In words: Let be the relation on Z given by a b if a b. (Note that we use language like in definitions, where if actually means if and only if.

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1 Reltions A binry reltion on set A is subset R Ñ A ˆ A, where elements p, bq re written s b. Exmple: A Z nd R t b bu. In words: Let be the reltion on Z given by b if b. (Note tht we use lnguge like in definitions, where if ctully mens if nd only if.) Exmple: A R nd R t b bu. In words: Let be the reltion on R given by b if b. Exmple: A Z nd R t b b pmod 3qu. In words: Let be the reltion on Z given by b if b pmod 3q. More exmples of (binry) reltions: 1. For A numbersystem,let b if b. R, S, T 2. For A numbersystem,let b if b. not R, not S, T 3. For A R, let b if b 0. not R, S, not T 4. For A set of people, let b if is (full) sibling of b. not R, S, T 5. For A set of people, let b if nd b spek common lnguge. R, S, not T A binry reltion on set A is... (R) reflexive if for ll P A; (S) symmetric if b implies b ; (T) trnsitive if b nd b c implies c, i.e. p b ^ b cq ñ c An equivlence reltion on set A is binry reltion tht is reflexive, symmetric, nd trnsitive. (Only #1)

2 Fix n P Z 0 nd define the reltion on Z given by Is is n equivlence reltion? b if b pmod nq. Check: we hve b pmod nq if nd only if b kn for some k P Z. reflexivity: symmetry: trnsitivity: 0 0 n X If b kn, thenb kn p kqn. X If b kn nd b c `n, then c p bq`pb cq kn ` `n pk ` `qn.x Yes! This is n equivlence reltion! Let A be set. Consider the reltion on PpAq by Is is n equivlence reltion? S T if S Ñ T Check: This is reflexive nd trnsitive, but not symmetric. So no, it is not n equivlence reltion. Is S T if S Ñ T or S Ñ T n equivlence reltion on PpAq? Check: This is reflexive nd symmetric, but not trnsitive. So still no, it is not n equivlence reltion. Is S T if S T n equivlence reltion on PpAq? Red: Why reflexivity doesn t follow from symmetry nd trnsitivity.

3 Let be n equivlence reltion on set A, ndlet P A. The set of ll elements b P A such tht b is clled the equivlence clss of, denoted by rs. Exmple: Consider the equivlence reltion on A t, b, cu given by Then, b b, c c, c, nd c. rs t, cu rcs, nd rbs tbu re the two equivlence clsses in A (with respect to this reltion). (We sy there re two, not three, since the equivlence clsses refers to the sets themselves, not to the elements tht generte them.) Let be n equivlence reltion on set A, ndlet P A. The set of ll elements b P A such tht b is clled the equivlence clss of, denoted by rs. Exmple: We showed tht b if b pmod 5q is n equivlence reltion on Z. Then r0s t5n n P Zu 5Z r1s t5n ` 1 n P Zu 5Z ` 1 r2s t5n ` 2 n P Zu 5Z ` 2 r3s t5n ` 3 n P Zu 5Z ` 3 r4s t5n ` 4 n P Zu 5Z ` 4 r5s t5n ` 5 n P Zu t5m m P Zu r0s r 5s r10s r6s t5n ` 6 n P Zu t5m ` 1 m P Zu r1s r 4s r11s. In generl, if x Prys, thtmensy x. So x y. So y Prxs. Clim: x Prys if nd only if rxs rys. We cll ny element of clss C representtive of C (since we cn write C rs for ny P C).

