In mathematics, Minkowski's theorem is the statement that every convex set in which is symmetric with respect to the origin and which has volume greater than contains a non-zero integer point (meaning a point in that is not the origin). The theorem was proved by Hermann Minkowski in 1889 and became the foundation of the branch of number theory called the geometry of numbers. It can be extended from the integers to any lattice and to any symmetric convex set with volume greater than , where denotes the covolume of the lattice (the absolute value of the determinant of any of its bases). Suppose that L is a lattice of determinant d(L) in the n-dimensional real vector space Rn and S is a convex subset of Rn that is symmetric with respect to the origin, meaning that if x is in S then −x is also in S. Minkowski's theorem states that if the volume of S is strictly greater than 2n d(L), then S must contain at least one lattice point other than the origin. (Since the set S is symmetric, it would then contain at least three lattice points: the origin 0 and a pair of points ± x, where x ∈ L \ 0.) The simplest example of a lattice is the integer lattice Zn of all points with integer coefficients; its determinant is 1. For n = 2, the theorem claims that a convex figure in the Euclidean plane symmetric about the origin and with area greater than 4 encloses at least one lattice point in addition to the origin. The area bound is sharp: if S is the interior of the square with vertices (±1, ±1) then S is symmetric and convex, and has area 4, but the only lattice point it contains is the origin. This example, showing that the bound of the theorem is sharp, generalizes to hypercubes in every dimension n. The following argument proves Minkowski's theorem for the specific case of L = Z2. Proof of the case: Consider the map Intuitively, this map cuts the plane into 2 by 2 squares, then stacks the squares on top of each other. Clearly f (S) has area less than or equal to 4, because this set lies within a 2 by 2 square.