Congruences and Modular Arithmetic

Transcription

1 Congruenes and Modular Aritheti a is ongruent to b od n eans that n a b. Notation: a = b (od n). Congruene od n is an equivalene relation. Hene, ongruenes have any of the sae properties as ordinary equations. Congruenes provide a onvenient shorthand for divisibility relations. Definiton. Let a, b, and be integers. a = b (od ) (read a equals b od or a is ongruent to b od ) if any of the following equivalent onditions hold: (a) a b. (b) b a. () a = b+j (or a b = j) for soe j Z. (d) b = a+k (or b a = k) for soe k Z. is alled the odulus of the ongruene. I will alost always work with positive oduli. Note that a = 0 (od ) if and only if a. Thus, odular aritheti gives you another way of dealing with divisibility relations. Another way of saying this is: Mod any ultiple of is 0. Reark. Many people prefer to write a = b (od ). Sine equality od is an equivalene relation, sine = is a little less writing than =, and sine there isn t uh risk of onfusion, I ll write =. Exaple. (a) Redue 101 (od 3) to a nuber in the range {0,1,...100}. (b) Redue 101 (od 3) to a nuber in the range {0,1,...100}. (a) 101 = 2 (od 3), beause = 99. (b) 101 = 1 (od 3), beause = 102. Proposition. Congruene od is an equivalene relation: (a) (Reflexivity) a = a (od ) for all a. (b) (Syetry) If a = b (od ), then b = a(od ). () (Transitivity) If a = b (od ) and b = (od ), then a = (od ). Proof. I ll prove transitivity by way of exaple. Suppose a = b (od ) and b = (od ). Then there are integers j and k suh that a b = j, b = k. Add the two equations: a = (j +k). 1

2 This iplies that a = (od ). Theore. Suppose a = b (od ) and = d (od ). Then: (a) a+ = b+d (od ). (b) a = bd (od ). Note that you an use the seond property and indution to show that if a = b (od ), then a n = b n (od ) for all n 1. Proof. Suppose a = b (od ) and = d (od ). (a) a b and d, so by properties of divisibility, (a b)+( d) = (a+) (b+d). This iplies that a+ = b+d (od ). (b) a b and d iply that there are integers j and k suh that a = b+j and = d+k. Multiplying these two equations, I obtain a = (b+j)(d+k) a = bd+(dj +bk +jk) a bd = (dj +bk +jk) Hene, a bd, so a = bd (od ). Corollary. Suppose a = b (od ). Then: (a) a± = b± (od ). (b) a = b (od ). Proof. Apply the theore to the equations a = b (od ) and = (od ). Assue that the odulus is a positive integer. By the Division Algorith, every integer n an be written as n = q+r where 0 r <. Reduing this equation od, I have q = 0 (od ), so n = r (od ). Sine 0 r <, I have r {0,1,... 1}. In other words, od every integer an be redued to a nuber in {0,1,... 1}. This set is alled the standard residue syste od, and answers to odular aritheti probles will usually be siplified to a nuber in this range. Exaple. (a) What are the equivalene lasses under the relation of ongruene od 3? 2

3 (b) Construt an addition table for addition od 3. (a) Consider ongruene od 3. There are 3 ongruene lasses: {..., 3,0,3,6,...}, {... 4, 1,2,5,...}, {... 5, 2,1,4,...}. Eah integer belongs to exatly one of these lasses. Two integers in a given lass are ongruent od 3. (If you know soe group theory, you probably reognize this as onstruting Z 3 fro Z.) (b) When you re doing things od 3, it is if there were only 3 nubers. I ll grab one nuber fro eah of the lasses to represent the lasses; for sipliity, I ll use 0, 2, and 1. Here is an addition table for the lasses in ters of these representatives: Here s an exaple: 2+1 = 0, beause 2+1 = 3 as integers, and 3 s ongruene lass is represented by 0. This is the table for addition od 3. I ould have hosen different representatives for the lasses say 3, 4, and 4, but I would have gotten an equivalent table. The siplest thing is to use the nubers {0,1,2} fro the standard residue syste od 3. Exaple. (Reduing an expression od n) Redue (od 7) to an eleent in the standard residue syste {0,1,...,6}. 100 = 2 (od 7), so = 2 5 = 32 = 4 (od 7). Exaple. Siplify (od 1000) to a nuber in the range {0,1,...999}. Rather than deal with large positive nubers, I ll onvert the to sall negative nubers: 994 = 6 (od 1000), 996 = 4 (od 1000), 997 = 3 (od 1000), 998 = 2 (od 1000). So = ( 6)( 4)( 3)( 2) = 144 (od 1000). Exaple. (A odular binoial theore) Prove that if p is prie, then By the Binoial Theore, A typial oeffiient reain are x p and y p. ( ) p = i (x+y) p = x p +y p (od p). (x+y) p = p i=0 ( ) p x i y p i. i p! is divisible by p for i 0,p. So going od p, the only ters that i!(p i)! 3

