Linear Algebra Solutions 7

# Linear Algebra Solutions 7 - ab must be one of 1 a b...

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are distinct elements of F by virtue of the cancellation law for addition . For this law states that 1+ x = 1+ y x = y and hence x 6 = y 1+ x 6 = 1+ y . Hence the above four elements are just the elements 0 , 1 , a, b in some order. Consequently (1 + 0) + (1 + 1) + (1 + a ) + (1 + b ) = 0 + 1 + a + b (1 + 1 + 1 + 1) + (0 + 1 + a + b ) = 0 + (0 + 1 + a + b ) , so 1 + 1 + 1 + 1 = 0 after cancellation. Now 1 + 1 + 1 + 1 = (1 + 1)(1 + 1), so we have x 2 = 0, where x = 1 + 1. Hence x = 0. Then a + a = a (1 + 1) = a · 0 = 0. Next a + b = 1. For a + b must be one of 0 , 1 , a, b . Clearly we can’t have a + b = a or b ; also if a + b = 0, then a + b = a + a and hence b = a ; hence a + b = 1. Then a + 1 = a + ( a + b ) = ( a + a ) + b = 0 + b = b. Similarly b + 1 = a . Consequently the addition table for F is + 0 1 a b 0 0 1 a b 1 1 0 b a a a b 0 1 b b a 1 0 . We now ±nd the multiplication table. First,
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Unformatted text preview: ab must be one of 1 , a, b ; however we can’t have ab = a or b , so this leaves ab = 1. Next a 2 = b . For a 2 must be one of 1 , a, b ; however a 2 = a ⇒ a = 0 or a = 1; also a 2 = 1 ⇒ a 2-1 = 0 ⇒ ( a-1)( a + 1) = 0 ⇒ ( a-1) 2 = 0 ⇒ a = 1; hence a 2 = b . Similarly b 2 = a . Consequently the multiplication table for F is × 1 a b 1 1 a b a a b 1 b b 1 a . (b) We use the addition and multiplication tables for F : A = 1 a b a a b b 1 1 1 1 a R 2 → R 2 + aR 1 R 3 → R 3 + R 1 1 a b a a a b a 10...
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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