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Prove irrational numbers

  1. Sep 28, 2005 #1
    we can prove that √5 is irrational through contradiction and same applies for √3.
    but can we prove that √5 + √3 is irrational by contradiction?
     
  2. jcsd
  3. Sep 28, 2005 #2

    arildno

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    Show that if [tex]\sqrt{5}+\sqrt{3}[/tex] is rational, then so must [tex]\sqrt{5}-\sqrt{3}[/tex] be rational.
    Derive that, hence, [tex]\sqrt{5}[/tex] is rational, a contradiction.
     
  4. Sep 28, 2005 #3

    HallsofIvy

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    ??? Are you saying that if x+ y is rational then x- y must be rational? What if [itex]x= \sqrt{5}[/itex] and [itex]y= -\sqrt{5}[/itex]?
     
  5. Sep 28, 2005 #4

    arildno

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    Not at all, but:
    Let [tex]\sqrt{5}+\sqrt{3}=p[/tex] where p is rational.
    We therefore have:
    [tex]\frac{2}{\sqrt{5}-\sqrt{3}}=p[/tex], that is:
    [tex]\sqrt{5}-\sqrt{3}=m[/tex], m=2/p rational.

    That is, since the difference of squares of [tex]\sqrt{5}[/tex] and [tex]\sqrt{3}[/tex] is rational, and [tex]\sqrt{5}+\sqrt{3},\sqrt{5}-\sqrt{3}\neq{0}[/tex] my argument holds.
     
    Last edited: Sep 28, 2005
  6. Sep 28, 2005 #5
    Is it possible for two irrational numbers to be added together to get a rational number?
     
  7. Sep 28, 2005 #6
    I don't know about that, but if you divide pi by phi you get a rational (they share a common irrational factor).
     
  8. Sep 28, 2005 #7

    arildno

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    Definitely, HallsofIvy gave an example of that.
     
    Last edited: Sep 28, 2005
  9. Sep 28, 2005 #8

    Gokul43201

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    And what might that be ?
     
  10. Sep 28, 2005 #9
  11. Sep 28, 2005 #10

    Hurkyl

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    The most algorithmic approach I know for these types of problems is to find an integer polynomial with your algebraic number as a root...
     
  12. Sep 28, 2005 #11

    CRGreathouse

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    Are you saying that [tex]\frac{\pi}{\phi^2}\in\mathbb{Q}[/tex]?

    Since [tex]\phi=\frac{1+\sqrt5}{2}[/tex] is algebraic and algebraic numbers are closed under multiplication and division, that would mean that pi is algebraic (all rationals are algebraic). The paper you link to, though, admits that "[tex]\pi[/tex] is a transcendental number" and that trancendentals aren't algebraic (by definition).
     
  13. Sep 29, 2005 #12

    matt grime

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    yes, as has been shown, but more gnerally (I can't recall where i first read this wordy description) but if you are trying to decide if two nasties can make a nice then try thinking if a nice minus a nasty can be nasty. here, obviously the rational minus irrational must be irrational, so anything like

    1-sqrt(2) and sqrt(2) add together to give what you want.

    the same works for continuous and discontinuous functions, invertible and uninvertible matrices, rational and irrational numbers, algebraic and transcendental numbers and so on.
     
  14. Sep 29, 2005 #13

    HallsofIvy

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    I looked at the paper referred to above. There are no proofs at all. The author refers to φ2 as a "composite" number with no explanation of what "composite" is supposed to mean for a non-integer number. All he does is show that 15 decimal place computations come out the same, then spends the rest of the paper defining "fraction", "rational", "irrational", etc.
     
  15. Sep 29, 2005 #14
    proof: if n is rational and m is irrational set [tex]n + m = l[/tex] is rational then [tex]l + (-n) = m[/tex] is rational, a contradiction
     
  16. Sep 29, 2005 #15

    CRGreathouse

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    Yes, that was what most bothered me. Normally there's a certain semblance of internal consistency in these pseudoproofs, but this one just didn't define enough to come to a contradiction that I was patient enough to find.
     
  17. Oct 30, 2005 #16
    You could also argue by contradiction that if
    [tex] \sqrt5 + \sqrt3[/tex] is rational then so must [tex](\sqrt5 + \sqrt3)^2[/tex] and something bad will happen.
     
    Last edited: Oct 30, 2005
  18. Feb 14, 2011 #17
    show me how by contradiction
     
  19. Feb 14, 2011 #18

    uart

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    Assume [itex]\sqrt{5} = \frac{a}{b}[/itex] for some integers "a" and "b".

    Then

    [tex]a^2 = 5 b^2[/tex]

    The LHS of the above equation has prime factor "5" to an even power while the RHS has prime factor "5" to an odd power. This is an obvious contradiction.
     
  20. Feb 14, 2011 #19
    The contradiction arises when you try to proof that sqrt(15) is rational.
     
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