I am working out of a textbook for self study past what will be covered in a class I am taking, and I am wondering if someone could explain a couple of questions to that I am just not grasping.
The question asks;
For each of the following relations, say which of the properties of reflexivity, symmetry, and transitivity are satisfied and which fail.
a) ≤ on Z.
b) Disjointness on P(N).
c) R on Z defined by: mRn iff m + n is even.
d) R on R defined by xRy iff xy ≥ 0. e) ⊥ (perpendicularity) on the set, L, of all straight lines in the plane.
As I said, this is something I am trying to teach myself before I take an introductory analysis class in the fall, if anyone could walk me through these solutions they are not in the textbook that I have and I am stumped. Thanks!
Edit/Update - I have been able to solve the last two problems and will post my solutions, if any of you could clarify items 6 and 8 I would greatly appreciate that. Thanks!
(11) R on R defined by xRy iff xy ≥ 0. Let R = the set of all reals, with relation defined by xRy iff xy ≥ 0. Since mm ≥ 0 for all m belonging to R, thus mRm, so the relation is reflexive. Then, mRm ≥ 0 and then mn ≥ 0, so nRm true so R is symmetric. However the relation R is not transitive, since -2R0, 0R3, that is (-2)0 ≥ 0 and 0(3) ≥ 0 but (-2)(3) ≥ 0, Is not true, so -2 cannot be related to 3. So for the relation in 11, we can say that it is reflexive, symmetric, but not transitive.
(13) ⊥ (perpendicularity) on the set, L, of all straight lines in the plane. L = the set of all lines in the plane, with relation defined by mRm iff m and n are perpendicular. Since m is not perpendicular to itself, for all m belonging to L, thus mRm is not possible so the relation is not reflexive. Now mRn, then m and n are perpendicular and n and m are perpendicular, hence nRm is true, so we can say that R is symmetric. The relation R is not transitive, since mRn, nRp are perpendicular, and n and p are perpendicular implies m and p are parallel, so m and p are not perpendicular, so mRp is not possible.
I'll do the first one for you and give you a method of attack for the rest of them.
So recall that a relation $R$ is reflexive when $x R x$, it is symmetric when $x R y$ implies $y R x$, and transitive when $x R y$ plus $y R z$ tells you that $x R z$.
Since $x \le x$, we can say that $\le$ is reflexive. However, $2 \le 3$ while $3 \not\le 2$, so we can say that it is not symmetric. On the other hand, if $x \le y$ and $y \le z$, we do know that $x \le z$, so $\le$ is transitive.
To show that a property does hold, you need to argue in general. To show it does not, you only need to find an example where it does not.