Why does a body not rotate if force is applied on the centre of mass? The definition of centre of mass on Wikipedia is given as

This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.

How can I prove that such a point is the weighted average of the radius vectors of all discrete masses mathematically?
Also see this question.
 A: Because force is the time derivative of momentum, and momentum is linked to the motion of the center of mass.
If you consider a rigid body as a collection of particles glued together and their position split into the position of the center of mass $\boldsymbol{r}_{\rm COM}$ plus some other relative position $\boldsymbol{d}_i$, then
$$ \boldsymbol{r}_i = \boldsymbol{r}_{\rm COM} + \boldsymbol{d}_i $$
and by taking the weighted average of the positions
$$\require{cancel} \sum_i m_i \boldsymbol{r}_i = \left( \sum_i m_i \right) \boldsymbol{r}_{\rm COM} + \cancel{ \sum_{i} m_i \boldsymbol{d}_i } $$
which means that the center of mass is the point which the weighted average relative position is zero $\sum_i m_i \boldsymbol{d}_i = 0$.
Now consider the motion of each particle as the velocity of the center of mass, and a rotation about the center of mass
$$ \boldsymbol{v}_i = \boldsymbol{v}_{\rm COM} + \boldsymbol{\omega} \times \boldsymbol{d}_i $$
Use the above to consider linear and angular momentum

*

*Linear Momentum
$$\boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i = \left( \sum_i m_i \right) \boldsymbol{v}_{\rm COM} + \boldsymbol{\omega} \times \left( \cancel{ \sum_i m_i \boldsymbol{d}_i }\right) = m\, \boldsymbol{v}_{\rm COM} $$


*Angular Momentum about center of mass
$$ \begin{aligned} \boldsymbol{L}_{\rm COM} & = \sum_i \boldsymbol{d}_i \times (m_i \boldsymbol{v}_i) \\ &= \left( \cancel{ \sum_i m_i \boldsymbol{d}_i} \right) \times \boldsymbol{v}_{\rm COM} + \sum_i \boldsymbol{d}_i \times m_i ( \boldsymbol{\omega} \times  \boldsymbol{d}_i) \\ &=  \mathbf{I}_{\rm COM}\; \boldsymbol{\omega} \end{aligned}$$
The last part of the puzzle is equating net force $\boldsymbol{F}$ to the rate of change of linear momentum and net torque about the center of mass $\boldsymbol{\tau}_{\rm COM}$ to the rate of change of angular momentum.
The equations below are the standard equations of motion for a rigid body.
$$ \boxed{ \begin{aligned}
  \boldsymbol{F} &= \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = m\,\boldsymbol{a}_{\rm COM} \\ \boldsymbol{\tau}_{\rm COM} & = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_{\rm COM} = \mathbf{I}_{\rm COM} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \boldsymbol{L}_{\rm COM}\end{aligned} }$$
So consider a force $\boldsymbol{F}$ applied away from the center of mass, which causes a net torque $\boldsymbol{\tau}_{\rm COM} = \boldsymbol{d} \times \boldsymbol{F} \neq 0 $ to a body initially at rest. This means that $\boldsymbol{\alpha} \neq 0$ causing rotational acceleration.
In summary, although a force applied on a body with always accelerate the center of mass, only a force through the center of mass causes no net torque, which would keep the body for accelerating rotationally.
A: Let $\vec r_1$ denote the position of the centre of mass of an object of mass $m$, given by the formula below.
$$\vec r_1 = \frac{1}{m}\int \rho \vec r^\prime \mathrm{d^3} \vec r^\prime$$
If an object is not rotating, then all of its points must have the same acceleration. Therefore, if a single force applied to the object does not cause rotation, it must be uniformly distributed over the mass. Let $\vec r_2$ denote the point at which a force can be applied to not cause rotation (the centre of mass using the definition you gave).
Consider the torque resulting from a force $\vec F$ applied at $\vec r_2$. Since the force is applied at $\vec r_2$, it is uniformly distributed over every infinitesimal piece of mass $\mathrm{d} m = \rho \mathrm{d^3} \vec r^\prime$. The torque about $\vec r_2$ resulting from this is given below. Note that $\vec r_1$ from above appears in this formula.
$$\vec\tau = \frac{1}{m}\int (\vec r^\prime - \vec r_2) \times \vec F \rho\mathrm{d^3} \vec r^\prime = \frac{1}{m}\left(\int (\vec r^\prime - \vec r_2) \rho\mathrm{d^3} \vec r^\prime \right) \times \vec F = (\vec r_1 - \vec r_2) \times \vec F$$
Since this setup does not cause rotation, the torque must be zero, from which we can conclude that $$\vec r_1 = \vec r_2$$
