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1 MthCity.org Merging mn nd mths Exercise.8 (s) Pge 46 Textbook of Algebr nd Trigonometry for Clss XI Avilble Version: 3.0 Question # Opertion performed on the two-member set G = {0, is shown in the djoining tble. Answers the questions: (i) Nme the identity element if it exists? 0 (ii) Wht is the inverse of? 0 (iii)is the set G, under the given opertion group? 0 Abelin nd non-belin? s i) From the given tble we hve 0 0 = 0 nd 0 = This show tht 0 is the identity element. ii) Since = 0 (identity element) so the inverse of is. iii) It is cler from tble tht element of the given set stisfy closure lw, ssocitive lw, identity lw nd inverse lw thus given set is group under. Also it stisfies commuttive lw so it is n belin group. Question # The opertion s performed on the set {0,,,3 is shown in the djoining tble, shown tht the set is n Abelin group? Suppose G = { 0,,,3 0 3 i) The given tble show tht ech element of the tble is member of G thus closure lw holds. ii) is ssocitive in G. 3 0 iii) Tble show tht 0 is identity element w.r.t iv) Since 0 0 = 0, 3 = 0, = 0, 3 = = 0, = 3, =, 3 = v) As the tble is symmetric w.r.t. to the principl digonl. Hence commuttive lw holds. Question # 3 For ech of the following sets, determine whether or not the set forms group with respect to the indicted opertion. From bove tble solve these (i-v) options. (i) As 0 Q, multiplictive inverse of 0 in not in set Q. Therefore the set of rtionl number is not group w.r.t to. (ii) - Closure property holds in Q under becuse sum of two rtionl number is lso rtionl. b- Associtive property holds in Q under ddition. c- 0 Q is n identity element.

2 FSc-I /.8 - d- If Q then dditive inverse Q such tht ( ) = ( ) = 0. Therefore the set of rtionl number is group under ddition. (iii) - Since for, b Q, b Q thus closure lw holds. b- For, b, c Q, ( bc) = ( b) c thus ssocitive lw holds. c- Since Q such tht for Q, = =. Hence is the identity element. d- For Q, Q such tht = =. Thus inverse of is. Hence Q is group under ddition. (iv) Since Z = {0, ±, ±, ± 3,... - Since sum of integers is n integer therefore for, b Z, b Z. b- Since ( b c) = ( b) c thus ssocitive lw holds in Z. c- Since 0 Z such tht for Z, 0 = 0 =Z. Thus 0 n identity element. d- For Z, Z such tht ( ) = ( ) = 0. Thus inverse of is. (v) Since Z = {0, ±, ±, ± 3,... For ny Z the multiplictive inverse of is Z. Hence Z is not group under multipliction. Question # 4 Show tht the djoining tble represents the sums of the elements of the set { E, O. Wht is the identity element of this set? Show tht this set is belin group.. As E E = E, E O = O, O O = E E O Thus the tble represents the sums of the elements of set { E, O. E E O The identity element of the set is E becuse E E = E E = E & E O = O E = E. O O E i) From the tble ech element belong to the set { E, O. Hence closure lw is stisfied. ii) is ssocitive in { E, O iii) E is the identity element of w.r.t to iv) As O O = E nd E E = E, thus inverse of O is O nd inverse of E is E. v) As the tble is symmetric bout the principle digonl therefore is commuttive. E, O is belin group under. Hence { Question # 5 Show tht the set {,, multipliction. Suppose G = {,, 3, when = is n belin group w.r.t. ordinry

