Given a rational polyhedon, say $P=\{x|Ax\le b\}$, can every supporting hyperplane be chosen to be rational? i.e. if $H=\{x|c^{T}x= d\}$ , $c$ and $d$ are rational vectors.
I want to prove that every face $F$ of $P$ can be written in the form $F=\{x\in P:c^{T}x=d\}$ where $c$ integral vector and $d$ integer, so I thought that if $c$ and $t$ can be chosen to be rational, then by multiplying with the lcm of the divisors I have the proof.
Let $F$ be a face of $P=\{x \in \mathbb{Q}^n| Ax \leq b, A \in \mathbb{Q}^{m \times n}, b \in \mathbb{Q}^m\}$. Thus, there exists some valid inequality $a_i x \leq b_i$ with $F=\{x \in P: a_i^T x \leq b_i \}$ and $a_i, b_i \in \mathbb{Q}$. It is possible to multiply the inequality with a positive integer $\lambda$ such that $\lambda a_i \in \mathbb{Z}^n$. This $\lambda$ is, obviously, not unique. One natural choice for $\lambda$ would be the least common multiple of the denominators of the non integer entries of $a_i$. Our inequality now is $\lambda a_i \leq \lambda b_i$ with the left side having only integer entries. We repeat the same for the right side: there is a $\mu \in \mathbb{Z}_+$ such that $\mu b_i \in \mathbb{Z}$. For example, if we pick $\mu$ to be the denominator of $b_i$ we'll get what we want. We multiply our inequality by $\mu$. Since $\mu$ is an integer, the left hand side will remain an integer. As $\mu b_i \in \mathbb{Z}$, the right side $\lambda \mu b_i$ will also be an integer because $\lambda \in \mathbb{Z}$.
Hence, we can set $c=\lambda \mu a_i$ and $d=\lambda \mu b_i$ and we will obtain $F'=\{x \in P: c^T x = d, c \in \mathbb{Z}^n, d \in \mathbb{Z} \} (=F)$.