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**Lierre** Hi vivian6606,

Yes, it is true. Because the set of the solutions of your equation (Ax=0, x >= 0) is a rational polyhedral cone. In particular, its extremal edges are rational lines. Since this cone is not empty, it has extremal edges.

This is the short answer. If you do not believe me, here is a more direct proof.

Consider E the R-vector-space Ker A, and C the intersection of E with the set $\displaystyle \{ \forall i, x_i \geq 0 \} $.

The coordinates x_i gives linear forms on E and C is the set of all v in E such that $\displaystyle x_i(v) \geq 0$.

C is not empty, is is your hypothesis. Assume that C contains an element v with $\displaystyle x_i(v) > 0$ for all i. (Easy exercise : reduce to this case by adding equations in A.)

Then C contains a neighbourhood of v, since the x_i are continuous.

Since the rationnal points of E (points a with the x_i(a) rational) are dense (you did the proof), C contains a rational point.