3 i) A tble show tht ll the entries belong to G. ii) Associtive lw holds in G w.r.t. multipliction. e.g. ( ) = = = = FSc-I /.8-3 iii) Since =, = =, = = Thus is n identity element in G. iv) Since = =, = =, = = therefore inverse of is, inverse of is, inverse of is. v) As tble is symmetric bout principle digonl therefore commuttive lw holds in G. Hence G is n belin group under multipliction. Question # 6 If G is group under the opertion nd, b G, find the solutions of the equtions: x = b, x = b Given tht G is group under the opertion nd, b G such tht x = b As G nd G is group so G such tht x = b And for ( ) x = b s ssocitive lw hold in G. e x b = by inverse lw. x b = by identity lw. x = b x = b For G, ( ) x = b s ssocitive lw hold in G. x e = b by inverse lw. x = b by identity lw. Question # 7 Show tht the set consisting of elements of the form 3b (, b being rtionl), is n belin group w.r.t. ddition. Consider G = { 3 b, b Q i) Let 3 b, c 3d G, where, b, c & d re rtionl. 3b c 3d = c 3 b d = 3b G where = c nd b = b d re rtionl s sum of rtionl is rtionl. Thus closure lw holds in G under ddition. ii) For 3 b, c 3 d, e 3 f G G

4 FSc-I /.8-4 ( ) = ( 3 b) (( c e) 3( d f )) ( 3 b) ( c 3 d) ( e 3 f ) ( ( c e) ) 3 ( b ( d f )) (( c) e ) 3 (( b d) f ) = = As ssocitive lw hold in Q = ( c) 3( b d) ( e 3 f ) = ( 3 b) ( c 3 d) ( e 3 f ) Thus ssocitive lw hold in G under ddition. iii) G s 0 is rtionl such tht for ny 3b G ( 3 b) (0 3 0) = ( 0) 3( b 0) = 3b And (0 3 0) ( 3 b) = (0 ) 3(0 b) = 3b Thus is n identity element in G. iv) For 3b G where & b re rtionl there exit rtionl & b such tht ( 3 b) ( ) 3( b) = ( ) 3 b ( b) = & ( b ) b ( ) ( b b) ( ) 3( ) ( 3 ) = ( ) 3 ( ) = Thus inverse of 3b is ( ) 3( b) exists in G. v) For 3 b, c 3d G 3b c 3 d = ( c) 3( b d) = ( c ) 3( d b) As commuttive lw hold in Q. = ( c d 3) ( 3 b) Thus Commuttive lw holds in G under ddition. And hence G is n belin group under ddition. Question 8 Determine whether ( P( S), ), where stnds for intersection is semi group, monoid or neither. If it is monoid, specify its identity. Let A, B P( S) where A & B re subsets of S. As intersection of two subsets of S is subset of S. Therefore A B = A B P( S). Thus closure lw holds in P( S ). For A, B, C P( S) A ( B C) = A ( B C) = ( A B) C = ( A B) C Thus ssocitive lw holds nd P( S ). And hence ( P( S), ) is semi-group. For A P( S) where A is subset of S we hve S P( S) such tht A S = S A = A. Thus S is n identity element in P( S), is monoid. P S. And hence

5 FSc-I /.8-5 Question 9 Complete the following tble to obtin semi-group under Let x nd x be the required elements. By ssocitive lw ( ) = ( ) c = c x = b Now gin by ssocitive lw ( ) b = ( b) c b = x = c b c c b b b c x c x Question 0 Prove tht ll non-singulr mtrices over the rel field form non-belin group under multipliction. Let G be the ll non-singulr mtrices over the rel field. i) Let A, B G then A B = C G Thus closure lw holds in G under multipliction. ii) Associtive lw in mtrices of sme order under multipliction holds. therefore for A, B, C G A ( B C) = ( A B) C 0 iii) I = 0 is non-singulr mtrix such tht A I = I A = A Thus I is n identity element in G. iv) Since inverse of non-singulr squre mtrix exists, therefore for A G there exist A G such tht AA = A A = I. v) As we know for ny two mtrices A, B G, AB BA in generl. Therefore commuttive lw does not holds in G under multipliction. Hence the set of ll non-singulr mtrices over rel field is non-belin group under multipliction. Book: Exercise.8 (Pge 78) Text Book of Algebr nd Trigonometry Clss XI Punjb Textbook Bord, Lhore. Avilble online t in PDF Formt (Picture formt to view online). Pge setup: A4 (8.7 in.0 in). Updted: These resources re shred under the licence Attribution- NonCommercil-NoDerivtives 4.0 Interntionl Under this licence if you remix, trnsform, or build upon the mteril, you my not distribute the modified mteril.

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