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 March 7th, 2013, 10:01 AM #1 Newbie   Joined: Sep 2012 Posts: 17 Thanks: 0 Order relation I have the following relation: $R= \{(x,y) \in \mathbb{R} \times \mathbb{R} | |x| < |y| \vee x= y\}$ Show that R is partially ordered. Is it totally ordered? My solution: It's not a total order since $\forall x \in \mathbb{R}: ( \neg(|-x| < |x|) \wedge -x \neq x)$ A relation is partially ordered iff it is reflexive, antisymmetric and transitive. Reflexitivity: $|x|=< |x| \vee x= x$. This is a tautology, hence R is reflexive. Symmetry: $(|x| < |y| \vee x = y) \wedge (|y| < |x| \vee y = x) \Leftrightarrow x = y \vee (|x| < |y| \wedge |y| < |x|) \Leftrightarrow x = y$ This statement is true iff x = y. That means R is antisymmetric. Transitivity: $(|x| < |y| \vee x = y) \wedge (|y| < |z| \vee y = z) \text{case 1:} x = y \Rightarrow x=y \wedge (|y| < |z| \vee y = z) \Leftrightarrow x=z \vee |x| < |z|. \text{case 2:} x \neq y \Rightarrow |x| < |y| \wedge (|y| < |z| \vee y = z) \Leftrightarrow |x| < |z| \vee x = z$ Thus R is transitive. Since R is reflexive, antisymmetric and transitive, it is partially ordered. However I'm very unsure on the transitive part. Thanks for help.

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