# hw4 - ba = 1 Now use the property of homomorphism we see f...

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MATH 3360 HOMEWORK SOLUTION 4 (Some computational problems are omitted.) p121, E3 It is equivalent to prove that, if ax = b has one solution x 0 , then it must have another. Say ac = 0, then x 0 + c is also a solution of ax = b . Why is x 0 + c diﬀerent from x 0 ? Because c 6 = 0. p122, E8 Easy. 1,2,4,5,7,8. p123, E10 a × 0 = a × (1 - 1) = a × 1 - a × 1 = a - a = 0. This problem is NOT trivial. p123, E17 Given ar = as , we can multiply both sides by a - 1 , the inverse of a . p124, E18 0 = β 2 - 1 = ( β - 1)( β + 1), and R has no zero divisor, so either β - 1 = 0 or β + 1 = 0. p133, E5 Since a is a unit in R , there is a b such that
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Unformatted text preview: ba = 1. Now use the property of homomorphism, we see f ( b ) f ( a ) = f (1) = 1, so f ( b ) is the inverse of f ( a ) in S . p133, E7 Easy check. p133, E11 F has characteristic 2 means that x + x = 0 (don’t write 2 x here) for any x . So ( i ) easily holds, and ( ii ) follows from: ( a + b ) 2 = a 2 + ab + ba + b 2 = a 2 + b 2 (and don’t write 2 ab here). p138, E16 See hint. p141, E12 Note that n 5-n 5 and n 3-n 3 are integers by Fermat’s Theorem. 1...
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