If Ax=v has a real solution then it must have a rational solution Suppose that $A$ is a  ${m\times n}$ matrix with rational entries. Let $v$ be a $m\times 1$ vector in $\mathbb{Q}^m$. Then I want to conclude that if the system of equations $Ax=v$ admits real solution then it must also admit a rational solution.
My approach towards a solution is via row reduction. But I am not able to move forward. There is a simliar question on some other stackexhange page but the solution is not very explanatory.
The following seems to be an interesting aspect of the problem- 
It is clear that the solution set of the above system represents a lower dimensional affine space of $\mathbb{R}^n$, and to say that the system has a rational solution is to say that this affine space intersects $\mathbb{Q}^n$.
 A: $\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\R}{\mathbb{R}}$$\DeclareMathOperator{\rank}{rank}$If $A x = v$ has a solution $x \in \R^{n}$, then $\rank(A) = \rank(A \mid v)$, as $v$ is a linear combination (with coefficients in $\R$) of the columns of $A$.
Conversely, $\rank(A) = \rank(A \mid v)$ tells you that the vector subspace of $\Q^{n}$ generated by the columns of $A$ is the same as the vector subspace of $\Q^{n}$ generated by the columns of $A$ plus $v$. Therefore $v$ is a linear combination with coefficients in $\Q$ of the columns of $A$. In other words, $A x = v$ has a solution $x \in \Q^{n}$.
A: Cramer's Rule will give by your hypothesis real solutions, but the rule expresses the solutions as quotients of what here are rational determinants. 
A: If you can assume that $\mathbb{Q}$ is a vector space over $\mathbb{Q}$, then by definition of matrix multiplication:
$$
\sum_{j=1}^{n}a_{ij}x_j = v_i
$$
for $i \in \{1,...,m\}$.
The existence of a solution means that there exists some $(x_1,...,x_m) \in \mathbb{R}^m$ such that these equations hold.
Since these equations are valid, and the $a_{ij}$'s and $v_i$'s are in  $\mathbb{Q}$, it follows from the assumption that $\mathbb{Q}$ is a vector space over $\mathbb{Q}$ that the $x_j$'s are also in $\mathbb{Q}$ (or in the case with infinitely many solutions, there exists some choice of $(x_1,...,x_m)$ that is rational).
