What are naturally occuring examples of relations that satisfy two of the following properties, but not the third: symmetric, reflexive, and transitive.
Hello, terr13!
So far, I have two of the examples . . .
What are naturally occuring examples of relations that satisfy two of the
following properties, but not the third: Reflexive, Symmetric, and Transitive?
Let $\displaystyle \circ$ = "is a brother of"
. . (The first is male, and both parties are children of the same parents.)
Reflexive: .$\displaystyle x \circ x$
. . $\displaystyle x\text{ is a brother of himself . . . True.}$
Symmetric: .$\displaystyle \text{If }x \circ y\text{, then }y \circ x$
. . $\displaystyle \text{If }x\text{ is a brother of }y\text{, then }y\text{ is a brother of }x.$
Not necessarily true . . . $\displaystyle y\text{ could be a {\bf sister} of }x.$
Transitive: .$\displaystyle \text{If }x \circ y\text{ and }y \circ z\text{, then }x \circ z$
. . $\displaystyle \text{If }x\text{ is a brother of }y\text{ and }y\text{ is a brother of }z\text{ , then }x\text{ is a brother of }x\text{ . . . True}$
The relationship $\displaystyle \circ$ is reflective and transitive, but not symmetric.
Let $\displaystyle \star$ = "knows" (is acquainted with).
Reflexive: .$\displaystyle x \star x$
. . $\displaystyle x\text{ knows himself.}\quad\hdots$ .True.
Symmetric: .$\displaystyle \text{If }x \star y\text{, then }y \star x$
. . $\displaystyle \text{If }x\text{ knows }y\text{, then }y\text{ knows }x\quad\hdots$ .True
Transitive: .$\displaystyle \text{If }x \star y\text{ and }y \star z\text{, then }x \star z$
. . $\displaystyle \text{If }x\text{ knows }y\text{ and }y\text{ knows }z\text{, then }x\text{ knows }z.\quad\hdots$ .not necessarily true
The relationship $\displaystyle \star$ is reflexive and symmetric, but not transitive.
Thanks for the fast reply, but the one I still have the most trouble with is finding one that is not reflexive. For not symmetric, I was thinking of using $\displaystyle \leq$. The problem I have with non reflexive is if we say the relation is !, and we have x!y and y!x, if x!y, and y!z, then x!z. But if we look at those two, we can use the symmetric relation in the transitive one and say if x!y, and y!x, then x!x, which proves reflexiveness.
Let $\displaystyle \circ$ = "is a brother of"
. . (The first is male, and both parties are children of the same parents.)
Reflexive: .$\displaystyle x \circ x$
. . $\displaystyle x\text{ is a brother of himself . . . True.}$
This is not standard English usage. If a mother and father have three children, all male, and you ask one of them "How many brothers do you have?", he will answer "Two" not "Three".
The last sentence is fallacious. Maybe there is an x for which x!y is false for all y. Example: the usual definition of "divides" on all integers requires that m|n if there is a unique integer q for which n = qm. The result is that m|m for every nonzero integer, but 0 does not divide 0, so the relation is not reflexive.
An easier way to come up with counterexamples is to look at small finite relations. For this problem, try the relation R on {1,2} defined by 2R2 (and nothing else).
This sort of relation is called a "partial equivalence relation" and is a big deal in theoretical computer science. You can start learning about it from Wikipedia here:
Partial equivalence relation - Wikipedia, the free encyclopedia