Ax=B has no solution In the textbook it says that no solution would mean
$$
r<m
$$
If $\mathbf{A}$ is an $m$ by $n$ matrix. I didn't quite get that. 
We know that
$$
r\le n
$$
for all times right, but isn't $r$ always larger than $m$? since $r$ is the rank of the matrix, it is both the degree of the row space and the column space. if $r$ is the degree of the row space it is the degree of the column space of transpose of the matrix $\mathbf{A}$ 
The transpose of the matrix $\mathbf{A}$ is $n$ by $m$, which means it has $m$ columns. since $r$ is still the degree this would imply that 
$$
r\le m
$$
So my question is why is it a requirement for
$$
r<m
$$
to have no solutions. there could easily be found cases where$$r<m$$ and has a solution. 
 A: Suppose the equation:  $Ax=b$, has no solutions for some particular $b$.  (having no solutions for all $b$ is just silly since $b=0$ one would always have at least one solution of $x=0$).
This tells us that $Ax=b$ is an inconsistent system and that $\text{rref}(A|b)$ has a row of $[0,0,\dots,0\mid 1]$.  (possibly also some rows of all zeroes too at the bottom).
This tells us that $\text{rref}(A)$ has a row of all zeroes.
One of the interpretations of $rank(A)$ is the number of pivots present in the rows of $\text{rref}(A)$.  We know it can not be $m$ (the number of rows) since there is at least one row of all zeroes where no pivot can exist.  (If it were $m$, then that would contradict that statement).
Thus, $rank(A) < m$

Note: here, the hypothesis is that given a vector $b$, if $Ax\neq b$ for all $x$ then...
Although there are always some pair of $b$ and $x$ that will always give a solution (e.g. both zero), we are not interested in searching for a $b$, we are only interested in searching for an appropriate $x$.
Further, we are not interested in proving the converse (which is not logically equivalent).  $Ax=b$ has no solution $\Rightarrow r<m$ is what we want to prove.  We do not want to prove $r<m \Rightarrow Ax=b$ has no solution.  Certainly, there exist cases of $b$ with $r<m$ and solutions to $Ax=b$ existing.  (again, take $b$ to be $0$ for trivial example)
