Equivalence Relations - Reflexive, Symmetric, and Transitive

[@katbutler33]

it feels like you're kindly yelling math at me which is probably the only way I'd learn this, so thank you

[@Heccinchonker12]

the fact that you are writing all this backwards so we can read it well behind the glass is very much appreciated

[@AdityaPoonia-2006]

bruh its a illusion board its called light board

[@chrismcbride5779]

Great explanation, great channel. ..is your hand pink?

[@GabrielTobing]

I just noticed.

[@hello_namoo]

imagine my professor can explain this simple topic within 5 mins

[@blairdelarosa494]

Thanks for explaining, im struggling in abstract algebra and this cleared up alot!

[@vamunki5013]

Caught red handed at being a good explainer

[@liana1063-i4y]

thank you for existing

[@merlinpunnoose8462]

You are a brilliant teacher...God bless you

[@fonzie2668]

Thank you so much for the explanation. Great job

[@Lena-of7wd]

Great explanation this was helpful, thank you!

[@solomonirailoa]

Thank you sir. Do you consider making an Abstract Algebra playlist?

[@AojiCode]

Thank you so much

[@MineCrafterCity]

Is there a video about antisymmetric relationships?

[@Bobintheb0x]

dude, this is very helpful. Thanks!

[@sowmyakalyan2004]

Excellent explanation

[@MathCuriousity]

Can you explain what you mean by the equality operator representing something different in the mod arithmetic example compared to the integers example? I don’t see how the equality operator is any different in mod arithmetic? Please crystallize this hidden knowledge you have attained!

[@lovestonthebeat]

Thanks,, RUSHING OFF TO THE EXAM IN 30M

[@mirzaaghaalikhan183]

Simply the best!

[@moeyk97]

so well explained!! thank you

[@indian7098]

Great Explaination. ,

[@oximas]

what's an example of non equivalent relations?

[@DrTrefor]

Love. I may love my wife, but that doesn't mean she loves me. Not symmetric.

[@xobk]

In my book we have questions like: x is related to y if y ≥ x+1. The domain is the set of all positive real numbers. Is this Transitive or Not Transitive? And the answer is: If y ≥ x+1 and z ≥ y+1, then z ≥ x+2 which implies that z ≥ x+1. Therefore the relation is transitive. I feel like I get the concept, but the problems aren't making sense. I cannot figure out how I am supposed to assume the value of z, or that it even exists. They do not pose that "z ≥ y+1" in the question. So if I am testing values for x, y, and z, I could set x=5, y=6, and z=1, then would it still be transitive?

[@MichaelGoldenberg]

Shouldn't it be: if y≥ x + 1 and x + 1 ≥ z, then y ≥ z ?

[@happypandaface710]

isn't your second example with modular arithmetic just explaining the equality operator again? using this explanation, you can use the equality operator to prove that any other operator is an equivalence relation... which seems wrong.

[@DrTrefor]

Well sort of. The point is that the symbol = does mean something different in modular arithemetic than it does with normal integers. But nevertheless, those differences still make it an equivalence relation which is why we feel embolded to just call it "equality" again. Otherwise, we might want to use a different symbol and mean something different by it.

[@aner7550]

Hi :) I'm a math student, and I'm trying to understand for really long time, why if a relation has those 3 conditions (ref, sem and trans..), he called as "equivalnce relation"? Why are we saying that if xRy so x is equivalent to y ? Where that name came from ? For me thay are just relate to each other but why they are equivalent ? Thank you.

[@donald6749]

This is a stumbling block for me as well. Seems arbitrary and esoteric

[@simonasb1208]

Excellent video.

[@MathCuriousity]

I also have a second pressing question if you are kind enough to answer: if we say 3=6/2 is that an example of reflexive property of equality or is it not since it’s technically not the equality relation of the object 3 over itself but of 3 on 6/2? I do think this would satisfy the symmetrical part though since that’s if ArB is true then bRa is true. Right?

[@trigfunction]

relations are defined on cartesian products like Z^2 or R^2, not on rational numbers like 6/2. What do you mean by 'the equality relation of the object 3 over itself but of 3 on 6/2'? I think you may need to study the definition of a relation more

[@MathCuriousity]

@@trigfunction ok here is my question what is the structure being preserved with homomorphisms? Can you give me an example or better two? I also don’t understand why a function that takes affine space to affine space is not a homomorphism.

[@rameshthoke8297]

nice explanation boss😎

[@lukemiller867]

This shit makes no sense I’m failing my final rn

[@DeezNardsX]

gg no re

[@lukemiller867]

@@DeezNardsX I somehow got a b in this class. Couldn’t tell you how

[@DeezNardsX]

@@lukemiller867 pray for me I have my final in 3 weeks

[@lukemiller867]

@@DeezNardsX good luck bro that final was so hard

[@eyzhie9096]

Is nobody going to talk about how he is actually writing backwards?

[@victorsoderu9491]

Flipped the video while editing

[@brendonong6649]

A hero

[@Shiva-xr6uz]

Sir your explanation is very good 👍

[@DrTrefor]

Thanks!

[@Shiva-xr6uz]

Can I ask you some questions?? Plz ☺️

[@MathCuriousity]

Hey Trev, May I pose a question: let’s say we have an equivalence relation aRb. Why can’t I represent this within set theory as set T comprising subset of Cartesian product of a and b, mapped to a set U which contains true or false? Thanks so much!!

