1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Prove that of sum square root of 2 and square root of 3 is not rational
Loading...