Cogruece Modulo - 03 The [ ] equivalece classes refer to the Differece of quares relatio ab if a -b o defied as Theorem 3 - Phi is Periodic, a, [ a ] [ a] The period is Let ad a We must show ( a ) a ice, ( a ) a a a a a ( a ) we have that ( a ) a ad hece ( a ) a Theorem 5 - Phi is Eve, a, [ a] [ a] Needs proof Theorem 7 - Half Period of Phi (whe is double eve), a, 4 [ a ] [ a] If is divisible by 4, the the period is / Needs proof Our Two Mai Fuctios - Commets o Theorems 3, 5, ad 7 The fully exteded Diagram for ca be thought of as the graph of the fuctio that takes each iteger to its equivalece class uder If we represet each class by its smallest oegative member ad use those represetatios to label the y axis, the it s this graph that s thought of as Periodic ad Eve We ca write this fuctio as from :, ( a ) [ a ] oto where is the set of equivalece classes of (( Artist s ote: trictly speakig, we are graphig the fuctio :, ( a ) the smallest o egative iteger that relates to a
We artistically choose to maitai the ambiguity sice all alog we have implicitly chose to represet each class by its smallest o-egative member The other artistic choice here is the to use )) The set of equivalece classes is the partitio of iduced by The Discrete Math exercise that started this ivestigatio was listig the elemets of the partitio for 4 As it tured out i the exercise, 4 {[0],[1]} with [0] {eves} ad [1] {odds} We ca give a separate proof of this (*todo) The first questio i our ivestigatio was How may equivalece classes are there for each? This is the same questio as What s the size of for each? We created the fuctio that gives that value ad called it f( ), but let s reame it ow (( Artist s ote: u is the Greek letter for which stads for umber )) Defie : ( ) umber of equivalece classes for size of the set From Theorems 3 ad 5 about the symmetry ad periodicity of ( 1) /, odd ( ) +1, eve Theorem 7 the yields a further refiemet of Theorem 8 - ymmetry of Phi ( 1) /, odd ( ) +1, eve but 4 4 +1, 4, we deduce that Argue usig periodicity ad symmetry from Theorems 3, 5, ad 7 A challegig proof? Let s defie a ew fuctio who value is oe less tha the maximum value of i Theorem 8 Defiitio ( 1) /, odd ( ), eve but 4 4, 4 Theorem 8 says that ( ) ( ) 1 Defiitio
We say that the diagram for is complete, or is complete, if ( ) ( ) 1 Whe is complete, has the maximum umber of equivalece classes possible by Theorem 8 The equivalece classes for are represeted by the cosecutive umbers 0,1,, ( ) {[0], [1],, [ ( )]} The diagram for will ot have ay dropouts The graph of will have its full saw-tooth shape that was first evisioed followig the observatio that a b ( a b)( a b) Theorem 9 - Prime If is prime, the is complete Has t bee proved yet Corollary If is prime ad, the ( ) ( 1) / Follows directly from Theorem 9 UTheorem 10 - If 4, the is complete Dis Do the Diagrams bear this out? Nope, 16 is ot complete sice [0] [4] ad 4 4 or 4 (16) What s complete? Are there ay more sufficiet coditios for beig complete? What do the Diagrams idicate? Dropouts If is ot complete, the it has dropouts Dropouts are where the saw-tooth shape is violated All that remais is to see if the dropouts have predictable patters Theorem 6 says whe there will be dropouts to [0] Theorem 6 - Characterizatio of [0] - oops, ot true The zero class [0] for is the set of all iteger multiples of the product of the prime factors of Use the lemma Lemma, a, a if ad oly if a is divisible by every prime factor of
Use the Prime Factorizatio Theorem (aka The Fudametal Theorem of Arithmetic) Corollary [0] {the multiples of } if ad oly if has o multiple prime factors Follows sice would be its ow product of the prime factors of Corollary There are [0] dropouts if ad oly if has multiple prime factors It follows from the symmetry that there are [0] dropouts if ad oly if [0] icludes more tha just the multiples of Corollary If has multiple prime factors, the is ot complete is ot complete sice there will be [0] dropouts Corollary to Lemma If has o multiple prime factors, the a if ad oly if a Follows sice a would be divisible by the product of every prime factor of which would be itself This corollary icludes the famous theorem a is eve if ad oly if a is eve as well as 3 a if ad oly if 3 a ad 6 a if ad oly it 6 a Theorem 6 - Characterizatio of [0] The zero class [0] for is the set of all multiples of the smallest umber whose square is divisible by We eed a ew fuctio for Theorem 6 spr stads for square-root prime-factor reductio spr:, m1 m m k 1 m1 m spr( ) p p p k whe p1 p p k m k spr( ) is the smallest umber whose square is divisible by spr( ) is the smallest umber a such that spr( ) spr( ) a if ad oly if is prime or simple-composite if ad oly if is multi-composite a if ad oly if a is divisible byspr( ) spr ( ) Theorem 6 ca ow be restated, spr( ) a
