M A T H F A L L CORRECTION. Algebra I 2 1 / 0 9 / U N I V E R S I T Y O F T O R O N T O


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1 M A T H F A L L HOMEWORK ASSIGNMENT #1 CORRECTION Alger I 2 1 / 0 9 / U N I V E R S I T Y O F T O R O N T O
2 1. Suppose nd re nonzero elements of field F. Using only the field xioms, prove tht 1 1 is multiplictive inverse of. Stte which xioms re used in your proof. Let, F such tht 0 0. First, we hve, F F ecuse field is closed under multipliction. Moreover, Therefore, since is nonzero element of field, there exists multiplictive inverse of elonging to this field y xiom F 4 of field. Let denote y 1 the identity element of multipliction of this field. Then: ( ) ( 1 1 ) = ( ) ( 1 1 ) y F 1 : commuttive property of multipliction = ( 1 ) 1 y F 2 : ssocitivity of multipliction = (1) 1 y F 4 : definition of multiplictive inverse = (1 1 ) y F 2 : ssocitivity of multipliction = 1 y F 3 : definition of identity elements = 1 y F 4 : definition of multiplictive inverse Finlly, since we hve ( ) ( 1 1 ) = 1, 1 1 is multiplictive inverse of. 2. Prove tht if nd re elements of field F, then 2 = 2 if nd only if = or =. First, we know tht: 2 = = 0 (1) Moreover: ( ) ( + ) = ( ) + ( ) y F 5 : distriutive lw = + ( ) + + ( ) y F 5 : distriutive lw = 2 + ( ) + + ( ) y F 1 : commuttive property of multipliction = 2 + ( ) + ( ) y F 5 : distriutive lw = ( ) y F 4 : definition of dditive inverse = ( ) Demonstrted in clss: F, 0 = 0 = 2 + ( ) y F 3 : definition of identity elements So we hve: ( ) ( + ) = 2 + ( ) (2) We lso hve: 2 + ( ) = + ( ) = ( ) y F 5 : distriutive lw = 0 y F 4 : definition of dditive inverse = 0 Demonstrted in clss: F, 0 = 0 So ( ) is n dditive inverse of 2. In ddition, 2 is lso n dditive inverse of 2. As we demonstrted in clss, the inverse for ddition is unique. Therefore we hve: 2 = ( ). Finlly, eqution (2) ecome: ( ) ( + ) = 2 2 (3) By comining (3) nd (1) we get: ( ) ( + ) = 0 2
3 As demonstrted in clss (12 th property of the theorem on sic properties of fields), we hve: c, c F, c c = 0 c = 0 c = 0 Therefore: ( ) ( + ) = 0 ( ) = 0 ( + ) = 0 = = 3. Let F 4 = {0, 1,, } e field contining 4 elements. Assume tht = 0. Prove tht = 1 = 2 = + 1. First, let us prove tht: = 1 Since F 4 is field, exists nd elongs to F 4 ecuse field is closed under multipliction. Therefore, there re 4 possile vlues for : 0,1, or.  Suppose tht = 0. Then = 0 = 0 = 0 F 4 = {0,1, } F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence 0.  Suppose tht =. Then = = 1 y cncelltion F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence.  Suppose tht =. Then = = 1 y cncelltion F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence. Finlly, the only possile vlue for is 1. Hence: = 1 = 1 = 2 Since F 4 is field, 2 = exists nd elongs to F 4 ecuse field is closed under multipliction. Therefore, there re 4 possile vlues for 2 : 0,1, or.  Suppose tht 2 = 0. Then = 0 = 0 F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence Suppose tht 2 = 1. Then = 1 = 1 = ecuse the multiplictive inverse is unique F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence Suppose tht 2 =. Then = = 1 y cncelltion F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence 2. Finlly, the only possile vlue for 2 is. Hence: = 1 = 2 = + 1 Since F 4 is field, + 1 exists nd elongs to F 4 ecuse field is closed under ddition. Therefore, there re 4 possile vlues for + 1: 0,1, or.  Suppose tht + 1 = 0. Then + 1 = 0 = 1 ecuse = 0 nd ecuse the dditive inverse is unique F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence Suppose tht + 1 = 1. Then + 1 = 1 = 0 y cncelltion F 4 = {0,1, }. This contrdicts the fct tht F 4 is field contining 4 elements. Hence Suppose tht + 1 =. Then + 1 = 1 = 0 y cncelltion. This contrdicts the fct tht in field the identity element for ddition ( 0 ) is different from the dditive element from multipliction ( 1 ). Hence
