Unformatted text preview: θ [ a ] 6 = [ a ] 3 , where a ∈ Z is arbitrary. Suppose [ a ] 6 = [ b ] 6 . Then 6  ( ab ) , hence 3  ( ab ) and we deduce that [ a ] 3 = [ b ] 3 . Thus if we write [ a ] 6 in different ways, we always get the same result when we apply the given formula for θ . Therefore θ is welldeﬁned. (b) Let a , b ∈ Z . Then θ ([ a ] ⊕ [ b ]) = θ [ a + b ] = [ a + b ] = [ a ] ⊕ [ b ] = θ [ a ] ⊕ θ [ b ] . (c) [ a ] ∈ ker θ if and only if [ a ] 3 = [ ] ; that is if and only if 3  a . Therefore ker θ consists of the elements of the form [ a ] 6 where 3  a , in other words { [ ] , [ 3 ] } ....
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 Fall '08
 PARRY
 Math, Algebra, Konrad Zuse, The Elements, ungraded homework

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