Prove that if f : A -> B and g : B -> C are one-to-one functions, then gof is one-to-one.

A function f : A -> B is one-to-one if and only if for every pair of distinct elements a, a’ in A, f(a) and f(a’) are distinct in B.

Similarly, a function g: B -> C is one-to-one if and only if for every pair of distinct elements b, b’ in B, g(b) and g(b’) are distinct in C.

Now, let us consider the composite function gof. Suppose that there exist two distinct elements x and y in A such that (gof)(x) = (gof)(y). Then, by the definition of the composite function, we have g(f(x)) = g(f(y)).

Since g is one-to-one, this implies that f(x) and f(y) must be the same element in B.

But since f is also one-to-one, this means that x and y must have been the same element in A, which is a contradiction.

Therefore, we have shown that if f: A -> B and g: B -> C are one-to-one functions, then gof: A -> C is also a one-to-one function.