Discrete Mathematics with Graph Theory (3rd Edition) 59

Discrete Mathematics with Graph Theory (3rd Edition) 59 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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

Ask a homework question - tutors are online