There is a very interesting story behind your question. In the early 1900's (after Special Relativity had been introduced) the solution to wave equation in vacuum:
$a\exp(i(kx-wt)) + b\exp(-i(kx-wt)) $
where $k$ is the wave vector and $w/k=v$ and v is the phase velocity.
De Broglie was the first to notice that the phase factors of the equation at every event remain invariant under Lorentz transformations if $(k,w)$ is considered a 4-vector. This meant invariant amplitude at every event. This is because the scalar product of the two 4-vectors $(k,w)$ and $(x,t)$ remains invariant under Lorentz transformations. From this remarkable piece of insight, he deduced that $(k,w)$ could represent the 4-momentum,
$(p,E)$, of a massive particle. This is how wave-particle duality was first discovered. Special Relativity gave birth to quantum mechanics in its proper form!
The commutation relation discovered subsequently
$[p,x]=-i\hbar$
on solving gives (setting the arbitrary phase factor in p to 1): $p=-i\hbar \frac{∂}{∂x} $. The uncertainty relation in momentum is derived from here.
The fundamental form of Schrodinger's equation
$i\hbar \frac{∂T}{∂t} = ET$
Where $E$ is the Hamiltonian and $T(t_0,t)$ is the Unitary time evolution operator has its origins in the relationship between $x$ and $p$ being extended to $t$ and $E$.
When introducing the Schrodinger equation in his book, Dirac points out that its derivation comes mainly from considerations of relativity. In fact Schrodinger's original equation was actually relativistic, where $E$ was the relativistic energy. Schrodinger wasn't sure whether to take the positive or negative root of $E^2$. So he discarded it in favor of its widely known non-relativistic form.
So in fact, other than the discovery of quantized energy levels in blackbody radiation and the photoelectric effect, every breakthrough in QM owes it's existence to special relativity.
EDIT: I earlier said w/k = c for a massive particle, this is incorrect. w/k is equal to the phase velocity, which is proportional to 1/u. u is the group velocity which is equal to the classical velocity of a massive particle. I have fixed the offending sentences.
The uncertainty principle is a simple consequence of the idea that quantum mechanical operators do not necessarily commute.
In quantum mechanics, you find that the state which describes a state of definite value of an observable $A$ is not the state which describes a state of definite value for an observable $B$ if the commutator of both observables $[A,B]$ is not zero. (Formally, the two operators are not simultaneously diagonalizable.)
You just write down the definition of the standard deviation of the operator $A$ on a state $\psi$,
$$ \sigma_A(\psi) = \sqrt{\langle A^2\rangle_\psi - \langle A\rangle^2_\psi}$$
where $\langle\dot{}\rangle_\psi$ is the expectation value in the state $\psi$ and with a bit of algebraic manipulation (done e.g. on Wikipedia) we find that
$$ \sigma_A(\psi)\sigma_B(\psi)\ge \frac{1}{2}\lvert \langle[A,B]\rangle_\psi\lvert$$
Now, the standard deviation (or "uncertainty") of an observable on a state tells you how much the state "fluctuates" between different values of the observable. The standard deviation is, for instance, zero for eigenstates of the observable, since you always just measure the one eigenvalue that state has.
Plugging in the canonical commutation relation
$$ [x,p] = \mathrm{i}\hbar$$
yields the "famous" version of the uncertainty relation, namely
$$ \sigma_x\sigma_p\ge \frac{\hbar}{2}$$
but there is nothing special about position and momentum in this respect - every other operator pair likewise fulfills such an uncertainty relation.
It is, in my opinion, crucial to note that the uncertainty principle does not rely on any conception of "particles" or "waves". In particular, it also holds in finite-dimensional quantum systems (like a particle with spin that is somehow confined to a point) for observables like spin or angular momentum which have nothing to do with anything one might call "wavenature". The principle is just a consequence of the basic assumption of quantum mechanics that observables are well-modeled by operators on a Hilbert space.
The reason how "waves" enter is that the uncertainty relation for $x$ and $p$ is precisely that of the "widths" of functions in Fourier conjugate variables, and the Fourier relationship we are most familiar with is that between position and momentum space. That the canonical commutation relations are equivalent to such a description by Fourier conjugate variables is the content of the Stone-von Neumann theorem.
However, it is the description by commutation relations and not that by Fourier conjugacy that generalizes to all quantum states and all operators. Therefore, it is the commutation relation between the operators that should be seen as the origin of their quantum mechanical uncertainty relations.
Best Answer
While uncertainties can be interpreted as statements about ensembles of identically-prepared states, that's not how they are defined. You take a state $\psi$ and then the uncertainty $\sigma_\psi(A) = \sqrt{\langle A^2\rangle - \langle A\rangle^2}$ for any observable $A$ is simply a property of that state.
A state that has a definite momentum of zero would have to be an eigenstate of momentum. You can straightforwardly show that $\sigma_\psi(A) = 0$ for an eigenstate of $A$. So any state that has a well-defined $\sigma_\psi(x)$ cannot be an eigenstate of momentum, since the uncertainty principle implies $\sigma_\psi(p)\neq 0$. Note that this argument shows that in general you cannot have eigen"states" of position or momentum if both $\sigma_\psi(x)$ and $\sigma_\psi(p)$ exist.
"It can't have zero momentum" is not supposed to mean it's impossible to measure 0 momentum - just that it's not an eigenstate where the only possible result for momentum is zero (or indeed any other value of momentum).
In somewhat sloppy phrasing, "It can't have zero momentum" might also refer to the expectation value of the magnitude of momentum: If $\sigma_\psi(p)\neq 0$, then since $\sigma_\psi(p)\leq \sqrt{\langle p^2\rangle}$ we have that $\langle p^2\rangle$ is also non-zero.