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Prove that sqrt2 + sqrt6 is irrational

  1. Sep 19, 2003 #1
    Prove that sqrt2 + sqrt6 is irrational.
    Where do I start with this?
     
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Sep 19, 2003 #2

    Hurkyl

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    Assume otherwise!

    Or alternatively, consider the integer polynomial that has that number as one of its roots.
     
  4. Sep 19, 2003 #3
    I thought about the first one. It doesn't suit me well.
    Suppose it is rational. Then,
    sqrt2 +sqrt6 = p/q
    With some other rational-irrational proofs, I squared both sides. So,
    2 +2sqrt12 +6 =p2/q2
    4(sqrt3+4)=
    (2q)2(sqrt3+4)=p2
    Now what?

    As for the second part, I'm not sure I know what you mean by "integer" polynomial. Do you mean a polynomial whose domain is the integers?

    (x - (sqrt2 + sqrt6))(x - r)=0
    x^2 - x(sqrt2 +sqrt6) -xr +r(sqrt2 + sqrt6) = 0
    (sqrt2 + sqrt6)(x -r) = x(x-r)
    Let's hope x does not = r.
    Then sqrt2 + sqrt6 = x
    ....
    This of course makes perfect sense, since x doesn't equal r, it must equal it's other root. What pointless work! Certainly, I don't understand what you mean.
     
  5. Sep 19, 2003 #4

    Hurkyl

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    Solve for sqrt(3)!




    I meant one whose coefficients are integers...

    x = sqrt(2) + sqrt(6)
    x^2 = 8 + 4 sqrt(3)
    x^2 - 8 = 4 sqrt(3)
    x^4 - 16 x^2 + 64 = 48
    x^4 - 16 x^2 + 16 = 0

    And depending on how much you remember about solving polynomials, the result is trivial from here.
     
  6. Sep 20, 2003 #5
    Ok, wow, you're going to think I'm really incompetent. I really don't understand what to do for either instance. Sure, I can solve for sqrt3, but then what?
    You can square both sides and you get something quite similar to your second method.
    3=(p/2q)^4 - 8(p/2q)^2 +16
    or 0=(p/2q)^4 - 8(p/2q)^2 +13
    I could solve this, but the result is obvious. I'd just be undoing the square I just applied to both sides.
    (p/2q)^2=4+sqrt3
    So that's not right...

    Clearly, you want me to solve the polynomial you wrote. But then I get,
    x2=(16+8sqrt3)/2=8+4sqrt3=4(2+sqrt3)
    So, x=2sqrt(2+sqrt3)
    What nonsense is this!

    Am I getting anywhere near a proof of anything besides my ignorance?
    Please help.
     
  7. Sep 20, 2003 #6
    Here's another one I can't get.
    If a is rational and b is irrational is a+b necessarily irrational? I say yes, but suppose not. Then,
    a+b=p/q for some p and q in J
    and a=c/d for some c,d in J
    then b=p/q-c/d=(pd-cq)/(pd)
    Now if we can believe that the integers are closed under * and +, then we have derived a contradiction since pd-cq will be an integer and pd will be an integer. But how can I show they in fact are closed under * and +?

    What if a and b are both irrational?
    Not necessarily because
    pi+(-pi)=0
    Does that suffice?
    How can I know that the additive inverse of an irrational is also irrational?
     
  8. Sep 20, 2003 #7

    Hurkyl

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    Nah, this is one of those things that you usually have to see a few times before the method sinks in.


    When you solved for sqrt(3), you got:

    sqrt(3) = (p/(2q))^2 - 4

    We assumed that sqrt(2) + sqrt(6) is rational, and from that assumption we've proven that there is some integers p and q so the above equation holds... the left hand side of this equation is an irrational number, but what about the right hand side?


    As for the integer polynomial, there is a theorem that if p/q is a rational root of an integer polynomial in lowest terms, then p divides the constant term and q divides the leading coefficient. For example, if the equation

    2x^3 - 3x + 6 = 0

    has any rational roots, they must be among these possibilities:

    1/1, -1/1, 2/1, -2/1, 3/1, -3/1, 6/1, -6/1, 1/2, -1/2, 3/2, -3/2

    so you just have to exhaust all the possibilities to prove that this polynomial has no rational roots.

    So try this theorem with the polynomial

    x^4 - 16 x^2 + 16 = 0

    for which we know sqrt(2) + sqrt(6) is a root.
     
  9. Sep 20, 2003 #8

    Hurkyl

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    As for the integers being closed under + and *, you should be able to assume that to be true without proof.
     
  10. Sep 21, 2003 #9
    Guess what? Problem number 18 in my text asks me to prove that very theorem about integer polynomials. It wasn't assigned so I might not get around to it. Sounds tough.

    But how about this one:
    prove or disprove: If n is a natural number, then
    41 + 2 + 4 + 6 + ... + 2n is prime.

    My main problem is understanding what the pattern is. To me this is like asking me to prove that any sum of numbers is prime. Can you see a pattern? Maybe the first term is supposed to be one (not 41)? Then the pattern would be 1+2+4+6+8+10+12+...+2n, right?
     
  11. Sep 21, 2003 #10

    Hurkyl

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    If I had to guess, I think they mean:

    41 + Σ2n
     
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