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Math Help - Function with at least one rational component implies non-negative?

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    Function with at least one rational component implies non-negative?

    Suppose that the function f: \mathbb {R} ^n \rightarrow \mathbb {R} is continuous and that f(u)>0 if the point  u \in \mathbb {R} ^n has at least one reational component. Prove that  f(u) \geq 0 for all points  u \in \mathbb {R} ^n

    Proof so far:

    Suppose that u = (u_1,u_2,...,u_k,...,u_n) with  u_k \in \mathbb {Q} , so that f(u)>0

    I want to show that f(v) \geq 0 \ \ \ \ \ \forall v

    How should I process?
    Last edited by tttcomrader; March 27th 2009 at 10:15 AM.
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    Quote Originally Posted by tttcomrader View Post
    Suppose that the function f: \mathbb {R} ^n \rightarrow \mathbb {R} is continuous and that f(u)>0 if the point  u \in \mathbb {R} ^n has at least one reational component. Prove that  f(u) \geq 0 for all points [tex] u \in \mathbb {R} ^n

    Proof so far:

    Suppose that u = (u_1,u_2,...,u_k,...,u_n) with  u_k \in \mathbb {Q} , so that f(u)>0

    I want to show that f(v) \geq 0 \ \ \ \ \ \forall v
    Given any point p= (x_1,x_2,\cdot\cdot\cdot,x_n), there exist a point p= (u_1,u_2,\cdot\cdot\cdot,u_n) arbitrarily close with any given component rational.

    How should I process?
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