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Let R be a binary relation on the set of all positive integers such that R = { (a,b) | a- b is an odd positive integer } Is R reflexive, symmetric, antisymmetric, transitive? Is R an equivalence relation? A partial ordering relation

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(b) R = \{ (a,b) | a-b \ \text{is an odd positive integer }\}

  • Reflexivity. For all x we have (x, x) \notin R because x-x=0 is not a positive integer. The realtion is NOT reflexive.

Symmetry. Let (x, y) \in R. Then x-y is a positive integer and y-x = -(x-y) is negative. Therefore, (y, x) \notin R and the relation is NOT symmetric because the following is always true:

(x, y) \in R \wedge (y, x) \in R \implies x=y

Antisymmetry. Let (x, y) \in R and (y, x) \in R. Then x-y is a positive integer and y-x is a positive integer from the definition. Therefore x-y>0 and y-x>0. There are no such numbers x, y that satisfy x>y and y>x . Therefore, (x, y) \in R \wedge (y, x) \in R is a contradiction and the relation is antisymmetric because the following is always true:

(x, y) \in R \wedge (y, x) \in R \implies x=y

  • Transitivity. Let (x, y) \in R \wedge (y,z) \in R. and x \subseteq y \wedge y \subseteq x. Then for alla \in x we have a \in y, and therefore a \in z. Thus x \subseteq zThen x-y is an odd positive integer and y-z is a odd positive integer from the definition. x-z = (x-y)+(y-z) - 2y. The difference of two even numbers is even, therefore (x,z) \notin R. The relation is NOT transitive.

Answer. R can be neither an equivalence relation nor a partial ordered set because it is at least not reflexive.


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