The Division Algorithm for ™ And
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+ß we have <œ+,; ! .
To see that <,À if <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 ; Subtracting these equations and rearranging gives ,Ð; ;ww Ñ œ < <ß and therefore ,lÐ; ;ww Ñl œ l< Also, we know from Ð‡Ñ that , < Ÿ 0 ß and adding this inequallity to the inequality ! Ÿ 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,; (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,; There are various ways to state the division algorithm for ™ . In this version, we require that the divisor ,!ß so 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 ; +œ,; Then +œ,Ð;Ñ Using this equation: Case i) If Let ;œ ;ß<œ!Þwww Then +œ,Ð;Ñ< œ,;< where !Ÿ<, Case ii) If , Let ;œ ;ww " and <œ,<Þ Then + œ ,Ð ;ww Ñ < œ ,Ð ; w Ñ , < w , œ,Ð;ww "ÑÐ,<Ñ œ,;<ß where !, So, in both cases, we can find integers ;< and for which +œ,; 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"Ñ namely, with and The division algorithm in ™ (in the form stated above, requiring the divisor ,! ) with ,œ$ and +œ ( says that we can write (œ$;<ß where !Ÿ<$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!Þ Then +œ,;< and !Ÿ<" If ,"ß let ;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 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+Ñ is 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