Let A∈Mm×n(Q) and b∈Qm. Suppose that the system of linear equations Ax=b has a solution in Rn. Does it necessarily have a solution in Qn?and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine.I considered the form of the equality asA(1)x1+⋯+A(n)xn=bwhere A(i) is a column vector of A. I then noticed that for xi∈R∖Q=T then, and this is where I think I'm doing something forbidden, each x has the represenationx1=k11τ1+⋯+k1pτpx2=k21τ1+⋯+k2pτp⋮xn=kn1τ1+⋯+knpτp,where τi is a distinct irrational number, kij∈R, and p is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with p and m. I feel this method can lead me to the answer, but I'm not sure where to go from here.I end up getting something like this, I believe, after substitution:A(k(1)τ1+⋯+k(p)τp)=bHere, k(i) is the vector((k1i),(…),(kni)).I think there is no discrepancy with p and m because A∈Mm×n(Q),K∈Mn×p(R), and τ∈Mp×1(T), so(m×n)⋅(n×p)⋅(p×1)=m×1.

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Let A∈Mm×n(Q) and b∈Qm. Suppose that the system of linear equations Ax=b has a solution in Rn. Does it necessarily have a solution in Qn?and I thought I&#039;d give an interesting, possibly wrong, approach to solving it. I&#039;m not sure if such things can be done, if not maybe you can help me refine.I considered the form of the equality asA(1)x1+⋯+A(n)xn=bwhere A(i) is a column vector of A. I then noticed that for xi∈R∖Q=T then, and this is where I think I&#039;m doing something forbidden, each x has the represenationx1=k11τ1+⋯+k1pτpx2=k21τ1+⋯+k2pτp⋮xn=kn1τ1+⋯+knpτp,where τi is a distinct irrational number, kij∈R, and p is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with p and m. I feel this method can lead me to the answer, but I&#039;m not sure where to go from here.I end up getting something like this, I believe, after substitution:A(k(1)τ1+⋯+k(p)τp)=bHere, k(i) is the vector((k1i),(…),(kni)).I think there is no discrepancy with p and m because A∈Mm×n(Q),K∈Mn×p(R), and τ∈Mp×1(T), so(m×n)⋅(n×p)⋅(p×1)=m×1.

Rebekah Irvine · Accepted Answer

Yes, if the coefficients of A and b are rational and the system has at least one solution in reals, it also has at least one rational solution.  The elementary row operations performed during Gaussian elimination only use existing entries from the matrix and their inverses, thus keeping the matrix (and the vector b) rational all the time. Once the matrix is in reduced row-echelon form, some of the unknowns will be &quot;free&quot; to be set to any value (this happens if the rank of the matrix is smaller than the number of unknowns) and the others will be completely determined by them and the vector b. Setting the &quot;free&quot; ones to rational numbers yields completely rational solution to the original system.  More generally, if a system of linear equations over some field F has a solution in its extension E, it also has a solution in F. Moreover, if we know there is at least one non-rational (or non-F solution in the general version) solution, the cardinality of the set of solutions in F will not be smaller than |F|, since there will be at least one &quot;free&quot; unknown.

Let A \in M_{m\times n}(\mathbb{Q}) and b\in \mathbb{Q}^m. Suppose that the system of l

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