# fylladio5-2005 - FULLADIO 05 Grammik 'Algebra SHMMU 2005-06...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: FULLADIO 05 Grammik 'Algebra SHMMU 2005-06 1. H isìthta x ∗ y = x + y + x2 y 2 orÐzei mÐa prxh ∗ sto R. Na breÐte to oudètero stoiqeÐo thc 1 prxhc ∗ kai na deÐxete ìti kje stoiqeÐo x ∈ R∗ me x < √ èqei dÔo summetrik stoiqeÐa, en¸ 3 4 1 kje x ∈ R me x > √ den èqei summetrikì stoiqeÐo, wc proc thn prxh aut . Ta stoiqeÐa 3 4 1 0, √ èqoun summetrik kai poi? 3 4 2. JewroÔme to sÔnolo a −b M= : a, b ∈ R , b a me prxeic thn {+} prìsjesh pinkwn kai {·} pol/smì pinkwn. Na deÐxete ìti: i) H algebrik dom (M, +) eÐnai antimetajetik omda. ii) H algebrik dom (M ∗ , · ) eÐnai omda. iii) H algebrik dom (M, +, · ) eÐnai s¸ma. Katìpin na lÔsete sto M thn exÐswsh 2 + I = O. X 3. 'Estw h omda (G, · ). i) Na deÐxete ìti gia kje a, b ∈ G isqÔei: (a · b)−1 = b−1 · a−1 , ìpou me x−1 sumbolÐzoume to summetrikì (antÐstrofo) tou x ∈ G. ii) An isqÔei h sqèsh (a · b)2 = a2 · b2 gia kje a, b ∈ G, na deÐxete ìti h omda G eÐnai antimetajetik . 4. O daktÔlioc (∆, +, · ) lègetai daktÔlioc tou Boole an isqÔei: x2 = x gia kje x ∈ ∆. Na deÐxete ìti kje daktÔlioc tou Boole eÐnai antimetajetikìc kai isqÔei x + x = 0 gia kje x ∈ ∆. 5. 'Estw (G, · ) mÐa omda kai H ⊆ G, H = ∅. To H lègetai upoomda thc G an to H eÐnai omda wc proc thn prxh · thc G. Na deÐxete ìti èna uposÔnolo H = ∅ thc omdac G eÐnai upoomda thc G an kai mìno an isqÔei x · y −1 ∈ H gia kje x, y ∈ H , ìpou y −1 eÐnai to summetrikì tou y ∈ G. S. Karansioc m r fy h hg x cge  c d ygd y ¤g swy!ief y r isAsy c # m ey fy h hg s  esyyx   egwj e r d  ffdce h f ey d f cy d a q ff a h f   y Asken Au ekdjy¤r t ¢sAsy c  m h d r fy h hg x ef f  ey y 6y  esyy¤ £sy c gd l c h xsyge u Y m h fy l a l f ef f   y dl¢ekeodxc r ey¤r t ¢sy c  m d x g e a f f  l e  lm   x  iec e  kd#yh g u Wxsxdxa   {djwz r r Ase¦kdjgc  f h ¡eygf a hfd a kdf t r t  m  hy es ge gy r u iEd  syyy wY v m f  g c  fy h hg x e y x ey fd a ly a fey x eyfge ey d yc  ApdedgTesg v m s u e } y hec  fda y  ly f r  Tm Apr u g cg!deQee  sge m wY v m n  e|ywkhgqd  gec xsgd  e u {de u   ir  u d  a fey x yfg y x fy zdy fycd f fd a l c e a e eyx y ssy cV VUS oW¨TR n  wec  ¦ v fy  x fd a   d f ca df d  xsc y  