Table of Contents
Is the relation R on the set Q of rational numbers?
Let R be a relation on the set Q of all rationals defined by R={(a,b):a,binQ” and “a-binZ}. Show that R is an equivalence relation. Given:R={(a,b):a,b∈Q and a-b∈Z}. So, R is reflexive.
How do you find the total number of reflexive relations?
Reflexive Relation Formula The number of reflexive relations on a set with the ‘n’ number of elements is given by N = 2n(n-1), where N is the number of reflexive relations and n is the number of elements in the set.
What is set Q0?
Q0 means set of rational number except zero.
How identity relation is an equivalence relation?
Definition An equivalence relation on a set A is one which is reflexive, symmetric, and transitive. This is because an equivalence relation behaves like the identity relation (the equality relation) on A. It lets things be similar without being equal.
When a relation R on a set A is said to be reflexive?
In Maths, a binary relation R across a set X is reflexive if each element of set X is related or linked to itself. In terms of relations, this can be defined as (a, a) ∈ R ∀ a ∈ X or as I ⊆ R where I is the identity relation on A.
What is a reflexive relation in sets?
Reflexive relation on set is a binary element in which every element is related to itself. R is set to be reflexive, if (a, a) ∈ R for all a ∈ A that is, every element of A is R-related to itself, in other words aRa for every a ∈ A.
How do you find the relation between xRy and xRz?
The relation R on the real numbers given by xRy iff number x − y ∈ Q. This is what I did. y − x ∈ Q thus yRx. Let xRy be x − y ∈ Q, then let yRz be y − z ∈ Q. Thus x − z ∈ Q and in conclusion xRz showing transitive.
Why is the square root of a rational number always non-negative?
The square of a rational number is always non-negative and therefore 1+xy is greater than 0. Hence for any a, (a,a) satisfies the relation. Thus it’s reflexive Thus if (x,y) belongs to the relation, then so does (y,x). Hence it’s symmetric ⟹ x and z need to have the opposite sign. Let’s take equation B, 1+yz>0.
What is the equivalence class of X in this equation?
The equivalence class of x is the set of all y such that x R y. So we are looking for all y such that y − x = q is rational. Solving this equation, we’re looking for all numbers of the form x + q, for q rational.
Is it possible to prove that a relation is not transitive?
Since y is negative and z is positive, that makes yz negative. To summarise, as long as we can find x,y and z such that x < 0, y < 0, z > 0 and | x z | > 1 , | y z | < 1 we have our combination of x,y and z which can prove that this relation ain’t transitive. Is this possible – yes it is. For example, let’s say x = -1/3, y=-1/6, z=5.