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Math Help - Spivak's Calculus - Proof

  1. #1
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    Spivak's Calculus - Proof

    (d) For which numbers a is it true that x^2 + axy + y^2 > 0 whenever x and y are not both 0?

    (x + \frac{ay}{2})^2 + (\frac{\pm\sqrt{-((ay)^2 - 4y^2)}}{2})^2 > 0 for (ay)^2 - 4y^2 < 0,
       \Rightarrow  (x + \frac{ay}{2})^2 + (\frac{\pm\sqrt{-(ay)^2 + 4y^2)}}{2})^2 > 0,
     \Rightarrow (x + \frac{ay}{2})^2 + \frac{-(ay)^2 + 4y^2}{4} > 0,
     \Rightarrow (x + \frac{ay}{2})^2 + \frac{-(ay)^2}{4} + \frac{4y^2}{4} > 0,
     \Rightarrow (x + \frac{ay}{2})^2 + \frac{-(ay)^2}{4} + y^2 > 0,
     \Rightarrow x^2 + axy + \frac{(ay)^2}{4} + \frac{-(ay)^2}{4} + y^2 > 0,
     \Rightarrow x^2 + axy + y^2 > 0 for (ay)^2 - 4y^2 < 0.

    (ay)^2 - 4y^2 < 0,
     \Rightarrow (ay)^2 < 4y^2,
     \Rightarrow a^2y^2  < 4y^2,
     \Rightarrow a^2 < 4,
     \Rightarrow a < \pm2.
    Therefore, x^2 + axy + y^2 > 0 for a < \pm2.


    Is this proof valid? When I sub in a value less than -2 for a, I end up with a result contrary to the conclusion of my proof. For instance:

    x^2 + axy + y^2 = (-5)^2 + (-100)(-5)(-10) + (-10)^2 = -4,875 > 0.
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  2. #2
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    Re: Spivak's Calculus - Proof

    This is what I might do. Divide by x^2 so

    \dfrac{y^2}{x^2} + a \dfrac{y}{x} + 1 > 0

    so

    \left( \dfrac{y}{x} + \dfrac{a}{2}\right)^2 > \dfrac{a^2}{4} - 1

    To guarantee this for all x and y not zero, then we would require that \dfrac{a^2}{4} - 1 < 0 so -2 < a < 2.
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    Re: Spivak's Calculus - Proof

    Your proof seems fine, but when I sub in a number for a that is less than -2, such as -2.1, I get the following:

    x^2 + axy + y^2 = (-5)^2 + (-2.1)(-5)(-10) + (-10)^2 = 20 > 0.

    So although the statement above is true, it can't be true that the statement is only true for -2 < a < 2.
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  4. #4
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    Re: Spivak's Calculus - Proof

    But don't you want your statement to be true for all x and y not zero. For example, if you choose a = -2.1 (like you said) then what you want is

    x^2 - 2.1xy + y^2 > 0 but if you choose x = y = 1 then

    1 - 2.1 + 1 = -.1 \ngtr 0
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    Re: Spivak's Calculus - Proof

    Quote Originally Posted by Danny View Post
    But don't you want your statement to be true for all x and y not zero. For example, if you choose a = -2.1 (like you said) then what you want is

    x^2 - 2.1xy + y^2 > 0 but if you choose x = y = 1 then

    1 - 2.1 + 1 = -.1 \ngtr 0
    Yes, the statement must be true for all xand y not 0, as the problem requests. But the problem mainly asks for which numbers a is the statement true. Of course, the statement is true if -2 < a < 2, but clearly the statement isn't only true for those numbers. And, if I interpreted the question correctly (please correct me if I'm wrong), what is required is that I find all a for which the statement is true. If I could just state a portion of that, I could easily say that the statement is true for one arbitrary number, such as 1, and that alone would qualify as a complete solution.

    I'll be logging off now, but I'll be back later.

    I've been trying to solve this problem for quite some time now. :/
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    Re: Spivak's Calculus - Proof

    Quote Originally Posted by RogueDemon View Post
     \Rightarrow a^2 < 4,
     \Rightarrow a < \pm2.
    ehm...
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    Re: Spivak's Calculus - Proof

    Quote Originally Posted by RogueDemon View Post
    Yes, the statement must be true for all xand y not 0, as the problem requests. But the problem mainly asks for which numbers a is the statement true. Of course, the statement is true if -2 < a < 2, but clearly the statement isn't only true for those numbers. And, if I interpreted the question correctly (please correct me if I'm wrong), what is required is that I find all a for which the statement is true. If I could just state a portion of that, I could easily say that the statement is true for one arbitrary number, such as 1, and that alone would qualify as a complete solution.

    I'll be logging off now, but I'll be back later.

    I've been trying to solve this problem for quite some time now. :/
    Are you saying that the following statement is true?

    x^2-2.1xy+y^2>0 whenever x and y are not both zero.
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