Let $a, b$ and $c$ be numbers with $0 < a < b < c$. Which of the following is
impossible?
$\mathrm{(A)} \ a + c < b  \qquad \mathrm{(B)} \ a \cdot b < c \qquad \mathrm{(C)} \ a + b < c \qquad \mathrm{(D)} \ a \cdot c < b \qquad \mathrm{(E)}\frac{b}{c} = a$
According to the given rules, every number needs to be positive. Since $c$ is always greater than $b$, adding a positive number ($a$) to $c$ will always make it greater than $b$.
Therefore, the answer is $\boxed{\textbf{(A)}\ a+c<b}$