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Proving Tips

  1. Oct 11, 2004 #1
    I was just wondering, since i m kind of weak in doing proofs, what is the best way of understanding on how to do proofs. What is the best way to master, if one can, on doing proofs? or even if not master, but to be able to do proofs without "thinking", like sometimes my teacher says he just does it without thinking. Plz give me some ideas and using different examples would be appreciated. For example, n E N and a is positive reals (a1+...+an)(1/a1+...+1/an)>=n^2, n E N and 1+2+...+n^2<=n^3, etc.
     
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  3. Oct 11, 2004 #2

    Tide

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    The "not thinking" part only comes with experience and it takes plenty of hard work and persistence to reach that stage but there are challenges even beyond that. There are still many (and the list grows daily!) theorems and conjectures in mathematics that no one has yet been able to prove. And many of the greatest minds have thought long and hard about them.

    I recommend working on as many proofs as you can. The greatest source of them is right before your eyes - all the stuff you do daily and take for granted! Question what you do and why it's right - or not.
     
  4. Oct 11, 2004 #3
    how would you go about proving that pi is irrational?
    or 2^(1/2) ?
     
  5. Oct 11, 2004 #4

    Tide

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    I'd love to see a proof that [itex]\pi[/itex] is irrational - without using calculus!
     
  6. Oct 11, 2004 #5

    Galileo

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    Here's a proof that [itex]\sqrt{2}[/itex] is irrational. From Rudin's Principles of Mathematical Analysis.

    We show that the equation
    [tex]p^2=2 \qquad(1)[/tex]
    is not satisfied by any rational [itex]p[/itex]. If there were such a [itex]p[/itex], we could write [itex]p=m/n[/itex] where [itex]m[/itex] and [itex]n[/itex] are integers that are not both even. Let us assume this done.
    Then (1) implies
    [tex]m^2=2n^2 \qquad (2)[/tex]
    This shows that [itex]m^2[/itex] is even (if [itex]m[/itex] were odd, [itex]m^2[/itex] would be odd), and so [itex]m^2[/itex] is divisible by 4.
    It follows that the right side of (2) is divisible by 4 so that [itex]n^2[/itex] is even, which implies that [itex]n[/itex] is even.
    The assumption that (1) holds thus leads to the conclusion that both [itex]m[/itex] and [itex]n[/itex] are even, contrary to our choice of [itex]m[/itex] and [itex]n[/itex]. Hence (1) is impossible for rational [itex]p[/itex].

    I like this proof. The result is profound, but the proof so simple.
     
    Last edited: Oct 11, 2004
  7. Oct 11, 2004 #6

    Hurkyl

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    Hrm. Can you even define pi without calculus? Even the traditional definition, the ratio of the circumference of a circle to its diameter, is calculus in disguise!
     
  8. Oct 11, 2004 #7

    Tide

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    Wow! Leibnitz and Newton beaten to the punch by the ancient Greeks! ;-)

    Seriously, there is no problem defining pi without calculus. You just can't do a very good job at finding numerical approximations for its value without calculus.
     
  9. Oct 11, 2004 #8

    Hurkyl

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    Ack, I'm hijacking the thread! I'll take it to PM
     
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