**Proof:** Since d\a and d\b, we know that d\(ax + by) for all integers x, y. Thus every member of T is a multiple of d. On the other hand, d may be written as d = ax_{0} + by_{0} for suitable integers x_{0} and y_{0}.* So that any multiple nd of d is of the form nd = n(ax _{0} +by_{0}) = a(nx_{0}) + b(ny_{0})*

Hence, nd is a linear combination of a and d, and by definition lies in T. (Proved)