely assuming that, then you do have a problem, and you have merely proved:
    If sqrt(3) is rational, then it cannot be expressed as a fraction in lowest terms​
    (or something equivalent)

    But you don't have to merely assume that a rational number can be written in lowest terms, do you?
     
  4. Aug 10, 2011 #3
    So this isn't really an assumption, per se? This is more of a 'without loss of generality' statement?
     
  5. Aug 10, 2011 #4

    Hurkyl

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    Right, although I wouldn't have chosen that phrasing.
     
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