text: As  and  we have  for i = 0 and 1. The relation follows by induction for all :
  Thus  and  are Bézout coefficients.
  Consider the matrix
  The recurrence relation may be rewritten in matrix form
  The matrix  is the identity matrix and its determinant is one. The determinant of the rightmost matrix in the preceding formula is −1. It follows that the determinant of  is  In particular, for  we have  Viewing this as a Bézout's identity, this shows that  and  are coprime. The relation  that has been proved above and Euclid's lemma show that  divides , that is that  for some integer . Dividing by  the relation  gives  So,  and  are coprime integers that are the quotients of  and  by a common factor, which is thus their greatest common divisor or its opposite.
  To prove the last assertion, assume that a and b are both positive and . Then, , and  if , it can be seen that the s and t sequences for (a,b) under the EEA are, up to initial 0s and 1s, the t and s sequences for (b,a). The definitions then show that the (a,b) case reduces to the (b,a) case. So assume that  without loss of generality.