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Data Str & Algorithm HW Solutions 5

Data Str & Algorithm HW Solutions 5 - 2.2 In...

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2 Mathematical Preliminaries 2.1 (a) Not re f exive if the set has any members. One could argue it is sym- metric, antisymmetric, and transitive, since no element violate any of the rules. (b) Not re f exive (for any female). Not symmetric (consider a brother and sister). Not antisymmetric (consider two brothers). Transitive (for any 3 brothers). (c) Not re f exive. Not symmetric, and is antisymmetric. Not transitive (only goes one level). (d) Not re f exive (for nearly all numbers). Symmetric since a + b = b + a , so not antisymmetric. Transitive, but vacuously so (there can be no distinct a , b ,and c where aRb and bRc ). (e) Re f exive. Symmetric, so not antisymmetric. Transitive (but sort of vacuous). (f) Re f exive – check all the cases. Since it is only true when x = y ,i t is technically symmetric and antisymmetric, but rather vacuous. Like-
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Unformatted text preview: 2.2 In general, prove that something is an equivalence relation by proving that it is re f exive, symmetric, and transitive. (a) This is an equivalence that effectively splits the integers into odd and even sets. It is re f exive ( x + x is even for any integer x ), symmetric (since x + y = y + x ) and transitive (since you are always adding two odd or even numbers for any satisfactory a , b , and c ). (b) This is not an equivalence. To begin with, it is not re f exive for any integer. (c) This is an equivalence that divides the non-zero rational numbers into positive and negative. It is re f exive since x ˙ x > . It is symmetric since x ˙ y = y ˙ x . It is transitive since any two members of the given class satisfy the relationship. 5...
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