(see instructions for Exercise 6 in Section 2.2): PROOF EVALUATION a) "THEOREM": For any sets A, B, and C, if A UB=AU C, then B = C. "Proof": We will prove this by contradiction. Suppose that A UB and A UC are not equal. Then there is some object x that is in one and not the other. We proceed by looking at two cases: First look at the case where xEA UB and not in A U C. Then x is not in A (be- cause, if it were, it would also be in A U C). So x must be in B (otherwise it wouldn't be in A U B). Also, x is not in C (because, if it were, it would also be in A U C). Therefore x is in B and not in C, which contradicts the condition B = C. The case where x EAU C and not in A U B is done by a parallel argument.
(see instructions for Exercise 6 in Section 2.2): PROOF EVALUATION a) "THEOREM": For any sets A, B, and C, if A UB=AU C, then B = C. "Proof": We will prove this by contradiction. Suppose that A UB and A UC are not equal. Then there is some object x that is in one and not the other. We proceed by looking at two cases: First look at the case where xEA UB and not in A U C. Then x is not in A (be- cause, if it were, it would also be in A U C). So x must be in B (otherwise it wouldn't be in A U B). Also, x is not in C (because, if it were, it would also be in A U C). Therefore x is in B and not in C, which contradicts the condition B = C. The case where x EAU C and not in A U B is done by a parallel argument.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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