Your other question was marked as a duplicate of this one, so even though I don't believe they're the same question I'll answer that one here.
For reference, the new question was:
Mathematically show that Quantum Mechanics and General Relativity are inconsistent
Whilst it can't be expressed purely mathematically (it relies on dimensionality arguments), it can be shown mathematically. The trouble is, the actual maths itself is pretty hard to follow at the best of times. I'll try to explain what mathematically goes wrong when you move GR into QM, but it might not look like there's much maths because you just won't understand if I start throwing around propagation integrals.
Firstly, though, you need to realise that quantum mechanics isn't a physical theory in the same way that general relativity is. QM is just a mathematical framework in which one can formulate a theory. Without any physical grounding, QM is just the abstract mathematics of infinite dimensional complex Hilbert spaces. The mathematical framework used for general relativity is the one of classical fields, which is the same framework used for classical electromagnetism.
The approach that has traditionally worked with moving a theory formulated in the framework of classical fields into the framework of QM is called 'canonical quantisation', and you end up with a theory formulated in terms of quantum fields (which is just a sub-framework of QM). When you do this you find that to calculate any probability amplitude (which are the things that a measurable in QM) you have to do a Taylor expansion.
This is because in quantum mechanics time evolution is given by exponentiation of the Hamiltonian (group theoretically, the Hamiltonian is the generator of time evolution). So for some initial state, $|i\rangle$:
$$
|i\rangle \to |i(t)\rangle = U(t)|i\rangle = e^{-i\int_{0}^{t}H(t')dt'}|i\rangle,
$$
The definition of such a thing is its Taylor series:
$$
e^{A} \equiv I + A + \frac{A^2}{2} + \cdots + \frac{A^k}{k!} + \cdots
$$
In QFT you need to be able to calculate this so you can work out the probability that your initial state evolves into some final state $|f\rangle$:
$$
\mathcal{P}(|i\rangle \to |f\rangle) = \langle f|U(t)|i \rangle^2
$$
Typically, the Hamiltonian ($H$) looks something like:
$$
H(t) = \int d^3x\ g\phi_1(t)\phi_2(t)\cdots
$$
where the $\phi_n(t)$ are different fields all interacting and $g$ is coupling constant, which describes the strength of their interaction. This means each term in the expansion looks like:
$$
U_n \sim g^n\ \left(\int d^4x\ \phi_1(t)\phi_2(t)\cdots\right)^n
$$
Now, $U(t)$ as a whole must be dimensionless (because it is an exponentiation). This means each and every term in the expansion must be dimensionless. Therefore if $g$ has energy dimension $[g]$ (we're in natural units so all units can be expressed as powers of energy), the integral in the above expression must have dimensions $-[g]$. In general, we don't know if these integrals are going to converge, so we introduce a cutoff high energy scale, $\Lambda$, above which we don't bother integrating. (This is equivalent to choosing a small distance cutoff $L\propto\frac{1}{\Lambda}$ below which we don't integrate.) Then we can examine the behaviour as $\Lambda\to\infty$. The finite cutoff means that the integral will scale like $\Lambda^{-[g]}$.
This means that $U_n \sim g^n \Lambda^{-n[g]}$, and so the rate of change of $U_n$ with respect to $\Lambda$ is given by:
$$
\frac{dU_n}{d\Lambda} \sim -n[g] g^n \Lambda^{-n[g]-1}
$$
This rate of change absolutely cannot be positive for quantum fields to be an effective framework to describe the theory. If it is positive, every single term in the Taylor expansion will diverge and you scattering amplitude just becomes infinite. Since $g$, $n$, and $\Lambda$ are all positive, this means:
$$
[g] \geq 0
$$
Note that everything I've said so far is generally true about the entire framework of quantum fields. Therefore, in any theory described in terms of quantum fields, all the coupling constants must have non-negative energy dimension. Note that [g] being zero is also not brilliant: in this case you're highly dependent on the specifics of the theory to save you.
Now let's look at general relativity. The field in question for GR is the metric tensor for a 4-dimensional Lorentzian manifold, $g_{\mu\nu}$ (nothing to do with $g$ the coupling constant). The 'free' case of GR is special relativity, in which case:
$$
g_{\mu\nu} = \eta_{\mu\nu}
$$
where $\eta_{\mu\nu} = \pm\mathrm{diag}(-1,1,1,1)$ is the standard Minkowski metric for a flat spacetime. Let's only consider small perturbations, $\delta_{\mu\nu}$ around the free case:
$$
g_{\mu\nu} = \eta_{\mu\nu} + \delta_{\mu\nu}
$$
The action (time integral of the Lagrangian) for general relativity is given by the Einstein-Hilbert action:
$$
S = \frac{1}{2G}\int d^4x\ \sqrt{-g}R
$$
where $G$ is Newton's gravitational constant, $g$ is the determinant of $g_{\mu\nu}$, and $R$ is the Ricci scalar. Expanding this out you get:
$$
S = \frac{1}{2G}\int d^4x\ \left((\partial\delta)^2 + (\partial\delta)^2\delta + \cdots\right)
$$
For this to fit in with the existing framework for QFT we need to rescale $\delta$, so we say:
$$
\delta \to \delta' = \frac{1}{\sqrt{G}}\delta
$$
which gives:
$$
S = \frac{1}{2}\int d^4x\ \left((\partial\delta')^2 + \sqrt{G}(\partial\delta')^2\delta' + \cdots\right)
$$
Then our Hamiltonian for quantum mechanics is given by:
$$
H = \int d^3x\ \left(\sqrt{G}(\partial\delta')^2\delta' + \cdots\right)
$$
Which means our coupling constants are successive powers of $G^{1/2}$, which has energy units of $-1$. Oh dear, all of our coupling constants have negative energy dimension! This is precisely what we wanted to avoid, as it means that the theory is more dependant on higher energies than lower ones; hence, we can't simply ignore high energies. $G \sim [\mathrm{energy}]^{-2}$ because the gravitational potential, $V$, is dimensionless and depends on mass:
$$
V = -\frac{GM}{r}
$$
So ultimately general relativity doesn't work as a quantum field theory because its strength scales up with mass and energy, which makes it impossible to apply a high energy cutoff.
That's why quantum fields aren't a good mathematical framework for general relativity. This is telling us that we need a different way to move GR into the framework of quantum mechanics as this naive direct approach yields the wrong theory. Or, maybe we need to develop a new framework that we can move all the theories currently described by QM into along with GR? In any case, such a framework must have a finite 'smallest possible length scale' that provides a natural cutoff for our integral (hello string theory).