Math 2020 Spring 2011 Solving Discrete Problems P. Achar Summary: Math 2020 Spring 2011 Solving Discrete Problems P. Achar Notes on Chapter 6: Division with Remainder Theorem (Division with Remainder). Let n N and m Z. There exist unique integers q, r Z such that m = nq + r and 0 r < n. () Proof. The proof is two parts: (1) existence of q and r such that () is true, and (2) uniqueness of q and r. We'll start with existence. Consider the set S = {m + an : a Z} Z0. Step 1a. S = . Proof of Step 1a. We'll consider two cases: m 0 and m < 0. If m 0, let's take a = 1. Then, since n > 0, we have m + an = m + n 0, so m + an is an element of S. This shows that S = . If m < 0, then let's take a = -m. In this case, we have m + an = m - mn = m(1 - n). Since n 1, we know that 1 - n 0. We also have m < 0 by assumption, so it follows that m(1 - n) 0. This shows that m + an S, so again, S = . Note that for all c S, we have 0 c. Therefore, applying Proposition 2.33 (a variant of the Well- Ordering Principle) to the set S and the integer 0, we learn that S has a smallest element. Let r = the smallest element of S. Since r S, there is some a Z such that r = m + an. Let q = -a. From these definitions, it follows that m = nq + r. To complete the existence part of the proof, we must Collections: Mathematics