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Homework Help: Prove that of sum square root of 2 and square root of 3 is not rational

  1. Nov 22, 2008 #1
    prove that the square root of 2 plus the square root of 3 is not rational?

    does always the sum of two not rational numbers is a not rational number?



    i know the proof 2 = a^2/b^2
    i separately proved that square root of 2 and square root of 3 are irrational

    how two prove that the sum of two such numbers is irrational too?

    whats the formal equation proof?
     
    Last edited: Nov 22, 2008
  2. jcsd
  3. Nov 22, 2008 #2

    Mark44

    Staff: Mentor

    Try a proof by contradiction. Suppose that [tex]\sqrt{2} + \sqrt{3} = \frac{a}{b}[/tex]
    where a and b are integers with no common factors.

    For your second question [tex]\sqrt{5} + (-\sqrt{5}) = 0 [/tex]
    Those are both irrational numbers, but their sum is rational.
     
  4. Nov 22, 2008 #3

    nicksauce

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    Well clearly two irrational numbers can sum to give a rational number. For example (2+sqrt(2)) + (2 -sqrt(2)).

    To show that the sqrt(2) + sqrt(3) is irrational, you can start with
    a = sqrt(2) + sqrt(3)
    Manipulate to get
    a^4 - 10a^2 + 1 = 0

    Then you can use the rational roots theorem, that shows that the only rational roots of this equation can be +- 1. http://en.wikipedia.org/wiki/Rational_root_theorem
     
  5. Nov 22, 2008 #4
    Somehow, that just seems like cheating, but it does fit the problem we were given.
     
  6. Nov 22, 2008 #5

    nicksauce

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    Any mathematical statement can be disproved by a counterexample. The simpler the counterexample, the better.
     
  7. Nov 22, 2008 #6
    ok i will try to prove by contradiction:
    suppose
    (2)^0.5 + (3)^0.5 is a rational number
    (if we multiply a rational number by a rational number we will get a rational number "h")
    5+2*(2)^0.5 * (3)^0.5=h
    24=h^2 -10*h +25

    h^2 -10*h +1=0
    what to do now?
     
  8. Nov 22, 2008 #7

    nicksauce

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    As I suggested earlier, make use of the rational roots theorem.
     
  9. Nov 22, 2008 #8
    by this rational roots theorem
    the possible roots is +1 and -1

    not one of the represent the actual roots of h^2 -10*h +1=0

    what is the next step in the prove?
     
  10. Nov 22, 2008 #9

    Mark44

    Staff: Mentor

    What did you set out to do in your post #6?
     
  11. Nov 22, 2008 #10
    You don't need anything but parity to prove this.

    Prove



    [tex]\sqrt{2}+\sqrt{3} = (p/q)[/tex]
    [tex]2 \sqrt{6}+5 = p^{2}/q^{2}[/tex]

    You know that the rationals form a field, which implies all the terms in the sum must be a rational.

    Now all that remains is to show [tex]\sqrt{6}[/tex] is irrational. So just show that if it equals p/q, p and q must both be even.
     
    Last edited by a moderator: Jun 22, 2009
  12. Nov 23, 2008 #11
    the root test showed me that there is no rational roots for this equation.
    is it fair to consider "h" irrational because of it?

    i was told that if we dont have a rational roots it doesnt mean that
    "h" is always irrational,"h" could be a complex number.

    ??
     
  13. Nov 23, 2008 #12

    Mark44

    Staff: Mentor

    If a number is complex, it certainly isn't a rational number. If h^2 - 10h + 1 = 0 doesn't have any rational roots, then there is no solution for h that is rational.
     
    Last edited: Nov 23, 2008
  14. Nov 23, 2008 #13
    you meant "irrational" (in the end) right?
     
  15. Nov 23, 2008 #14

    Mark44

    Staff: Mentor

    I edited my reply. The last part should have said "there is no solution that is rational."
     
  16. Jun 22, 2009 #15
    how did you arrive at a^4 - 10a^2 + 1 = 0?
    need help
     
  17. Jun 22, 2009 #16

    Mentallic

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    Well since we've already summoned this thread from the grave...
    [tex]a=\sqrt{2}+\sqrt{3}[/tex]

    [tex]a^2=(\sqrt{2}+\sqrt{3})^2=2+2\sqrt{2}\sqrt{3}+3=5+2\sqrt{6}[/tex]

    [tex](a^2-5)^2=(2\sqrt{6})^2[/tex]

    [tex]a^4-10a^2+25=24[/tex]

    [tex]a^4-10a^2+1=0[/tex]
     
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