To see that r is reflexive suppose that f is an

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To see that R is reflexive, suppose that f is an element of the set A. Then f:[-1,1] is a function. Since we have that f(0) - f(0) = 0 and 0 ε , (f,f) ε R. Since f was arbitrary, it follows that ( f)( f ε A (f,f) ε R ). Thus R is reflexive. To show that R is symmetric, pretend that f and g are members of A with (f,g) ε R. It follows from the definition of the relation that there is some real number C for which f(0) - g(0) = C. Now C ε implies -C ε . Since we have g(0) - f(0) = -C, it follows that (g,f) ε R. Consequently, we have shown that R is symmetric. Finally, to verify that R is transitive, let f, g, and h be elements of A with (f,g) ε R and (g,h) ε R. It follows from the definition of the relation R that there are real numbers, say C 1 and C 2 , such that f(0) - g(0) = C 1 and g(0) - h(0) = C 2 . Since C 1 + C 2 ε when C 1 and C 2 are in , and we have that f(0) - h(0) = f(0) - g(0) + g(0) - h(0) = C 1 + C 2 , (f,h) ε R. Consequently, R is transitive.// Question: How many equivalence classes does this relation have? One, for any two real-valued functions on [-1,1] are equivalent under this relation. Why?? _________________________________________________________________ TIMTOWTDI: The "induction step" in Problem 6 may be handled somewhat differently. If (a,b) ε R, then R reflexive implies that (b,b) ε R. The induction hypothesis, R R n , may then be used to imply that (b,b) ε R n . Then the definition of the composition may be applied to infer that (a,b) ε R n R = R n+1 . Yadda, yadda, yadda ....
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  • Fall '08
  • Graph Theory, Equivalence relation, Transitive relation, Tree traversal

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