[Discrete Mathematics] Relations Examples

In this video we do some proofs with reflexivity and introduce the notion of irreflexivity.

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[eimaldorani]

@2:55 x!=y and y!=z for distinct x, y and z is symmetric and transitive but not reflective.

[wtbruzm]

Let A = {#, 7, x} and let R be the relation on A given by R = {(x, x), (7, 7), (#, #)}. Prove or disprove reflexivity, symmetry, and transitivity of R.

[ferdiedsouza]

@3:50 . Symmetry + transitivity works only when its defined for all elements in the set , else the reflexive relation might fail .

[JamisonOwenShow]

Do you have any videos that go more in depth on equivalence relations and transitive closure?

[DD]

2:32 I don't understand the symmetricity part for question i).

Using the definition of symmetricity:

forall x, y in A ( ((x, y) in R) then ((y, x)) in R) )

If we take x = 2 and y = 3 and try: (2 < 3) -> (3 < 2), antecedent is true, and consequence is false, so implication is false so thus it is not symmetric. However, if we take x = 3 and y = 2 and try: (3 < 2) -> (2 < 3), antecedent is false, and consequence is true, so implication is true so now it is symmetric? For relations including strict inequalities, is there a rule that x < y < z? I'm not quite sure where I'm messing up.

[yamatanoorochi]

In the first question, 0:00
It's implicit that R1 and R2 are from within the same set?

[Prenner]

In the last part ("Assume: ..."), you said that it implies xRx "by transitivity", but isn't it by symmetry?
Because also the example above could only be irreflexive because it was not symmetric, or do I get something wrong?

And thank you so much for your videos! They help a lot!

[vishnureddy]

Isn't the empty relation transitive, symmetric, and irreflexive?

[kaspinator]

Isn't it possible to have (x, x) in R1, and (y, y) in R2? This way both R1 and R2 are reflexive, but the intersection would not be, making the statement false?

[ikemblem]

Does anyone know what it means to "define the relation"? Been searching everywhere and can't seem to find it.