6gd  geyc f q feyc  he ef fd a xdy  gs  e i   gs  yi¨gec gx r gd u ly d  d Ap¦n  f  e t e v m n hp dhn yAq leh f iyyz egd l ff f h lgey dfd a y ie  egd t jgo u Ay q se u d i#x Ai  m leid  gey  rq ac e d y fd a l a hdcgfe f ly lyy f f f A u xp6c #ikdjieeidxAp#spA{kfyv m h d hdcfe ffdy  nygd xx  V  V U S 6yiegdxs ir  t pg¤Qx\$W¨TR m gef g u f u  gey a zg cy  h ad i&kdyApywegwh ieAe ir xc gx r gd u ldc  e fd a xdy  ly d Apn  f  e t  m n kdgyv v feh p d hn  ey yAqTx d y fdy a  xdy  ly d expc d  ix r id u Asn  f  e t  m n h d hdcfe f fdy  nygd  x  x  V ~ V U S 6wgegdxy gr  t pg6yW¨TR m m e q f } y y f ff  h c  x f ff c y f g dyf d x l xyf d a d z l a hdcgfe f f rq f e l c h p|ekgkdwbxk{dqx#d u ge q  u egd  6iwgsc #yjikdjiegdxxdlx Ai xxsyge u Y α α |z | =2 ⇔ α|z |2 + (α2 + z∈ : |z |2 + 1 α +1 1 1)|z | + α = 0 ⇔ |z | = α |z | = α (0, 0) 1 α α x, y ∈ R x + y, x2 y 2 ∈ R ⇒ x + y + x2 y 2 ∈ R ∗ R e ∗ ∀x∈R x = x ∗ e = x + e + x2 e2 ⇔ e(1 + e) = 0 (1) e=0 ∗ e=0 x x∈R x ∗ x = 0 ⇔ x + x + x2 x 2 = 0 (2) x=0 x ∆ = 1 − 4x3 √ −1 ± 1 − 4x3 3>0 • 1 − 4x x1,2 = 2 1 x3 < 1/4 ⇔ x < √ x x 1 , x2 3 4 √ 1 1 √ • x= √ x =−32 3 3 4√ 4 −32 1 • 1 − 4x3 < 0 ⇔ x > √ x 3 4 C∗ |w| |w|2 + 1 |z |2 |z | = +1 (x, y ) ∈ S ⇔ x3 = x2 (y + 2) ⇔ x2 (y − x + 2) = 0 ⇔ {x = 0 y − x + 2 = 0} yy y−x+2=0 [0], [1], ..., [ν − 1] (x, y ) ∈ S ⇔ x2 − y 2 ≤ 0 ⇔ (x − y )(x + y ) ≤ x − y ≥ 0, x + y ≤ 0} ⇒ x − y = (k1 + k2 ) ν + v1 − v2 ⇐⇒ v1 − v2 = 0 ⇔ v1 = v2 . (1) dege i kdf t m f ur yy  rq f a cyx cg nc sssx Ai kdekf¦Apoge  sgd  c eyx l a hdcgfe f xx fd a #ikdjiegdx6ec   d  v Z x − y = k ν (1) ≡ (mod ν ) k∈Z lh xwv v m 0 ⇔ {x − y ≤ 0, x + y ≥ 0 ν x = k 1 ν + v1 y = k2 ν + v2 x, y x − x = 0 = 0 ν ⇒ x ≡ x(mod ν ) x ≡ y (mod ν ) ⇒ x − y = k ν, k ∈ Z ⇒ y − x = (−k ) ν, −k ∈ Z ⇒ y ≡ x(mod ν ) x ≡ y (mod ν ) y ≡ z (mod ν ) x−y = kν (+) k, λ ∈ Z =⇒ x − z = (k + λ) ν, k + ν ∈ Z ⇒ x ≡ z (mod ν ) y − z = λν y ssy f gd  y x v bY wr u rt rp fd a Y VX V U S sqihgec b`6¦W¨TR PB QEDCI'B HGFEDCA97 164¢¢10(&%¦\$#!