Doubt about ray diagrams In a ray diagram, 2 rays are considered enough to locate the image of a point on a given object. But how can we say that the rays other than the one we drew     will meet at that same point? 
I guess we can justify this by saying that we get only one image of a given object by a single mirror/lens (right?). So every point on the object must correspond to only one point on the only image. Is this reasoning correct?
Also, can somebody provide a more "rigorous" proof ( maybe with some math involved)
Thanks 
 A: This concerns what assumptions we are making about our optical system. Consider making a rudimentary lens using a flat slab of glass with a prism glued on the side. A ray going through the center goes straight on through; a ray going through the prism will be bent, so these two rays will meet somewhere, but there is no reason why other rays will meet at that same point. On the other hand, if our lens is an ideal lens, then the rays will all meet. The definition of "an ideal lens" is that it is an optical component which has this property. Once we have agreed that definition, then the method of just picking two rays is obviously sufficient to locate the image.
Now you can if you like explore what properties will bring about such an ideal lens. One way to define it is to say the focal length is independent of where the ray passes through the lens, and the direction change is the same for all rays passing through a given point on the lens. To realise this with a realistic device, the easiest approach is to adopt the "paraxial approximation" in which all rays under consideration stay close to the optic axis in their entire journey through any lenses under consideration. In this case a thin lens with spherical surfaces will do the job to first approximation.
A: This is a direct result of paraxial optics. By paraxial, one means that all the rays are nearly parallel to the optical axis.
Let's make this claim more rigorous. Any given ray at some point is characterized by its height $x$ and angle $\theta$ in respect to the optical axis. In this scenario, nearly every optical element can be approximated as a linear transformation of the $(x,\theta)$ vector, since $\theta\ll 1$ is small. In other words, we can associate with every optical system a matrix, called ABCD matrix, such that
$$\left(\matrix{x^\prime\\ \theta^\prime}\right)=\left(\matrix{A&B\\C&D}\right)\left(\matrix{x\\ \theta}\right)$$
where the $\prime$ indicates the coordinates after the system. In particular $x^{\prime}=Ax+B\theta$. In the special case of $B=0$ we can assert that $x^{\prime}=Ax$, i.e. $x^{\prime}$ is independent of $\theta$. Thus all the rays from height $x$ before the system intersect at a point of height $x^{\prime}$ immediately after. In this sense $B=0$ is the condition for imaging. In the case of an ideal lens, this reduces to the famous imaging formula
$$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$
For more information you can refer to any undergraduate book on optics (Fundamentals of Photonics by Saleh and Teich for example), or simply to this Wikipedia page.
A: Rays can cross without forming an image. If the two rays have qualitatively different histories, then they don't form an image, and the fact that those two rays cross at that point is a coincidence. As an example, try drawing a ray diagram for an object at the center of an equilateral triangle formed by three mirrors. There are intersections inside the triangle, but they aren't images. The images are all outside the triangle.
Even for rays that do have the same history, in general they will not all cross at exactly the same point. This is a type of aberration. Aberrations exist in real optical devices, but we ignore them in the simplified and idealized models and approximations that we teach in freshman physics.
As a simple mathematical example, suppose parallel rays come along the x axis into a converging optical device such as a lens. As an idealization/approximation, we can model the lens by saying that the slope $y'$ of the ray after passing through the device is proportional to $y$. Then it's easy to show that the rays will all intersect.
A: 
I guess we can justify this by saying that we get only one image of a
  given object by a single mirror/lens (right?). So every point on the
  object must correspond to only one point on the only image. Is this
  reasoning correct?

You are correct it's the lens main property to create a single image point for each point source.

Also, can somebody provide a more "rigorous" proof ( maybe with some
  math involved)

Well if the property above wasnt fullfilled you wouldnt be able to get clear images because image points would be the sum of several source points (in reality this happens to some degree)
