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1 Module 2 Congruence Arithmetic pages 9 5 Here are some excellent websites that can help you on this topic: Each has a different view. Please check them out. Note, too, that the Cut-the-Knot site is huge and has lots and lots of games, exercises, and explanations. It s a wonderful resource! I ll begin with some background information that the authors assume you have. Now to Congruence Arithmetic. This is often called modular arithmetic and it is a Big Picture kind of mathematics. You will be teaching problems like = and (2) = and classroom to your young students. 5 = in your We ll be doing problems like these, too, but with a special point of view on the work. Merriam-Webster online dictionary defines both congruence and modulus Congruence : a statement that two numbers or geometric figures are congruent Modulus : the number (as a positive integer) or other mathematical entity (as a polynomial) in a congruence that divides the difference of the two congruent members without leaving a remainder compare RESIDUE b (2) : the number of different numbers used in a system of modular arithmetic 1

2 We ll modify the congruence definition a bit to be about numbers. And we ll use modulus to mean divisor. Here s the new point of view: Two numbers are congruent mod b if they have the same remainder when divided by b, the modulus. If a and c are our numbers, and b is our divisor AND if they have the same remainder when divided by b, we denote this by writing: a c note this does NOT say they are equal! b Examples: 18 they re both multiples of three and the shared remainder is 0 The way to say this in words is 18 is congruent mod to when divided by, each has a remainder of 2 In words: 11 is congruent mod to 17 The way to say it algebraically is 18 = (5) + 0 and = (2) + 0 we actually show the zeros on this! 11 = (2) + 2 and 17 = (5) + 2 2

3 Now for some conventions: You know that when you see: There are NO understood moduli (the plural of modulus). x that s square root, the 2 is understood x the coefficient is 1 and the exponent is 1 and they are understood to be there if you need them Not so in congruence arithmetic. Each modulus will be explicitly displayed on the congruence symbol. Examples: 21 5 the modulus is 5; the remainder is the modulus is 10; the remainder is Now, just naturally there are some HUGE patterns to be seen is congruence arithmetic Let s start with mod I m first going to use the whole numbers (everybody chops modulus off to mod ) ({0, 1, 2,, }) Let s look at a non-negative number line and put on some information: 0 = (0) + 0 = (1) + 0 = (2) = () = (0) + 1 = (1) = (2) = () = (0) = (1) = (2) = () + 2 Now take a minute and look down the columns left to right what patterns do you see? I hope you see that the remainders follow a pattern AND that each whole number multiplier gets used times and it s then discarded!

4 Let s use a wrapping convention to show this on a number circle Now, it turns out that congruence arithmetic is really not based on the whole numbers it is based on the integers. So let s start putting the negative numbers into the pattern, too.

5 Negative numbers are a bit harder to see in the algebraic way than the positives that s why I waited a bit to show you them. 1 = ( 1) + 2 = ( 2) = ( 1) = ( 2) + 1 = ( 1) + 0 = ( 2) + 0 The problem is that the negatives can be written a couple of ways, but if you stick to the pattern you ll stay straight. Once we ve got all the numbers organized, we put them in special sets called congruence classes (also called equivalence classes, residue classes, or remainder classes!) These are sets and have special, special notation for them. [ 0 ] = { x where x is an integer} [ 1 ] = { x 1 where x is an integer} [ 2 ] = {x 2 where x is an integer} Notice that we ll use [ 2 ] or just 2, if the context is clear, to stand for any number that has a remainder of 2 when divided by. 5

6 Now let s shift to mod 7. The congruence classes start with [ 0 ] and go through [ ]. And now [ 0 ] = [ 1 ] = [ 2 ] = { 7x 2 such that x is an integer} = { 12, 5, 2, 9, 1 } Use the wrapping convention to see this

7 And let s do it with mod So the equivalence classes are: [ 0 ] [ 1 ] [ 2 ] [ ] 7

8 Now an interesting real life example of modular arithmetic is Clock Arithmetic Let s take a US clock with it s 12 equivalence classes (note a military or European clock has 2 equivalence classes) Now, let s do some arithmetic. Note that the context of Clock Arithmetic is clear so I dropped the square brackets. When in doubt, use them! = = = Note that you can count it out on the face of the clock or you can do base 10 arithmetic and divide by 12 to get the remainder to get the answer. How many equivalence classes are there? What are they? 8

