ds-t3-a

# A full 3 ary tree with 24 internal vertices must have

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A full 3-ary tree with 24 internal vertices must have V =( 3 )( 2 4 )+1=7 3 vertices. Since 24 are internal, there must bel=7 3-2 4=4 9 leaves. _________________________________________________________________ 8. (10 pts.) Suppose that A is the set consisting of all real- valued functions with domain consisting of the interval [-1,1] . Let R be the relation on the set A defined as follows: R = { (f,g) ( C)( C ε and f(0) - g(0) = C) } Prove that R is an equivalence relation on the set A. Proof. We must establish that R is reflexive, symmetric, and transitive to show that it is an equivalence relation. 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 2 , (f,h) ε R. Consequently, R is transitive.// Question: How many equivalence classes does this relation have?
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