[Discrete Mathematics] Relations Examples

In this video we do some proofs with reflexivity and introduce the notion of irreflexivity.
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Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.

[@JamisonOwenShow]

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

[@ferdiedsouza]

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

[@eimaldorani]

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

[@vishnureddy3977]

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

[@ikemblem2335]

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

[@Prenner2]

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!

[@Trevtutor]

if xRy and yRx, then xRx follows from transitivity, not symmetry.

[@Prenner2]

TheTrevTutor I get it now, thanks so much!

[@rudyeilabouni]

It's like xRy and yRz then xRz, that's transitivity... In this example, instead of z, you have x because of symmetry, but it's the same principle.

[@Prenner2]

Yeah, you're right. Thank you too!

[@yamatanoorochi3149]

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

[@Pages_Perfected]

where is comparison relation ?

[@D17D]

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.

[@kaspinator5265]

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?

[@Trevtutor]

But then there's nothing in the intersection, so it's trivially reflexive.

[@user-wt9br3uz7m]

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.

[@BolaAAyoub]

examinashiny :D

[@MuhammadIsmail-un3qd]

last question is not cleared .....