Unformatted text preview: Section 3.1 57 23. (~) Suppose I is onetoone and Xl =I X2. We must prove that X~ + XtX2 + X~ + a(Xl + X2) + b =I O. Since I is onetoone and Xl =I X2, we know that I(Xl) =I I(X2); thus, x¥ + ax~ + bXl + c =I x~ + ax~ + bX2 + c; equivalently, (xf  x~) + a(x~  x~) + b(Xl  X2) =I O. Since Xl  X2 =I 0, we may divide this last inequality by Xl  X2. Since x¥  x~ = (Xl  X2)(X~ + XlX2 + x~) and x~  x~ = (Xl  X2)(XI + X2), we obtain (x~ + XlX2 + x~) + a(xl + X2) + b =I 0 as required. (+) Suppose that Xl =I X2 implies x~ + XlX2 + x~ + a(xl + X2) + b =I O. We must prove that I is onetoone. Thus, we assume I(Xl) = I(X2) and prove Xl = X2. The equation I(Xl) = I(X2) says x¥ + ax~ + bXl + C = x~ + ax~ + bX2 + c; or, equivalently, as before, (xf  x~) + a(x~  x~) + b(Xl  X2) = O. If Xl =I X2, then we divide by Xl  X2 as before, obtaining x~ + XlX2 + x~ + a(xl + X2) + b = 0, a contradiction. Thus Xl = X2....
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 Summer '10
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 Graph Theory, Inverse function, Domain of a function, Injective function, X~

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