Introduction
In this article, we will analyze the relation "～" defined on the set of real numbers $R$. The relation is defined such that for any x, y ∈ $R$, x is related to y (denoted by x ～ y) when x · y 1. We will determine if this relation is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, serial, semi-connex, connex, or transitive.
Notation
In this context, "·" denotes multiplication.
Reflexivity
A relation is reflexive if every element is related to itself. For the relation x ～ y, we must check if for every x ∈ R, x ～ x.
[x ～ x text{ if and only if } x cdot x 1]
If x 0, then 0 cdot 0 eq 1, and if x ≠ 0, then to satisfy x cdot x 1, x must be 1 or -1. Therefore, not all x ∈ R satisfy x cdot x 1. Hence, the relation is not reflexive.
Irreflexivity
A relation is irreflexive if no element is related to itself. For the relation x ～ y, we must check if for every x ∈ R, x ～ x is false.
[x ～ x text{ if and only if } x cdot x 1]
As previously established, this only holds for x 1 or x -1. Therefore, the relation is not irreflexive.
Symmetry
A relation is symmetric if whenever x is related to y, y is also related to x. For the relation x ～ y, we must check if for every x, y ∈ R, x ～ y implies y ～ x.
[x ～ y text{ if and only if } x cdot y 1 text{ and } y ～ x text{ if and only if } y cdot x 1]
Since multiplication is commutative, (x · y 1) implies (y · x 1). Therefore, the relation is symmetric.
Asymmetry
A relation is asymmetric if there is no x such that x is related to y and y is related to x. For the relation x ～ y, this would mean there is no x, y such that x ～ y and y ～ x simultaneously.
[x ～ y text{ and } y ～ x text{ if and only if } (x cdot y 1 text{ and } y cdot x 1)]
Since multiplication is commutative, the condition (x · y 1) is the same as (y · x 1). Therefore, the relation is not asymmetric.
Antisymmetry
A relation is antisymmetric if for every x, y, if x is related to y and y is related to x, then x y. For the relation x ～ y, this would mean there is no x, y such that (x · y 1) and (y · x 1) and (x ≠ y).
[x ～ y text{ and } y ～ x text{ implies } x y text{ if and only if } x cdot y 1 text{ and } y cdot x 1 text{ and } x y]
Again, since (x · y 1) is the same as (y · x 1), and (x ≠ y) is not possible, the relation is not antisymmetric.
Seriality
A relation is serial if for every x, there exists a y such that x is related to y. For the relation x ～ y, this would mean for every x ∈ R, there exists a y ∈ R such that x ～ y.
[x ～ y text{ if and only if } x cdot y 1]
To satisfy x cdot y 1, y should be the multiplicative inverse of x, i.e., y 1/x. This inverse exists for all x ≠ 0, but for x 0, no such y exists. Therefore, the relation is not serial.
Semi-Connexivity
A relation is semi-connex if for every x, y ∈ R, either x is related to y or y is related to x. For the relation x ～ y, we must check if for every x, y ∈ R, either x ～ y or y ～ x.
[x ～ y text{ if and only if } x cdot y 1 text{ and } y ～ x text{ if and only if } y cdot x 1]
This is the same as checking if for every x, y ∈ R, either x cdot y 1 or y cdot x 1. If x 0, there is no y such that 0 · y 1, and if y 0, there is no x such that x · 0 1. Therefore, the relation is not semi-connex.
Connexity
A relation is connex if for every x, y ∈ R, either x is related to y or y is related to x, and also x y or x ≠ y. For the relation x ～ y, we must check if for every x, y ∈ R, either x ～ y or y ～ x, and also x y or x ≠ y.
[x ～ y text{ if and only if } x cdot y 1 text{ and } y ～ x text{ if and only if } y cdot x 1]
Similar to semi-connexity, this is the same as checking if for every x, y ∈ R, either x cdot y 1 or y cdot x 1. As before, for x 0, there is no y such that 0 · y 1, and for y 0, there is no x such that x · 0 1. Therefore, the relation is not connex.
Transitivity
A relation is transitive if for every x, y, z ∈ R, if x is related to y and y is related to z, then x is related to z. For the relation x ～ y, we must check if for every x, y, z ∈ R, if x ～ y and y ～ z, then x ～ z.
[x ～ y text{ and } y ～ z text{ implies } x ～ z text{ if and only if } (x cdot y 1 text{ and } y cdot z 1) implies (x cdot z 1)]
If x cdot y 1 and y cdot z 1, then we can multiply the two equations to get (x cdot y) cdot (y cdot z) 1 cdot 1 implies x cdot (y cdot y) cdot z 1 implies x cdot 1 cdot z 1 implies x cdot z 1. Therefore, the relation is transitive.