I agree with Peter K.'s answer, but I would like to add one important point: the two statements in the question are only true for causal systems. The most general statement about stability for LTI systems described by rational transfer function is:
An LTI system with a rational transfer function is stable if the region of convergence (ROC) of its transfer function includes the $j\omega$-axis (for continuous-time systems), or the unit circle $|z|=1$ (for discrete-time systems).
Since for causal systems the ROC is to the right of the pole with the largest magnitude (or outside the circle with radius equal to the largest pole radius in the $z$-plane), the above statement coincides with the statements in the question. However, non-causal systems can have poles in the right half-plane (or outside the unit circle of the $z$-plane) and still be stable.
Take as an example the discrete-time transfer function
which has one pole inside the unit circle and one pole outside the unit circle. You can't say that the corresponding system is unstable, because it depends on which system you mean. The transfer function (1) has 3 possible ROCs, corresponding to 3 different systems:
The first one corresponds to a causal system which is unstable, because one of its poles is outside the unit circle (the ROC does not include the unit circle). The second one corresponds to an anti-causal system which is also unstable, because it has one pole inside the unit circle (again, the ROC does not include the unit circle). Finally, the third ROC includes the unit circle, i.e. it corresponds to a stable system. It is a non-causal system with a two-sided impulse response which decays exponentially as the time index $n$ goes to $\pm \infty$.