Properties of+and· Proof(continued). (5)Suppose u0 and c1u=c2-u.Hence (c1-c2)u=0(**).We want to prove c1-c2 OinR.Suppose not.Then (c1-c2)-1ER exits.Multiply both sides of *by(c1-c2)-1 to get (c-c2)-1.(c1-c2)u)=(c1-c2)-1.0=0. LHS=(c1-c2)-1(c1-c2)u)=1u=u. But RHS=0 by (3)that we just proved.Hence u=0, contradicting our assumption that u0.Hence,our assumption that c1-c20 has led to a contradiction.Hence c1-c2 =0 and C1=C2. 4口+++左+4生+定QCProperties of + and · Proof (continued). (5) Suppose u 6= 0 and c1·u = c2·u. Hence (c1 − c2)·u = 0 (∗∗). We want to prove c1 − c2 = 0inR. Suppose not. Then (c1 − c2) −1 ∈ R exits. Multiply both sides of ∗∗ by (c1 − c2) −1 to get (c1 − c2) −1 ·((c1 − c2)·u) = (c1 − c2) −1 ·0 = 0. LHS = ￾ (c1 − c2) −1 (c1 − c2)·u
= 1·u = u. But RHS = 0 by (3) that we just proved. Hence u = 0, contradicting our assumption that u 6= 0. Hence, our assumption that c1 − c2 6= 0 has led to a contradiction. Hence c1 − c2 = 0 and c1 = c2