Theorem 6 - Characterizatio of [0] The zero class [0] for is the set of all multiples of spr( ) I other words, [0] {all multiples of spr( )} [0] will cotai more tha just the multiples of if ad oly if is multi-composite Therefore, Corollary to Theorem 6 has [0]-dropouts if ad oly if is multi-composite However, whe spr( ), the is multi-composite but oly because it s divisible by 4 The oly [0]-dropout is i this case is [ ] which wo t show o the half cycle chart Defiitio Here s the fuctio that uderlies the relatio s : s( a, b) a b? Is s Ijective? urjective? Ca you say aythig about the values i certai sectios of the table of values? The values are either eve multiples of eves or odd multiples of odds Here s some defiitios that will help us use Theorem 6 ad help us study the fuctio s Defiitio If is composite but has o multiple prime factors, the is simple composite Otherwise, if is composite with multiple prime factors, the is multi composite The prime factor reductio of a umber is the umber obtaied by takig the prime factorizatio of ad reducig the power of each prime factor to oe m1 m m pfr :, pfr( ) p1 p pk whe p k 1 p p k All itegers are oe of three thigs: prime, simple composite, or multi composite pfr( ) if ad oly if is prime or simple composite Theorem 6 ca be restated: The zero-class for is the set of all multiples of pfr( ) Defiitio A eve umber divisible by 4 is called double eve A eve umber ot divisible by 4 is called sigle eve If a umber is double eve, it s the eve multiple of a eve umber We do t defie double odd sice all odd umbers are already the product of two odds (Verify) All itegers belog i oe of three classes: odd{k 1}, sigle eve{4k }, or double-eve{4 k } We ca restate the defiitio of as Theorem 11 -
The umbers i the image of s are either odd or double eve There are o sigle eve umbers i the image of s hould be easy What s complete? Let s get back to the characterizig of the completeess of Theorem 1 - The Dropout Theorem The diagram for has a dropout if ad oly if there s a multiple of expressible as a product c d of two odds or two eves whose average is less tha ( ) I this case, the dropout is Needs a proof c d c d Theorem 1 gives a reliable procedure for computig ( ) ad ( ) for ay ------------------------------------------------------------------------------- The story First was the exercise The the observatio that a b ( a b)( a b) ad the first diagrams draw from the multiples of up The the saw tooth shape I the begiig, we did t thik about (we did t kow about) phi We did thik about our diagrams though Now that phi has become more importat, I m goig back ad editig the ames of Theorems 3, 5, 7, ad 9 Notice i particular whe reamig Theorem 7 sice we ca ow metio double eves ------------------------------------------------------------ The presetatio I m startig to evisio a parallel thig: you preset the material, the make a commet about the process that segues to me, the back to you agai ------------------------------------------------------------ Further work Is there a algebra o?
------------------------------------------------------------------------------- ------------------------------------------------------------------------------- spr( ) is a type of square-root for the itegers With a regular square root, you get sqrt ( ) spr ( ) With this you get spr ( ) ((called mu which is the greek letter right before u)) (maybe: The simple composite of a umber is the umber obtaied by takig the prime factorizatio of ad reducig the power of each prime factor to oe Do we eed a fuctio for this?) Yes, the prime factorizatio reductio fuctio pfr : where if the pfr( ) p1 p pk m1 m m p k 1 p p k The zero class [0] for reductio of greater or equal to (( Let s defie a ew fuctio called u-bar is the set of all iteger multiples of the smallest prime factor ( 1) /, odd ( ), eve but 4 4, 4 The bar is a artistic choice cosistet with the closure goig from to But wait, bad choice sice it s ot btw, trivial may be too strog, my artistic advice to him o use mu which is before u )) The equality i Theorem 8 holds for ( ) The Diagram for has a dropout where [ b] [ a] ad 0 a b ( ) Whoa maybe it s all about the multiplicities of the prime factors of!!?? 4 ) 4 1 : {0,1,,( 1) }
={equivalece classes of } Where Let s defie this fuctio for each positive iteger sod : {0,1,,( 1) } ={equivalece classes of } Where ---------------- Hey, what about Whereas 1 mod frac, pod exp frac log pod? Remember that from music/math stuff? =======================================================================