This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Section 3.1 57 23. (~) Suppose I is one-to-one and Xl =I- X2. We must prove that X~ + XtX2 + X~ + a(Xl + X2) + b =I- O. Since I is one-to-one 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 one-to-one. 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....
View Full Document
- Summer '10
- Graph Theory