9 Popper 0, Question 1 In clock arithmetic = A. 1 B. C. 9 Now, in the book, we re on about page 1. Where the authors are talking about the definition of being equivalent mod b Popper 0, Question 2 Which of the following are the equivalence classes mod? A. [ 1 ] [ 2 ] [ ] [ ] [ 5 ] [ ] B. [ 0 ] [ 1 ] [ 2 ] [ ] [ ] [ 5 ] Arithmetic modulo 2 Now let s take a moment to look at mod 2. I contend that you know lots and lots about mod 2 already! You just need to translate it to this context. The equivalence classes are [ 0 ] and [ 1 ]. [ 0 ] = { 2x where x is an integer} = {, 2, 0, 2,,, } [ 1 ] = { 2x 1where x in an integer} = { 5,, 1, 1,, 5, 7, } What do we usually call these? 9

10 [ 0 ] are the evens and [ 1 ] are the odds! We ll be back to this at the end of the video. Now let s look in the book on page 1. There s a theorem to illustrate and understand. Theorem: If you have two numbers in the same equivalence class for a given modulus, then the difference of the two numbers is divisible by the modulus. [First, remember that divisible means divides evenly with a zero remainder!] Next let s illustrate this theorem. Let s pick mod 7 first. 19 and 5 are in the same equivalence class: 19 = 7(2) now let s look at 19 5 = 1 = 7(2) + 0. YES! Now let s say it this way: this says the difference is a multiple of 7 Every number in [ 0 ] is a multiple of 7. Here s a mod 5 example. 28 and 1 are in the same equivalence class: [ ] mod modulus now let s look at 28 1 = 15 = 5() + 0, yes the difference is a multiple of the Which we ll say this way: And here s a last example: mod 17 2 they re both in [ 5 ] mod Subtract 2 from both sides: = 0 This says the difference is a multiple of the difference is and this is true! 10

11 Also on page 1 is a second very nice theorem: If you have an equivalence for a given modulus, you may multiply both sides of the equivalence by the same natural number* and get an equivalence. Illustration: 10 (they are both in [ 1 ]) * they say number but they mean natural number! Multiply both sides by 2 8 on the left and 20 on the right are they equivalent? Let s check 8 20 = 12 which is a multiple of. Yes! 8 20 They are each in [ 2 ]. Now note what the theorem did NOT say. It it NOT say you d stay in [ 1 ]. It said you d get an equivalence and you did. Illustration: 11 7 multiply both sides by and 12 are they equivalent? Popper 0, Question 12 = 21 which is a multiple of 7! They re both in [ 0 ]! Given 15. Multiply both sides by 2 to get 0. In which equivalence class are 0 and mod? A. 0 B. 1 C. 2 D. 11

12 Now the third theorem on page 1 is this one: If you have two equivalence relationships with the same modulus, you may multiply the left sides times each other and the right hand sides times each other and the result will be an equivalence. In variables this says: a and q q b c q d ac bd Illustration: is 5 91? Both are in [ 1 ] Now for the Big Picture look. (2) and 1(7) are really an element of [ ] times and element of [ 2]. So we found that [ ] ([ 2 ]) 5 [ 1 ] For example 8(12) = 9 yep just like in clock arithmetic we ll say (2) = 1 mod 5 When you re teaching kids, you pull individual numbers and get individual answers. We re saying use any number in [ ] and any number in [ 2 ] and you ll get a number in [ 1 ]. Here s another illustration: = 8 yep! So [ 1 ] ([ ]) [ ] or 1() = mod 12

13 Here s an example of this that leads to a nice theorem (not in the book) Let s look at the following problems = = 1 + = 9 See the pattern? Even plus Odd = Odd [ 0 ] + [ 1 ] 2 [ 1 ] Here it is in words: An even integer plus and odd integer is an odd integer. Let s prove it: An even integer looks like 2x where x is an integer. [ 0 ] = An odd integer is 2q + 1 where q is an integer. [ 1 ] = Note that I need two different variables! If I used x twice, I d be proving it only for ADJACENT evens and odds each number would have the same integer multiplier x. I want to prove it for evens and odds that are anywhere on the number line. Let s add them: 2x +2q + 1 = 2(x +q) +1 associative law and distributive law Now the result is odd; it s 2 times an integer plus 1. Definition of [ 1 ] Done! 1