4 Theorem. The equivlence clsses of A prtition A into subsets, mening 1. the equivlence clsses re subsets of A: rs Ñ A for ll P A; 2. ny two equivlence clsses re either equl or disjoint: for ll, b P A, eitherrs rbs or rsxrbs H; nd 3. the union of ll the equivlence clsses is ll of A: A PArs. We sy tht A is the disjoint union of equivlency clsses, written A ß PArs, L A TEX: \bigsqcup, \sqcup For exmple, in our lst exmple, there re exctly 5 equivlence clsses: r0s, r1s, r2s, r3s, nd r4s. Any other seemingly di erent clss is ctully one of these (for exmple, r5s r0s). And r0syr1syr2syr3syr4s Z. So Z r0s\r1s\r2s\r3s\r4s. Ok, so wht re numbers, nywy? Recll from the homework, the Zermelo-Frenkel xioms of set theory, which tells us how to compre sets, put sets in other sets, nd to tke unions, intersection, nd power sets of sets. X Set theoretic definition of the nturl numbers. (Z 0 ) Let 0 H. Given n, definethesuccessor to n s Spnq n Ytnu. (By successor to n we bsiclly men n ` 1.) Let N be the set of ll sets generted by 0 nd S. For exmple, 1 0 Yt0u HYtHu thu, 2 1 Yt1u thuytthuu th, thuu, 3 2 Yt2u th, thuu Y tth, thuuu th, thu, th, thuuu, nd so on. (Note tht we identified n with n.) Compute 4. Think: Given n, m P N, ren Y m nd/or n X m elements of N? If so, wht elements re they?

5 Set theoretic definition of the nturl numbers. (Z 0 ) Let 0 H. Given n, definethesuccessor to n s Spnq n Ytnu. LetN be the set of ll sets generted by 0 nd S. For exmple, 1 thu, 2 th, thuu, 3 th, thu, th, thuuu, 4 th, thu, th, thuu, th, thu, th, thuuuu, nd so on. (Note tht we identified n with n.) Addition: Define ` : N ˆ N Ñ N by, for ll, b P N, 1. ` 0 ; nd 2. ` Spbq Sp ` bq. For exmple, ` 1 ` Sp0q Sp ` 0q Spq; ` 2 ` Sp1q Sp ` 1q SpSpqq. Check tht ` 3 SpSpSpqqq S 3 pq. Think: ` b S b pq. Given n, definethesuccessor to n s Spnq n Ytnu. LetN be the set of ll sets generted by 0 nd S. For exmple, 1 thu, 2 th, thuu, 3 th, thu, th, thuuu, 4 th, thu, th, thuu, th, thu, th, thuuuu, nd so on. (Note tht we identified n with n.) Addition: Define ` : N ˆ N Ñ N by, for ll, b P N, 1. ` 0 ; nd 2. ` Spbq Sp ` bq. Think: ` b S b pq S `b p0q. Multipliction: Define : N ˆ N Ñ N by, for ll, b P N, 1. n 0 0; nd 2. Spbq p bq`. For exmple, 1 Sp0q 0 ` 0 ` ; Check tht ` 3 ` `. 2 Sp1q 1 ` `. Think: b S b p0q.

6 Addition: Define ` : N ˆ N Ñ N by, for ll, b P N, 1. ` 0 ; nd 2. ` Spbq Sp ` bq. Think: ` b S b pq S `b p0q. Multipliction: Define : N ˆ N Ñ N by, for ll, b P N, Properties: 1. n 0 0; nd 2. Spbq p bq`. Think: b S b p0q. 1. Addition is commuttive, i.e. ` b b ` for ll, b P N. 2. Addition is ssocitive, i.e. `pb ` cq p ` bq`c for ll, b, c P N. 3. Multipliction is commuttive, i.e. b b for ll, b P N. 4. Multipliction is ssocitive, i.e. pb cq p bq c for ll, b, c P N. 5. Multipliction is distributive cross ddition, i.e. pb ` cq p bq`p cq for ll, b, c P N. (These ll follow from the definitions, but we ll skip proofs for the ske of time.) Peno xioms The nturl numbers N re defined by the following xioms. 1. We hve 0 P N. (techniclly, 0 H) 2. There exists n successor function S : N Ñ N (nmely, if n P Z, thenspnq PN) stisfying (i) 0 R SpNq; (ii) S is injective (if Spnq Spmq, thenn m); nd (iii) if X Ñ N stisfies n 0 P X nd SpXq Ñ X, wehvex N. Note: () We hve not ssumed tht 0 is the only element tht is no one s successor (but it follows, in prt from 1(iii)). (b) By chnging 0 out for something else (like 1), or chnging Spnq to something else (like n 1), we cn generte other sets tht re bsiclly the nturl numbers ll over gin. This is why we re not fussy bout whether N is Z 0 or Z 0. (c) The lst xiom (1(iii)) is the bsis of proof by induction.

