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Unformatted text preview: xz = xy + yz . Therefore, if xy is divisible by 2 and yz is divisible by 2 then xz = xy + yz is divisible by 2. 1 This proves that x ∼ y is an equivalence relation. #13. It should be easy to see that g ◦ f is such that ( g ◦ f )( a ) = t , ( g ◦ f )( b ) = s , ( g ◦ f )( c ) = t , ( g ◦ f )( d ) = r . #15. We have: f ( x ) = sin x , g ( x ) = e 2 x , h ( x ) = x 3 + 1. Therefore, ( f ◦ g )( x ) = sin( e 2 x ). ( g ◦ f )( x ) = e 2 sin x . ( f ◦ g ◦ h )( x ) = f ( g ◦ h )( x ) = f ( e 2( x 3 +1) ) = sin( e 2( x 3 +1) ). ( h ◦ g ◦ f )( x ) = h ( g ◦ f ) = h ( e 2 sin x ) = ( e 2 sin x ) 3 + 1 = e 6 sin x + 1. 2...
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 Spring '11
 astina
 Math, Set Theory, Natural Numbers, Sin, Natural number, Equivalence relation, Prime number, Transitive relation

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