Reflexive, Symmetric, Transitive

• iHeartof12
In summary: In other words, find three elements x,y,z such that (x,y)R(z,w) and (z,w)R(u,v) but not (x,y)R(u,v).In summary, the relation on the set ℝxℝ is reflexive and symmetric, but it is not transitive. To show that it is reflexive, we need to show that (x,y)R(x,y) for all real x,y. For symmetry, we ask if (x,y)R(z,w) implies (z,w)R(x,y). And to prove that it is not transitive, we need to find a counterexample where (x,y)R(z,w) and (z,w)R(u,v)
iHeartof12
Indicate if the following relation on the given set is reflexive, symmetric, transitive on a given set.

R where (x,y)R(z,w) iff x+z≤y+w on the set ℝxℝ.

It is reflexive because any real number can make x+z=y+w.
It is not symmetric because if x+z≤y+w it's not possible for x+z≥y+w.
It is transitive

Thank you

To show something is reflexive, you need to show that a R a for all a in the set.

So, does (x,y) R (x,y) for all real x,y?When showing symmetry, do not reverse the sign. For symmetry, we ask: Given (x,y)R(z,w), is (z,w)R(x,y)? Well, (x,y)R(z,w) implies that x+z<=y+w. And (z,w)R(x,y) implies that z+x<=w+y. So knowing that x+z<=y+w, is z+x<=w+y?

(x,y)R(x,y) is true for all real x,y so the relation is reflexive.
z+x≤w+y so (z,w)R(x,y) and the relation is symmetric.

How would I show that the relation isn't transitive?

iHeartof12 said:
(x,y)R(x,y) is true for all real x,y
All you have done here is to write down the statement that you're supposed to prove or disprove. You also need to do the actual proof. Use the definition of R to find out if that statement is true or not.

iHeartof12 said:
z+x≤w+y so (z,w)R(x,y) and the relation is symmetric.
Why is z+x≤w+y? What are w,x,y,z anyway? If you're going to make a statement that involves a variable x, you need to do one of the following:

1. Assign a value to x before you make the statement.
2. Make it very clear that you're making a "for all x..." statement. (It's sufficient to say something like "let x be an arbitrary real number").
3. Make it very clear that you're making a "there exists an x..." statement.

Make a mental note of this. You need to apply this principle every time you try to prove something.

iHeartof12 said:
How would I show that the relation isn't transitive?
Find a counterexample.

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What is a reflexive relation?

A reflexive relation is a relation where every element is related to itself. In other words, for every element a in the set, (a,a) is a part of the relation. This is often represented as a diagonal line in a matrix.

What is a symmetric relation?

A symmetric relation is a relation where if (a,b) is a part of the relation, then (b,a) is also a part of the relation. In other words, the relation is bidirectional and both elements are related to each other. This is often represented as a mirrored image in a matrix.

What is a transitive relation?

A transitive relation is a relation where if (a,b) and (b,c) are both part of the relation, then (a,c) is also a part of the relation. In other words, if there is a chain of relationships between elements, the end points are also related. This is often represented as a chain in a matrix.

How are reflexive, symmetric, and transitive related?

These three properties are often used to describe the behavior of relations. A relation can have none, some, or all of these properties. A relation that is reflexive, symmetric, and transitive is called an equivalence relation.

Why are reflexive, symmetric, and transitive important in mathematics?

These properties are important because they help define and classify different types of relations. They also have many applications in fields such as graph theory, computer science, and social sciences. In addition, they are used to prove theorems and solve problems in mathematics.

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