14 Popper 0, Question How can I tell if two numbers are equivalent mod 11? A. divide both numbers by 11 and check the remainders B. rewrite each number as a multiple of 11 plus a number and check the numbers. C. subtract the numbers and see if the difference is divisible by 11. D. all of the above E. ask the numbers politely Now back to the Big Picture. Let s check out this statement: [ 1 ] + [ ] [ 0 ] First let s look at some examples: 5 + = = = 12 Ok, it seems to work. Note that I can use ANY number from [ 1 ] and ANY number from [ ] and the equivalence holds. Let s prove it. [ 1 ] = some x + 1 [ ] = some p + 1

15 Again, don t reuse x in the algebraic statement of the numbers use a different variable. x p + = (x + p) + ( 1 + ) associative law (x + p) + distributive law (x + p + 1) is a multiple of so it is in [ 0 ] Done! Pretty amazing! Popper 0, Question 5 [ 1 ] + [ 1] 2 [ 0 ] A. True B. False END FIRST VIDEO 15

16 Now we ll move to page 2 amazing! We ve got one page done and now we re on the second page. Well there s a nice theorem about exponents or powers. It says If you have an equivalence mod b, and you use the same power or exponent on each side you get an equivalence. In symbols: a b now let s use d as our exponent; note that d is a natural number c a d c b d Illustrations: 7 These are both in [ ]. So, let s square both sides! Yes, 9 9 = 0 a multiple of! They are both in [ 1 ]. Let s cube both sides of Let s check: 8 = (2) with a remainder of 2 The powered equivalence is from [ 2 ] Note that in our new arithmetic this says: 2 2 1

17 Popper 05, Question 1 Which of the following illustrates: A. B. C D. none of them On to page. An English mathematician: Dr. Cayley really organized congruence arithmetic. It was discovered or invented by Gauss (Swiss) in the early 1800 s and by the end of the century, with Cayley s contributions it was mainstream for math. Cayley took addition and subtraction and organized them into equivalence class tables. This allowed much deeper studies by modulus. Let s look at this example: When little kids are learning to add, they eventually learn that = 12. In our class we d put 5 and 7 in their proper equivalence classes and 12, too. Here s the same problem done in mod, mod, and finally mod 11. Each of these is = or or Note that the first 2 examples the answer is a multiple of AND a multiple of because 12 is a multiple of both. It is NOT a multiple of 11, though. 17

18 Cayley s brilliant insight was to see that by using congruence classes (residue classes, equivalence classes remember all the names!) he could make tables of the addition and multiplication outcomes. He did this for addition and multiplication because they are well behaved operations. Division, however, is a real problem for congruence arithmetic. Let s do a review, check out the problems with division, and then we ll move on to Dr. Cayley s tables. To date, we know that Adding the same natural number to both sides of an equivalence preserves equivalence. Multiplying the same natural number to both sides of an equivalence preserves equivalence. Powering both sides of an equivalence with the same natural number power preserves equivalence. This should sound a whole lot like working with equations! Adding and multiplying the same number to both sides of an equation maintains equality. Powering each side maintains equality. Unfortunately, dividing both sides of an equivalence by the same natural number only works when certain conditions are met! We ll get some examples of that after the popper Popper 05, Question 2 Given 5 8 Which of the following are true? A B. 2000() 5 8(2)(10 ) C. 5 5 (2 ) 5 D. all of them are true E. none of them are true x x x where 10 18

19 Let s look at division, page. What we want to say is dividing both sides by the same natural number will give you an equivalence. But look at this: They are both in [ 0 ] Divide both sides by 2 9 This is NOT true! The problem here is that (, 2) = 2. If the modulus and the divisor are relatively prime, THEN division works but if they re not, then it works sometimes and not others. In your homework you have to figure out which is which on division with a divisor and modulus that are not relatively prime. Let me give you a few more examples of that: 8 1 divide both sides by NOT true And (2, 8) = 2 Here s one that IS true even though (2, ) = 2 It s in mod and uses the same numbers as the preceding example. 1 divide both sides by 2 7 TRUE Weird but true. What s different between these examples? What important thing makes example work even though (modulus, divisor) is not 1? That s the homework problem. Now let s look at some examples where (modulus, divisor) = 1. These ALWAYS work! I divided both sides by (note that (, 7) = 1) TRUE Here s another (5, ) = 1. 5 is my modulus and is my divisor TRUE So division is quirky and requires a special extra condition to work. 19

20 On page, we start working with Cayley tables. Now in the tables, it is understood that we are using congruence classes for the numbers. It would be too messy and crowded to put square brackets on each number. So we eliminate them, but each number stands for the whole equivalence class. Here is a mod addition table. Let s fill it in together Let s look at adding in an elementary school and then adding in congruence arithmetic = = = Now let s check it on the table with the rows and columns. 20