¨¦¤¢ B B @8 ) 53 2  ) '   ¥ £ " £     ©§ ¥ £ ¡ m ly d e g c  d he fd a ey d d he fd a eyf Apsef r hgge u #lxyhie u 6fd  f r i¤c    v m ef r g6ec  qd   ge  d ey f  f dy f f fy f dy e x¤6xpykkf¦n  dxkfgxge u  m g t  ir  u Au Ap#ie u xdx g gd  gec  y cy le l  y f f fd a e nf i q  }  yz r kdf A q a u Y m ce rd f d f yz  kdf t  r gd   ·m r y  ge  d e£e u Eg dc f  r yx wY v h ey a cd ey f ld he f ld he ly fe ey d ex fd a  zg V ¶ V U S 6e  sxeysgT{d¨ª m ief r gwd  gf r ige u g!Apykhi¦cA u Tyh r c yTe  W¨TR m  hg  r  ygebAEµ } rs³p u  m f d a  e   h e y u u ¯ ® ¬ « ­ mc \$lgyy iee kda u± jr¬ yy}°u« g´ ju ~ hg c r t m hhge ewg a u  r f  e! u Wrs³pcApbgc q e ¨m r t f d  y ldy ² ¦ hhge eyi a u  r m dxeg u d!s³p&cApylgsc q e ¦ jwyQjgX y f d r   dy ² ±¬ « ° ¯®­ ¬ « hidc  fa r f yz  kd%ª  r fª V¨ VUS yz  {d¨1w©W¨TR m x∈H x −1 = 1 · x−1 ∈ H x∈H⇒x H H⊆G G 1 = x · x−1 ∈ H x, y ∈ H ⇒ x · y = x · (y −1 )−1 ∈ H << · >> H G H H=∅ x∈∆ (x + x)2 = x + x ⇒ (x + x) · (x + x) = x + x ⇒ +++ =x+x⇒x+x+x+x=x+x⇒x+x=0 x, y ∈ ∆ (x + y )2 = x + y ⇒ x2 + x · y + y · x + y 2 = x + y ⇒ x + x · y + y · x + y = x + y ⇒ x · y + y · x = 0 ⇔ x · y = −y · x (1) x x2 · y = −x · y · x ⇒ x · y = x x · y · x = −y · x2 ⇒ −x · y · x = y · x (3) −x · y · x x·y =y·x y · x + y · x = 0 ⇒ y · x = −y · x (5) x·y =y·x x2 x2 x2 hhge eyi x2 (a · b)−1 = b−1 · a−1 ⇔ (a · b) · (b−1 · a−1 ) = (b−1 · a−1 ) · (a · b) = 1 = a · [(b · b−1 ) · a−1 ] = a · (1 · a−1 ) = a · a−1 = 1 (b−1 · a−1 ) · (a · b) = 1 (ii) (a · b)2 = a2 · b2 ⇔ (a · b) · (a · b) = a2 · b2 ⇔ (a−1 · a) · [(b · a) · (b · b−1 )] = (a−1 · a2 ) · (b2 · b−1 ) ⇔ b·a=a·b x2 − y 2 = −1 2xy = 0 01 X= −1 0 kdgyv v m feh ⇔ = m lh xwv v m x −y y x n eydcf ieid ihc bY fda (i) x −y y x m (a · b) · (b−1 · a−1 ) V§VUS ¢W¨TR x = 0, y 2 = 1 y = 0, x2 = −1, −1 0 0 −1 0 −1 ⇔ x = 0, y = ±1 ⇔ X = 1 0 X 2 + I = O ⇔ X 2 = −I ⇔ ⇔ m yx6ec  dhxfda d e v m fy y ayd ¦ d ce A¢sdxApieid `m ef r Ayh h y\$lgdef q sy  ef\$i  y|ykhid£i  fefAwh¦wfc  d l fy zdy fyc l c  h d a fy f d he fy zdy fycd fd a d  ef r g!n  yykhic  } ihgec b\$y¦W¨TR fdaY V¥ VUS f  fd  a d  y¤ycdefexpyd q e  wY m  ey fy h hg x f   cy s  e|yy¤qr t As u d e 10 01 xy −y x 00 ∈M 00 (M, +, ·) 1 x −y ∈ M∗ ⇒ A−1 = 2 y x x + y2 ∈M O= x =0 ∈ M∗ (M, +) m (M, +, ·) A= I= x=0 ...
View Full Document

## This note was uploaded on 10/02/2009 for the course G 001 taught by Professor Shmmygr during the Spring '07 term at National Technical University of Athens, Athens.

Ask a homework question - tutors are online