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Unformatted text preview: !} and I has an inverse B A: To find a formula for 11 (x), let y = Il(x), x E B. Then x = I(y) = 2y3! l' so 2xy +x = 3y and y(2x  3) = x. Since x x E B, we know 2x  3 # 0; thus, y = 2x _ 3 = Il(x). Xl 3 X2  3 (d) Suppose I(Xl) = I(X2). Then 3 = 3' SOXlX2+3x l3x 29 = XlX23xl +3X 29, Xl + X2 + 6Xl = 6X2 and Xl = X2. Thus I is onetoone. Next y E rngl ++ y = I(x) for some x E A x3 ++ there is an x E A such that y = 3 . x+ ++ there is an x E A such that y(x + 3) = x  3 ++ there is an x E A such that x yx = 3y + 3 ++ there is an x E A such that x(1 y) = 3(y + 1) ++y#1. Thus rngl = B = {y E R I y # 1} and I has an inverse B A. Let y = Il(x) with x E B. Then x = I(y) = y  3, so x(y + 3) = y 3, xy + 3x = y 3, y(1 x) = 3(x + 1). Since y+3 ...J. 1 Il( ) _ _ 3(1 + x) x,, x y1 . x...
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 Summer '10
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 Graph Theory

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