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Unformatted text preview: x R we have: f 2 ( x ) + g 2 ( x ) =  [ f ( x ) + g ( x )] 22 fg   f ( x ) + g ( x )  2 + 2  fg  M 2 1 + 2 M 2 . This implies that  f ( x )  = p f 2 ( x ) p f 2 ( x ) + g 2 ( x ) q M 2 1 + 2 M 2 , and thus f is bounded. Similarly, g is also bounded. 3. No it is not a eld. The number 1, for instance, does not have a negative. 4. We showed in class that a b = 1, and we will use this fact here. The element a 2 must be either 0 , 1 ,a or b . We check each of the options: If a 2 = a a = 0, then from the properties of a eld we get that a = 0, which is impossible. If a 2 = a , we can multiply both sides by a1 to get a = 1, which is impossible. If a 2 = 1, then a 2 = a b , and we can multiply both sides by a1 to get b = 1, which is impossible. The only option left is therefore a 2 = b ....
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 Spring '10
 Math

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