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

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

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

Proof so far:

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

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

How should I process?

2. Originally Posted by tttcomrader
Suppose that the function $\displaystyle f: \mathbb {R} ^n \rightarrow \mathbb {R}$ is continuous and that $\displaystyle f(u)>0$ if the point $\displaystyle u \in \mathbb {R} ^n$ has at least one reational component. Prove that $\displaystyle f(u) \geq 0$ for all points [tex] u \in \mathbb {R} ^n

Proof so far:

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

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

How should I process?