>>3811
It's a direct mathematical consequence of how physical variables work in quantum mechanics.
In quantum mechanics, a physical variable is an operator on the wavefunction. Position operator is multiplication by coordinate, and momentum operator is derivative by coordinate (multiplied by -iħ). The two operators are Fourier transforms of each other; that is, you can formulate the momentum operator in an alternative way by performing the Fourier transform on the wavefunction, so you describe the particle not by its probability distribution in space but in wavenumber, which makes the momentum operator a simple multiplication by wavenumber. (Fourier transformation turns derivatives into multiplicators and vice versa)
If you assume a particle has a certain position, then you can describe it by Dirac delta function — a "function" whose value is zero almost everywhere, except in the point where the particle is, normalized so that the integral of such "function" is one (so you can imagine as if the value of the function in that one point is infinite). A Fourier transform of Dirac delta function is a function of form e^ikx (the constant k is not relevant here), and you probably know that this is a cosine wave in real and sine wave in imaginary component, and that it's absolute value is equal to one everywhere. So, if the position of a wavefunction is completely certain, then the wavefunction must be a Dirac delta function in space coordinates, but this means the momentum of that particle is completely undefined, as the absolute value of the wavefunction in wavenumber coordinates is a constant — every wavenumber is equally possible.
The Uncertainty principle is what you get if you assume that the wavefunction is a Gaussian distribution. Fourier transform of a Gaussian distribution is another Gaussian distribution, whose width is inversely proportional to that of the original one. So, the product of widths of the wavefunction in space coordinates and in wavenumber coordinates, both Gaussian distributions, must be a constant. There's also another mathematical proof that you cannot construct a function for which the product of standard deviations in space and wavenumber is smaller than that of a Gaussian distribution, hence the inequality in the Uncertainty principle.
tl;dr: pure maths