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## Here's the Solution to this Question

(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 all$a \in x$ we have $a \in y$, and therefore $a \in z$. Thus $x \subseteq z$Then $x-y$ 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.