4 Finlly, the only possile vlue for + 1 is. Hence: = 1 = 2 = Write the following complex numers in the for + i, with, R: i + 2i 5 i 1 2i + 2i 5 i = 1 2i 1 + 2i 5 i 1 = 1 2i 5 i 5 i + 2i 5 i 2i 2i = 5 i 10i 2i 2 + 4i2 10i 2i 2 = 5 i 10i i + 2 = 9 i 10i + 2 = 9 i 10i = 9 i 10i i 2 10i 2 = 90i 10i i 100i 2 4 = 92i = 92i = 8 92i 104 = ( ) i = ( ) i Finlly, we hve z = 1 + 2i = ( ) i with = 1 23, = nd, R. 2i 5 i (1 + i) 5 Notice tht: 1 + i C (1,1) C. Moreover, we hve demonstrted during clss tht: (, ) (c, d) = ( c d, c + d) Thus: (1,1) 5 = ((1,1) 2 (1,1) 2 ) (1,1) By ssocitivity of multipliction = ((0,2) (0,2)) (1,1) = ( 4,0) (1,1) = ( 4, 4) Finlly, we hve z = (1 + i) 5 = 4 4i with = 4, = 4 nd, R. 4
5 5. Fields: 5.1. Prove tht the set F 1 = { + 3, Q} (endowed with the ddition nd multipliction inherited from R) is field Closure under ddition nd multipliction First, we hve to prove tht F 1 is closed under the inry opertions + nd, i.e. : F 1 F 1 F 1 nd +: F 1 F 1 F 1. Let x 1, x 2 F 1 with x 1 = + 3, x 2 = c + d 3 nd,, c, d Q. Then: x 1 + x 2 = ( + 3) + (c + d 3) = ( + 3) + (d 3 + c) y commuttive property of ddition on R = + ( 3 + d 3) + c y ssocitivity of ddition on R = + ( + d) 3 + c ecuse ddition nd multipliction re distriutive opertions on R = ( + c) + ( + d) 3 y commuttive nd ssocitive property of ddition on R Thus, x 1 + x 2 = ( + c) + ( + d) 3 F 1 ecuse e = + c Q nd f = + d Q F 1 closed under ddition. x 1 x 2 = ( + 3) (c + d 3) = (d 3 + c) + 3 (c + d 3) ecuse ddition nd multipliction re distriutive opertions on R = d 3 + c + 3c + 3d 3 ecuse ddition nd multipliction re distriutive opertions on R = (d 3 + c 3) + c + d 3 3 y ssocitivity nd commuttive property of ddition nd multipliction on R = (d + c) 3 + c + 3d ecuse ddition nd multipliction re distriutive opertions on R = (c + 3d) + (d + c) 3 y commuttive property nd ssocitivity of ddition on R Thus, x 1 x 2 = (c + 3d) + (d + c) 3 F 1 ecuse e = c + 3d Q nd f = d + c Q F 1 closed under multipliction. F1: Commuttive property for ddition nd multipliction Let x 1, x 2 F 1 with x 1 = + 3, x 2 = c + d 3 nd,, c, d Q. Then: x 1 + x 2 = (, ) + (c, d) = ( + c, + d) As proved ove = (c +, d + ) y commuttive property of ddition on R = (c, d) + (, ) = x 2 + x 1 Since x 1 + x 2 = x 2 + x 1, ddition is commuttive on F 1. 5
6 x 1 x 2 = (, ) (c, d) = (c + 3d, d + c) As proved ove = (c + 3d, d + c) y commuttive property of multipliction on R = (c, d) (, ) = x 2 x 1 Since x 1 x 2 = x 2 x 1, multipliction is commuttive on F 1. F2: Associtivity for ddition nd multipliction Let x 1, x 2, x 3 F 1 with x 1 = + 3, x 2 = c + d 3, x 3 = e + f 3 nd,, c, d, e, f Q. Then: (x 1 + x 2 ) + x 3 = ((, ) + (c, d)) + (e, f) = ( + c, + d) + (e, f) = (( + c) + e, ( + d) + f) = ( + (c + e), + (d + f)) y ssocitivity of ddition on R = (, ) + (c + e, d + f) = (, ) + ((c, d) + (e, f)) = x 1 + (x 2 + x 3 ) Since (x 1 + x 2 ) + x 3 = x 1 + (x 2 + x 3 ), ddition is ssocitive on F 1. (x 1 x 2 ) x 3 = ((, ) (c, d)) (e, f) = (c + 3d, d + c) (e, f) = ((c + 3d) e + 3 (d + c) f, (c + 3d) f + (d + c) e) = (ce + 3de + 3df + 3cf, cf + 3df + de + ce) = (ce + 3df + 3de + 3cf, cf + de + 3df + ce) = (ce + 3df + 3de + 3cf, cf + de + 3df + ce) = ( (ce + 3df) + 3 (de + cf), (cf + de) + (3df + ce)) = (, ) (ce + 3df, cf + de) = (, ) ((c, d) (e, f)) = x 1 (x 2 x 3 ) ddition nd multipliction re distriutive on R ssocitivity of ddition on R ssocitivity of multipliction on R ddition nd multipliction re distriutive on R Since (x 1 x 2 ) x 3 = x 1 (x 2 x 3 ), multipliction is ssocitive on F 1. F3: Existence of identity elements for ddition nd multipliction Let x 1, 0 F1 F 1 with x 1 = + 3, 0 F1 = 0 = nd,, 0 Q. Then: x F1 = (, ) + (0,0) = ( + 0, + 0) = (, ) = x 1 y definition of 0 on R Therefore, 0 F1 = (0,0) F 1 is the dditive identity element for F 1 6
7 Let x 1, 1 F1 F 1 with x 1 = + 3, 0 F1 = 1 = nd,, 1,0 Q. Then: x 1 1 F1 = (, ) + (1,0) = ( , 0 + 1) = (, ) y definition of 0 nd 1 on R = x 1 Therefore, 1 F1 = (1,0) F 1 is the multiplictive identity element for F 1. Moreover, we hve (0,0) (1,0) 0 F1 1 F1. Finlly, 1 F1, 0 F1 F 1 such tht F 1, + 0 F1 = 1 F1 = with 0 F1 1 F1. F4: Existence of inverses for ddition nd multipliction Let x 1, x 2 F 1 with x 1 = + 3, x 2 = c + d 3 nd,, c, d Q. Then: x 1 + x 2 = 0 F1 (, ) + (c, d) = (0,0) ( + c, + d) = (0,0) + c = 0 + d = 0 c = x 2 = (, ) Moreover,, Q, Q x 2 F 1. Finlly, x 1 F 1, x 2 = (, ) F 2 such tht x 1 + x 2 = 0 F1. Let x 1, x 2 F 1 with x 1 = + 3, x 2 = c + d 3 nd,, c, d Q such tht x 1 0 F1 x 2 0 F1. Then: x 1 x 2 = 1 F1 (, ) + (c, d) = (1,0) (c + 3d, d + c) = (1,0) c + 3d = 1 d + c = 0 d + ( 1 3d 2 d + 32 d ) = 0 = 0 Assuming tht 0 WLOG (since x ) 2 d d = 0 2 d 3 2 d ( ) = 7
8 Moreover: = 0 2 = 3 2 = 3 = 3. But,, Q such tht = 3 = 3 if 0 0. Therefore,, Q such tht 0 0, { ( ) 1 3 ( c = 2 3 2) { ( ) 1 c = ( ( )) { ( ) c = ( ) { ( ) 2 c = ( ) { ( ) x 2 = ( ( ), ( ) ) Moreover,, Q Q Q x ( ) ( ) 2 F 1. Finlly, x 1 F 1, x 2 F 2 such tht x 1 x 2 = 1 F1. F5: Distriutive property of ddition nd multipliction Let x 1, x 2, x 3 F 1 with x 1 = + 3, x 2 = c + d 3, x 3 = e + f 3 nd,, c, d, e, f Q. Then: x 1 (x 2 + x 3 ) = (, ) ((c, d) + (e, f)) = (, ) (c + e, d + f) = ( (c + e) + 3 (d + f), (d + f) + (c + e)) = (c + e + 3d + 3f, d + f + c + e) ddition nd multipliction re distriutive on R = ((c + 3d) + (e + 3f), (d + c) + (f + e)) = (c + 3d, d + c) + (e + 3f, f + e) = (, ) (c, d) + (, ) (e, f) = x 1 x 2 + x 1 x 3 commuttive property nd ssocitivity of ddition on R 8
9 Finlly, x 1, x 2, x 3 Q, x 1 (x 2 + x 3 ) = x 1 x 2 + x 1 x 3 To conclude, F 1 is set endowed with two inry opertions : F 1 F 1 F 1 nd +: F 1 F 1 F 1, which hs two specil elements 0 F1, 1 F1 F 1 such tht 0 F1 1 F1 nd stisfies the five properties of field. Therefore, F 1 is field Is the set F 2 = { + 3, Z} (with the sme ddition nd multipliction) lso field? Suppose tht F 2 is field. Notice tht 1 F2 = (1,0) F 2 with 1,0 Z is the identity element for multipliction since x = (, ) F 2 with, Z, we hve: ( + 3) ( ) = ( + 3) 1 = ( + 3) x 1 F1 = x x F 2 Moreover, we lso note tht 0 F2 = (0,0) F 2 with 0 Z is the identity element for ddition since x = (, ) F 2 with, Z, we hve: ( + 3) + ( ) = ( + 3) + 0 = ( + 3) x + 0 F2 = x x F 2 Since 1 Z, x 1 = (1,1) F 1. Moreover, since x 1 Z, x 2 F 2 such tht x 2 = (, ) with, Z nd x 1 x 2 = 1 F2 such tht x 2 0 F2. Therefore, we hve: (1 + 3) ( + 3) = = ( + ) 3 = = 1 since, Z + = = 1 ecuse 0 0 = = 1 2 = = 1 2 = 1 2 Therefore, x 2 = ( 1 2, 1 2 ) is the inverse of (1,1). This contrdicts the fct tht, Z for x 2 = (, ) F 2. Thus, for x 1 = (1,1) F 2, x 2 F 2 such tht x 1 x 2 = 1 F2. Hence F 2 is not field. 9
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