The Division Algorithm for ™and Theorem (Division Algorithm for ) Suppose +, and are natural numbers and that ,Ÿ+Þ Then there is a natural number ; and a whole number < such that + œ ,; < and ! Ÿ < ,Þ Moreover, ;< and are unique. ÐWe usually call ;< the “quotient” and the “remainder” when +, is divided by .Ñ Proof Let E œ Ö= − À =, +×Þ There must be at least one such value of = Ðsee the “Archimedean Principle”, text, p. 105), so EÁgÞ By the Well-Ordering Principle (WOP), E contains a smallest element: call it 6. Since "ÂE, we know that 6" so ;œ6" is a natural number. Define <œ+,;Þ This makes +œ,;<ÞÐBut we still need to prove that !Ÿ<, Ñ. 6, +because 6 − E We know that Ð6 "Ñ, œ ;, Ÿ +because ; 6 and therefore ;  EÞ Since ;,Ÿ+ßwe have <œ+,; !. To see that <,Àif <œ+,; ,, we would have + ,,;œ,,Ð6"Ñœ6,, which is false. Therefore !Ÿ<, so the proof “there exist” ; and < with the desired properties is complete To prove ;<and are unique: Suppose +œ,;< where !Ÿ<,Ð‡Ñ and that + œ ,;ww < where ! Ÿ < w , Ї‡Ñ Subtracting these equations and rearranging gives ,Ð; ;ww Ñ œ < <ß and therefore ,lÐ; ;ww Ñl œ l< <lÞ Ð‡‡‡Ñ Also, we know from Ð‡Ñ that , < Ÿ0 ß and adding this inequallity to the inequality ! Ÿ <w , Ї‡Ñ gives ,<w <, so that l<w <l , Substituting this in Ї‡‡Ñ, we get ,lÐ; ;w Ñl ,Þ This means that l; ;ww l " and, since l; ; l is a integer, this implies that l; ; w l œ !ß that is, ;œ;Þw From this, and the equations +œ,;< +œ,;<ww we get that <œ<Þw Therefore ;ß< are unique. ñ (over Ä ) We can also state a division algorithm for ™. Theorem (Division Algorithm for ™) Suppose +, and are integers and that ,! Then there is an integer ;<+œ,;<!Ÿ<,Þ and an integer such that and Moreover, ;< and are unique. There are various ways to state the division algorithm for ™. In this version, we require that the divisor ,!ßso actually ,−, and that !Ÿ<,Þ Another version (you might try to prove it) allows , to be any nonzero integer and has !Ÿ<l,lÞ In all versions, the statement requires that the remainder <,/nonnegative: that fact is usually what's important when the Division Algorithm is used. Proof If +!, we get ; and < from the Division Algorithm in . If +œ!ßlet ;œ<œ! . If +!, then apply the Division Algorithm in for dividing + by ,. There are natural numbers ;<ww and for which +œ,;<ww where !Ÿ< w , Then +œ,Ð;Ñ<ww where , < w Ÿ!Þ Using this equation: Case i) If <w œ!À Let ;œ ;ß<œ!Þwww Then +œ,Ð;Ñ< œ,;< where !Ÿ<, Case ii) If , <w !À Let ;œ ;ww " and <œ,<Þ Then + œ ,Ð ;ww Ñ < œ ,Ð ; w Ñ , < w , œ,Ð;ww "ÑÐ,<Ñ œ,;<ß where !,<w œ<, (since , <w !Þ) So, in both cases, we can find integers ;< and for which +œ,;<!<,, with . The proof that ;< and are unique is left as an exercise (see proof of the previous theorem for ideas). ñ Example The division algorithm in : $( so we can write (œ$;<where !Ÿ<# Ð;œ#<œ"Ñnamely, with and The division algorithm in ™ (in the form stated above, requiring the divisor ,!) with ,œ$and +œ ( says that we can write (œ$;<ßwhere !Ÿ<$Þ Here the values that work are ;œ $and <œ#, and that's the only way to pick ; and < if you want !Ÿ<$Þ We proved the Division Algorithm for using WOP. Here's an alternate proof using PMI, doing “induction on ,.” Stated more formally, we want to prove: Ða, − ÑÐa+ − ÑÐb; − ÑÐb< − Ñ Ð+ œ ,; <Ñ • Ð! Ÿ < ,Ñ To prove this universal statement (working left to right), we pick any ,− Þ For this arbitrary but fixed natural number , we need to prove that Ða+ − ÑÐb; − ÑÐb< − Ñ Ð+ œ ,; <Ñ • Ð! Ÿ < ,Ñ This is a statement of the form Ða+− RÑTÐ+Ñ, where TÐ+Ñ is the statement Ðb; − ÑÐb< − Ñ Ð, œ +; <Ñ • Ð! Ÿ < +Ñ We use induction on the natural number +Þ This looks just a little odd, but +ß=9à is a natural number induction is OK if it makes you more comfortable, change “+8Þ ” everywhere to “ ” Proof Base case: Suppose +œ". If ,œ", let ;œ" and <œ!ÞThen +œ,;< and !Ÿ<" If ,"ßlet ;œ! and <œ"Þ Then +œ,;< and !Ÿ<"Þ So TÐ"Ñis true. Induction step: Suppose TÐ+Ñ is true for some particular value of +. Thus, we are assuming (for this value of +) that there are natural numbers ;ww and < for which +œ,;< ww and !Ÿ< w ,Þ We need to prove that TÐ+"Ñ is true, that is, that there exists natural numbers ; and < for which +"œ,;< , where !Ÿ<,Þ Case i: If <wwww œ,"À +œ,;< œ,;,", so +"œ,;ww , œ,Ð; "Ñ! Let ;œ;w " and <œ! Then +"œ,;<, where !Ÿ<, Case ii: If <,"Àwww +œ,;< so +"œ+,Ð<"Ñw Let ;œ;ww and <œ< " Then +"œ,;<ß where !Ÿ<,Þ In both cases, we can find the necessary ;<ÞTÐ5"Ñ and So is true. By PMI, Ða+ − Ñ T Ð+Ñis true (for the particular , we chose). Since the argument works no matter which ,− we choose, we conclude that Ða, − ÑÐa+ − ÑÐb; − ÑÐb< − Ñ Ð+ œ ,; <Ñ • Ð! Ÿ < ,Ñ ñ.