Iterating the same argument, "r""N"−1 divides all the preceding remainders, including "a" and "b". None of the preceding remainders "r""N"−2, "r""N"−3, etc. divide "a" and "b", since they leave a remainder. Since "r""N"−1 is a common divisor of "a" and "b", "r""N"−1 ≤ "g". In the second step, any natural number "c" that divides both "a" and "b" (in other words, any common divisor of "a" and "b") divides the remainders "r""k". By definition, "a" and "b" can be written as multiples of "c" : "a" = "mc" and "b" = "nc", where "m" and "n" are natural numbers. Therefore, "c" divides the initial remainder "r"0, since "r"0 = "a" − "q"0"b" = "mc" − "q"0"nc" = ("m" − "q"0"n")"c". An analogous argument shows that "c" also divides the subsequent remainders "r"1, "r"2, etc. Therefore, the greatest common divisor "g" must divide "r""N"−1, which implies that "g" ≤ "r""N"−1. Since the first part of the argument showed the reverse ("r""N"−1 ≤ "g"), it follows that "g" = "r""N"−1. Thus, "g" is the greatest common divisor of all the succeeding pairs:  For illustration, the Euclidean algorithm can be used to find the greatest common divisor of "a" = 1071 and "b" = 462. To begin, multiples of 462 are subtracted from 1071 until the remainder is less than 462. Two such multiples can be subtracted ("q"0 = 2), leaving a remainder of 147:  Then multiples of 147 are subtracted from 462 until the remainder is less than 147. Three multiples can be subtracted ("q"1 = 3), leaving a remainder of 21:  Then multiples of 21 are subtracted from 147 until the remainder is less than 21.