21 Now let s do algebra with it: Write it for kids: x 1 x 2 And now from elementary school to our class write this in congruence classes mod : 9 + x = 7 x = 2 Now let s look at multiplication:

22 And again: 8 (11) = 88 Congruence classes mod : Now multiplication has some challenges that addition does not have. Look at the rows across from 2 and not every congruent class shows! This means that if you re asked to solve: x 2 5 there s no solution check the row! While x 0 has 2 solutions: [ 0] and [ ] check the row! This is unexpected but really is true: Let s look at some examples with numbers: 2x = 11 x = 5.5 Oops! Not a natural number, no solution in our system x = x = 9 [ ] x = 0 x = 12 [ 0 ] You get two types of solutions, those divisible by and those with a remainder of. Let s review some from the tables: + 5 mod 5 () mod 22

23 Popper 5, Question In mod, x = 2 has A. 2 solutions B. 1 solution C. no solutions On page is a nice and not obvious theorem: Every even power of an odd number is congruent mod 8 to 1. What is this saying? Let s illustrate it: is an odd number, is an even number This is a literal translation of the theorem and not quite an illustration. 8 1 Let s show that it s true (10) 1 yes it s true Another illustration: (21) 1 true again. Here s 2 more illustrations: () (00) 1 2

24 Popper 05, Question A. True B. False Popper 05, Question A. True B. False END VIDEO 2 2

25 Now, let s get a bit into what congruence arithmetic used to be used for by mathematicians and why it s still taught to them. To find out some facts about a big number, you use substitution in a clever way with congruences. I ll start with a small problem: Is 2 congruent to 2 mod? Well 2 is a pretty small number so it would be easy to divide and find out. Here s how we ll do it with congruence arithmetic. First note that Now note that 1 and REPLACE the s in the above equation with 1 s NOTE: I switched from equals to equivalent when I did the substitution! But, yeah, there it is. Historically congruence arithmetic was used for finding out if a number was prime or not. Currently we teach it to math majors to improve their thinking skills and hone their mathematical intuition about numbers. Let s look at example 1 on page. It shows a congruence arithmetic way to find out if 999,999 is divisible by 7 (and, thus, not prime). So we want to show that 999, If it is, then it s not prime. Now the cleverness of the proof is skill based and it s a skill that can be learned. Start with 999,999 and note that it is ,999 = 10 1 now 10 7 and 10 and all by earlier theorems ,999 = note: no more = 25

26 Now 2 ( ) by exponent properties from algebra 2 ( ) = ,999 = = notice how I m using smaller numbers! Now taking this last result and substituting it ,999 = = Collapsing the interim work we have Now 8 is equivalent to 1 mod 7 so the right hand side is 1 1 =0 999, WOW! Perfect! It s not prime. Catch your breath. There s more. 2

27 Let s refer back to Chapter, page and look at Fermat s formula for primes. Fn n This worked up to n = 5 which is composite. Let s look at how they figured that out. 2 F5 2 1=, 29, 97,297 = 1 (, 700, 17) That s billion and some So now let s walk through the proof that which is to say that 2 2 1is divisible by 1 We need to do some preliminary work: Note that And let s factor 0 to primes 0 = (10) = 5 (128) = So now I ve got 0 = 7 5(2 ) 1 1** 7 5(2 ) Additionally let s note that 7() = 28 and that 25 = 5 and that and, moreover, 2 1 ( ) Back to the equation ** above, but leaving off 0 and working from the middle: 7 5(2 )

28 (5(2 )) ( 1) (2) = (2 ) 1 1 7() = (2 ) 1 1 Further note that Substituting on the left: (2 ) 1 1 Multiply both sides by 1 and combined the powers of Add 1 to both sides Wow! Again. Yes it s composite. 28

29 Popper 0, Question 1 Is this true: A. Yes B. No I ve got an example of proving numbers composite by congruences to show you. This are not in the book. I think it will be helpful for you to see a few more of them. Is 15, 2 divisible by? Well: = 18 which is divisible by Yes. Show a congruence proof of this. We want to show that 15, 2 0. One thing to note is the 15, is 15, 25 which is easy to factor. It s 5. So let s start with 15, 2 = 15, 25 = 5 1. Note that ( 1) 1 So subtract 1 from both sides: Yes. This one was MUCH shorter wasn t it? 29

