- Let \(X\) be a set and \(R\) is a relation \(X\). Define \(xRy\) if \(x|y\). Is \(R\) an equivalence relation, partial ordering relation or totally ordering relation?
Solution:
(1) equivalence?
- Reflexivity:
\(x\mid x\) is true for any \(x\in X\), so \(R\) is reflexive. - Symmetry:
If \(x\mid y\), it does not necessarily imply \(y \mid x\). For example, if \(x = 2\) and \(y = 4\), \(2 \mid 4\), but \(4 \nmid 2\). Thus, \(R\) is not symmetric. - Transitivity:
If \(x\mid y,~y\mid z\Rightarrow x\mid z\), so \(R\) is transitive.
Since \(R\) is not symmetric, it is not an equivalence relation.
(2) partial ordering relation?
-
Reflexivity:
As shown earlier, \(R\) is reflexive.
-
Antisymmetry:
If \(x \mid y\) and \(y \mid x\), then \(x = y\) (since \(x \mid y\) and \(y \mid x\) imply \(x\) and \(y\) have the same absolute value). Thus, \(R\) is antisymmetric. -
Transitivity:
As shown earlier, \(R\) is transitive.Since \(R\) satisfies reflexivity, antisymmetry, and transitivity, it is a partial ordering relation.
(3) totally ordering relation?
- For some \(x, y \in X\), neither \(x \mid y\) nor \(y \mid x\) may hold. For example, if \(x = 2\) and \(y = 3\), \(2 \nmid 3\) and \(3 \nmid 2\). Thus, \(R\) is not a total ordering relation.