# Reflexive, Symmetric, Transitive

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

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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?

Fredrik
Staff Emeritus
Gold Member
(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.

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.

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

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