30 Let me go a step further and discuss how I built this problem to show you. I started with Note that IS divisible by so then I hid the and subtracted the 1 and asked: Is 15, 2 divisible by? Popper 0, Question 2 Given that is a true statement, is a true statement? A. yes because 7 B. no because of the 1 0

31 Now back to the book, page Fermat s Theorem one of many he proved If a is NOT divisible by P, a prime number, then P 1 a p 1. As always, let s illustrate this theorem: a = 11 P = This is true 11 is an odd number Another illustration: a = 1 P = Is 19 equivalent to 1 mod? 19 = (5) + 1 see that + 1 That s the name of the congruence class. TRUE. Another illustration: a = 12 P = Well, now () 1 5 Yes. That works 1

32 Popper 0, Question Is 12 1a true statement? 7 A. No, it s equivalent to 1 B. Yes, it s an illustration of Fermat s Theorem And in another seemingly random subject change, on page 7 we shift our attention to Pascal s Triangle. A very famous, well-studied object. It was known in India and China long before Pascal s great-great-great-grandparents even met but he got the credit for it in the books. Let s look at the pattern for building the triangle:

33 And let s look at the usual tie-in to polynomials ( x y) 0 ( x y) 1 ( x y) x 2xy y ( x y) x x y xy y 2 2 Row 5 coefficients of (x + y) Factette #1 If you want to find a specific coefficient use the combinations formula with carefully chosen n and k. Suppose you want the coefficient for the 7 th term in (x + y) 1 Use 1 for n and (not 7!) for k n C k n! k!( n k)! C 1 1! 1! ! 00!(1 )!!(8!) 5 2(8!) Popper 05, Question 5C 2= A. 10 C. 20 E. B. 15 D. 5

34 Here s another mathy type question that some folks thought up and then answered: Is there a pair of adjacent coefficients with the first one being 2/ of the second one? Well, there s an infinite number of these pairs. In fact, you can pick any ratio you want that s a proper fraction and find pairs that work. Let s solve the problem for 2/ The first coefficient will be k and the next one will be k + 1, and we want k 2 ( k 1) First let s look at the triangle to see if we can spot any BINGO! = 2/ of Ok. So here s a proof for a = any fraction. We start by evaluating the combination formula for k, k + 1 and a the multiplier. And since they ll be in the same row we know they re both in row n. k a( k 1) n! a( n!) k!( n k)! ( k 1)!)( n ( k 1))! distribute the negative sign n! a( n!) k!( n k)! ( k 1)!)( n k 1)!

35 First let s divide both sides by n! n! a( n!) k!( n k)! ( k 1)!)( n k 1)! this cancels it Note that k! = k(k 1)! on the left denominator. Now multiply both sides by (k 1)! 1 a kk ( 1)!( n k)! ( k 1)!)( n k 1)! 1 a kn ( k)! ( n k 1)! Now note that ( n k 1)! = (n k +1)(n k)! Replace it on the right and multiply both sides by (n k)! 1 a kn ( k)! ( n k 1)( n k)! Now we re down to: 1 a k ( n k 1) Cross multiply and you get (n k + 1) = ak Solve for n to get n = ka +k 1 This will tell you which row they re in. You need to be sure that ka is a natural number! 5

36 Now all this slinging things factorial around may be new to you. Let me show you an example that has it all in it but is simpler: Simplify: ( k 1)! ( k 1)! ( k 1)! ( k 1) k( k 1)! ( k 1)! ( k 1)! you get k 2 +k Let s look at it with numbers: k = ! ! 257(25) 255! 255! Now on to some Homework Hints: For Problem 7 As an intro, let s look at primes mod for primes > Primes mod for primes > You rewrite [ 0 ] as [ ] and [ 1 ] as [ 7 ] and [ 2 ] as [ 8 ] and [ ] as [ 9 ] and make a list that looks like [ ] [ 5 ] [ ] [ 7 ] [ 8 ] [ 9 ] smallest number in the equivalence class other numbers in the equivalence class It looks like the primes are clustered in [ 5 ] and [ 7 ]. Why would that happen?

37 Well [ ] = all n + numbers and these are EVEN primes larger than 2 all all ODD none can be in here Ditto [ ] these are all multiples of and EVEN ditto [ 8 ]. But what about [ 9 ] those are odds but they re composite odds: [ 9 ] = n + 9 = (2n + ) so they re all composites This rules out of the equivalence classes! There s only 2 left where primes can show up! Problem 8 is an induction proof! Popper 05, Question 5 True or False: hint check the subscripts! This is a vicious and cruel one! 11 A. True B. False 7

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