4 For exaple (x+y) 2 = x 2 +y 2 (od 2) and (x+y) 3 = x 3 +y 3 (od 3). The result is not true if the odulus is not prie. For exaple, (1+1) 4 = 0 (od 4), but = 2 (od 4). Exaple. Prove that if x Z, then 4x 2 +x+3 is not divisible by 5. The phrase not divisible by 5 leads one to think of doing things od 5. Every integer is equal to one of 0, 1, 2, 3, or 4 od 5. Make a table: x (od 5) x 2 +x+3 (od 5) Fro the table, 4x 2 +x+3 0 (od 5) for all x Z, so 5 4x 2 +x+3 for all x Z. Exaple. Give a ounterexaple to show that a = b (od n) does not iply that x a = x b (od n), for a,b,n,x Z. For instane, 7 = 4 (od 3), but (od 3) (sine (od 3)). Exaple. Solve the ongruene 6x+1 = 2(x+2) (od 7). The odular aritheti properties allow e to solve this equation the way I would solve a linear equation, up to a point. I ultiply out the left side, then get the x s on one side: 6x+1 = 2(x+2) (od 7) 6x+1 = 2x+4 (od 7) 4x = 3 (od 7) If this were an equation overthe real nubers, you ould divide both sides by 4 equivalently, ultiply both sides by 1 4. What would 1 4 ean od 7? This is the ultipliative inverseof 4, whih we write as 4 1 (in odular aritheti you don t use fration notation). This eans: What nuber ultiplied by 4 gives 1 od 7? Sine there are only 7 nubers od 7, I an do this by trial and error, ultiplying 4 by 0, 1,...until I get 1. I find that 2 4 = 8 = 1 (od 7). So for this odular equation, I ultiply both sides by 2: 2 4x = 2 3 (od 7) 8x = 6 (od 7) x = 6 (od 7) 4

5 You an see that finding ultipliative inverses od n an be useful in solving ongruenes. Soeties they an be found by ore refined trial and error than siply trying all the nubers od n. Sine ultiples of n equal 0 od n, I an soeties use this to find inverses od n. 1 = 1+n = 1+2n = 1+3n = (od n). Exaple. (a) Use trial and error to find 5 1 (od 7). (b) Use trial and error to find 45 1 (od 89). () Prove that 25 does not have a ultipliative inverse od 30. (a) I take ultiples of 7 and add 1, stopping when I get a nuber whih is divisible by 5: 7+1 = 8, but = 15. and Sine 3 5 = 15, I have 5 1 = 3 (od 7). (b) I take ultiples of 89 and add 1, stopping when I get a nuber whih is divisible by 45: 89+1 = 90, and Sine 2 45 = 90, I have 45 1 = 2 (od 90). () Suppose that 25x = 1 (od 30) (so x = 25 1 (od 30)). Then 25x = 1 (od 30) 6 25x = 6 1 (od 30) 150x = 6 (od 30) 0 = 6 (od 30) This ontradition shows that there is no suh x, so 25 does not have a ultipliative inverse od 30. The previous ethod still has its liitations, as you an see by trying to use it to find 47 1 (od 61). And as you saw, soe eleents don t have ultipliative inverses od n. The following theore says whih eleents have ultipliative inverses, and how to find the if they exist. Theore. has a ultipliative inverse od n if and only if (,n) = 1. Proof. Suppose has a ultipliative inverse od n. This eans that a = 1 (od n) for soe a. Then Hene, (,n) = 1. Conversely, if (,n) = 1, then a+bn = 1 for soe b Z. a+bn = 1 for soe a,b Z. 5

6 Reduing od n, I get a = 1 (od n), whih eans that has a ultipliative inverse od n. n. As the proof shows, you an find a 1 (od n) by applying the Extended Eulidean algorith to a and Exaple. (Finding eleents whih have ultipliative inverses) Whih eleents of {0,1,2,...,11} have ultipliative inverses od 12? The nubers in {0,1,2,...,11} whih are relatively prie to 12 and 1, 5, 7, and 11. Hene, 1, 5, 7, and 11 have ultipliative inverses od 12. Exaple. Find 47 1 (od 61). Apply the Extended Eulidean Algorith to 61 and 47: Write the linear obination, then redue od 61: ( 10) = = 1 (od 61) Hene, 47 1 = 13 (od 61). Proposition. If a = b (od ), then Proof. Write Then (Notie that and right side, but it s relatively prie to Hene, a = b ( od ). a b = k, where k Z. (a b) = k. are integers, sine and.) Now. Therefore, it ust divide k: k = j for soe j Z. (a b) = j a b = j 6 divides the

7 Therefore, a = b ( od ). Notie that you divide the equality by, but you divide the odulus by. Exaple. (Solving a ongruene with anellation) Solve 12x = 30 (od 38). 12x = 30 (od 38) 6 2x = 6 5 (od 38) By the previous result, if I anel the fators of 6, I ust divide the odulus by (6,38) = 2. This akes the odulus 19: 6 2x = 6 5 (od 38) 2x = 5 (od 19) Now (2,19) = 1, so I an solve this ongruene by ultiplying by 2 1 (od 19). Noting that 2 10 = 10 = 1 (od 19), I see that I need to ultiply by 10: 10 2x = 10 5 (od 19) 20x = 50 (od 19) x = 12 (od 19) (If you didn t see that 2 1 = 10 (od 19) by trial, you d use the Extended Eulidean algorith as before.) The original ongruene was od 38, so I want all solutions in the range {0,1, I have one: x = 12. To get others, I add ultiples of 19 until I exeed 37. Thus, x = = 31 is the other solution. All together, the solutions are x = 12 (od 38) and x = 31 (od 38). Exaple. (A ongruene with no solutions) Show that the following ongruene has no solutions: 4x = 5 (od 14). Suppose that x is a solution. Multiply the equation by 7: 4x = 5 (od 14) 7 4x = 7 5 (od 14) 28x = 35 (od 14) 0 = 7 (od 14) This ontradition shows that the equation has no solutions. These exaples show that linear ongruenes ay have solutions or ay be unsolvable. We an understand better what is happening by relating the to linear Diophantine equations by Brue Ikenaga 7