Proving Rational Solutions in Linear Systems with Rational Coefficients

[e(ho0n3]

I need some closure on the following,

Question:
Prove that if a linear system of equations with only rational coefficients and constants has a solution then it has at least one all-rational solution. Must it have infinitely many?

My Solution:
If the RHS of the equations is a rational number, then the sum of the terms of the LHS must be rational, so the terms must be rational and hence the solution must be rational right? The only reason this argument wouldn't work is if the RHS is equal to 0. Must if have infinitely many rational solutions? No, but it can though (if the form of the solution includes some free variables, picking rational numbers for the free variables will produce a rational solution).

 

[fresh_42]

If $Ax=b$ with rational $A,b$ has a rational solution which is not rational, then we would have a $\mathbb{Q}-$linear combination of at least one non rational number, w.l.o.g. $x_1$ with $a_{11}\neq 0$ and $a_{11}x_1+\ldots +a_{1n}x_n= b_1$. However, this is only possible, if other terms eliminate $a_{11}x_1$ which makes the new solution rational.

Now we have a rational equation $Ax_b=b$, which leads to $Ay=0$ with new variables $y=x-x_b$. We can now restrict $A$ onto its image and get a regular matrix $A'=\left.A\right|_{\operatorname{im}A}$ and an equation $A'y'=0$ which means $y'=0$. So we need to consider the kernel of $A$. With the same argument as above we get a rational basis of this kernel. All combined, we have only rational solutions of $Ax=b$.