Prove if a and b are natural numbers with b < = a then there exists a q and r in the natural numbers such that a=qb +r.(Division Algorithm)

Prove if a and b are natural numbers with b < = a then there exists a q and r in the natural numbers such that a = qb + r and 0 < = r < b. Where q and r are the quotient and remainder. This is a proof of the division algorithm.

First we let T equal a number s in the natural numbers where a < sb. By the well ordering principle T has a smallest element which we can call w. Now we let q = w - 1 and r = a - qb since a > = b , w > 1 , q is in the natural numbers. Thus q < w , a > qb and r > = 0. By definition of r a = qb + r. Now suppose if r > = b. Then a - (w - 1)b > = b. so a >= wb, which contradicts the fact that a < wb. Therefore, r < b. In other words, the remainder should never be larger than the divisor. So r < b.