7 The nturl numbers N re defined by the following xioms. 1. We hve 0 P N. 2. There exists n successor function S : N Ñ N (nmely, if n P Z, thenspnq PN) stisfying (i) 0 R SpNq; (ii) S is injective (if Spnq Spmq, thenn m); nd (iii) if X Ñ N stisfies n 0 P X nd SpXq Ñ X, wehvex N. Order on N. For, b P N, wesy b if b SpSp Spq qq. Properties: (i) For ll, b P N, wehve b or b. (ii) If b nd b, then b. (iii) If b nd b c, then c. (iv) If b then ` c b ` c. (v) If b then c bc. (These ll follow from the xioms, but we ll skip proofs for the ske of time.) Integers Now tht we hve N, we cn define Z by formlly letting N t n n P Nu, where 0 0; nd Z N Y N. Extend S : Z Ñ Z by defining Sp q for ny P N t0u s Sp q b, where Spbq. We cn define the predecessor function P : Z Ñ Z by P pxq y whenever Spyq x. Letting p xq x, thissystht Spxq y if nd only if P pyq x. We cn lso extend our definitions of ` nd to Z. Properties: 1. Addition nd multipliction still stisfy commuttivity, ssocitivity, nd distributivity. 2. We still hve ` 0 (dditive identity) nd 1 (multiplictive identity) forll P Z. 3. We lso hve tht `p q 0 (prove). (dditive inverses) We cll ny number system tht hs n ddition nd multipliction tht stisfy ll these properties ring (modern lgebr).

8 Integers Now tht we hve N, wecndefinez by formlly letting N t n n P Nu, where 0 0; nd Z N Y N. ExtendS : Z Ñ Z by defining Sp q for ny P N t0u s Sp q b, where Spbq. We cn define the predecessor function P : Z Ñ Z by P pxq y whenever Spyq x. Letting p xq x, thissystht Spxq y if nd only if P pyq x. We cn lso extend our definition of order to Z. The only modifiction is: (i) For ll, b P N, wehve b or b. (ii) If b nd b, then b. (iii) If b nd b c, then c. (iv) If b then ` c b ` c. (v) If b nd c P N, then c bc. These properties mke Z n ordered ring. Rtionl numbers Let Q Z ˆpZ t0uq, nd define n equivlence reltion on Q by p, bq px, x bq for ll x P Z t0u. Under this equivlence reltion, write rp, bqs. b Then rtionl numbers re! ) Q ˇ, b P Z,b 0. b (Note tht we get lzy, nd write 1.) Define ` : Q ˆ Q Ñ Q nd : Q ˆ Q Ñ Q by b ` c d d ` b c nd b d b c d c b d.

9 Let Q Z ˆpZ t0uq nd define n equivlence reltion on Q by p, bq px, x bq for ll x P Z t0u. Under this equivlence reltion, write b rp, bqs (writing 1 ). Then rtionl numbers re! ) Q ˇ, b P Z,b 0. b Define ` : Q ˆ Q Ñ Q nd : Q ˆ Q Ñ Q by b ` c d d ` b c nd b d b c d c b d. Agin Addition nd multipliction still stisfy commuttivity, ssocitivity, nd distributivity. 2. We still hve x ` 0 x (dditive identity) ndx 1 x (multiplictive identity) forllx P Q. 3. We lso hve tht x `p xq 0. (dditive inverses) So Q is lso ring. In ddition, for ll {b P Q, b b 1 (multiplictive inverses). This mkes Q field (gin, modern lgebr). Order on Q Define b b (you cn show b á b ). We define on Q by the following: for, b, c, d P N, wehve 1. b c d 2. b c d ;nd whenever d b c; 3. b c d whenever c d b. Then, gin, (i) For ll, b P N, wehve b or b. (ii) If b nd b, then b. (iii) If b nd b c, then c. (iv) If b then ` c b ` c. (v) If b nd 0 c, thenc bc. This mkes Q into n ordered field.

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