[@suyash8264]

It seems like you are trying to represent an equivalence relation within set theory. The idea of using a subset of the Cartesian product of elements a and b, mapped to a set containing true or false, is on the right track, but there are some details that need clarification. An equivalence relation � R on a set � A typically satisfies three properties for all elements � , � , � a,b,c in � A: Reflexivity: � � � aRa for all � a in � A. Symmetry: If � � � aRb, then � � � bRa. Transitivity: If � � � aRb and � � � bRc, then � � � aRc. Now, let's try to represent this using set theory. The idea is to represent the relation � R as a set of ordered pairs satisfying these properties. A common way to represent an equivalence relation is using equivalence classes. An equivalence class [ � ] [a] of an element � a is the set of all elements related to � a under � R. In set-builder notation, it can be written as [ � ] = { � ∈ � ∣ � � � } [a]={b∈A∣aRb}. To represent the equivalence relation � R as a set, you can use the following approach: Define the set: Let � A be the set on which the equivalence relation is defined. Define the relation set � T: � = { ( � , � ) ∈ � × � ∣ � � � } T={(a,b)∈A×A∣aRb} Define the equivalence class set � U: � = { [ � ] ∣ � ∈ � } U={[a]∣a∈A} where [ � ] = { � ∈ � ∣ � � � } [a]={b∈A∣aRb}. So, � T is the set of ordered pairs related under � R, and � U is the set of equivalence classes. Remember that in set theory, you're dealing with sets and their elements, and the ordered pairs in � T represent elements related under � R. The elements of � U are sets of elements related to each other, forming equivalence classes. The mapping to true or false might not be necessary in this context, as the focus is on capturing the relationships between elements in the equivalence classes.

[@suyash8264]

It seems like you are trying to represent an equivalence relation within set theory. The idea of using a subset of the Cartesian product of elements a and b, mapped to a set containing true or false, is on the right track, but there are some details that need clarification. An equivalence relation � R on a set � A typically satisfies three properties for all elements � , � , � a,b,c in � A: Reflexivity: � � � aRa for all � a in � A. Symmetry: If � � � aRb, then � � � bRa. Transitivity: If � � � aRb and � � � bRc, then � � � aRc. Now, let's try to represent this using set theory. The idea is to represent the relation � R as a set of ordered pairs satisfying these properties. A common way to represent an equivalence relation is using equivalence classes. An equivalence class [ � ] [a] of an element � a is the set of all elements related to � a under � R. In set-builder notation, it can be written as [ � ] = { � ∈ � ∣ � � � } [a]={b∈A∣aRb}. To represent the equivalence relation � R as a set, you can use the following approach: Define the set: Let � A be the set on which the equivalence relation is defined. Define the relation set � T: � = { ( � , � ) ∈ � × � ∣ � � � } T={(a,b)∈A×A∣aRb} Define the equivalence class set � U: � = { [ � ] ∣ � ∈ � } U={[a]∣a∈A} where [ � ] = { � ∈ � ∣ � � � } [a]={b∈A∣aRb}. So, � T is the set of ordered pairs related under � R, and � U is the set of equivalence classes. Remember that in set theory, you're dealing with sets and their elements, and the ordered pairs in � T represent elements related under � R. The elements of � U are sets of elements related to each other, forming equivalence classes. The mapping to true or false might not be necessary in this context, as the focus is on capturing the relationships between elements in the equivalence classes.

[@MathCuriousity]

@@suyash8264 Hey that was AMAZING! My follow up question is more--deep and fundamental! Hopefully you can grasp my question: inspired my whole wish to figure out if set theory can within itself have relations which map certain a set of propositions (for instance the proposition “set A is a subset of Set B) to to a set of truth values (true or false). So in your opinion, what’s the big problem with what I want to do? Basically use relations in set theory to state a proposition is true or false? Or perhaps I’m asking too much of this mapping? Meaning I’m assuming the mapping means “this proposition is true” which is on a meta level actually and Not what the mapping of some proposition to “true” is actually saying?!!!

[@NeverLuckyRubberDucky]

@@suyash8264 chatgpt much?

[@suyash8264]

@@MathCuriousity In classical logic and set theory, propositions can be encoded as sets, and truth values true or false can be represented by the sets {1} and {}, respectively. For example, you might represent the proposition "Set A is a subset of Set B" as a set, and if this set is non-empty, it is considered true; otherwise, it is false. One potential challenge you might encounter is the issue of self-reference and paradoxes. Set theory, particularly when dealing with propositions about sets (like Russell's paradox), can lead to logical contradictions. For instance, the set of all sets that do not contain themselves leads to a paradox. In formalizing propositions within set theory, Godel's incompleteness theorems also come into play. These theorems suggest that there will always be true mathematical statements that cannot be proven within a given formal system, introducing limitations on the completeness and consistency of such systems. Additionally, the level of abstraction involved in dealing with propositions about sets might make it challenging to directly map all propositions to truth values within the set theory framework. Some propositions may refer to concepts outside the formal system, leading to potential limitations. while using set theory to encode propositions and truth values is a well established and valuable approach in mathematical logic, it's essential to be aware of potential pitfalls related to self reference, paradoxes, and the limits imposed by Godel's incompleteness theorems.

[@randomkeyboardsmash]

Great Explanation!

[@archanavarun6843]

I'm still scratching my head, how could you write, mirror language, so consistently??

[@halaammar1833]

He just flipped the video

[@Arvanart]

What about acyclic relations

[@sowmyakalyan2004]

Can you clear a doubt for me?

[@olliesasiancollectables]

is m+n=8 relfexive?

[@albertmargaryan8390]

ignore all the good comments👐👐🗣🗣🗣

[@hardXcoreminecraft]

why are you yelling

[@tomkhxde3495]

wtf honestly :D if the relation is mod(%) and we have lets say (a,b) and (b,a) symmetry, that would imply a%b = b%a in your case 10%3 = 3%10 which is not tru, same with the transitivity :D

[@tomkhxde3495]

okay my bad, did some research and %3 is a relation itself, thus i correct myself, a%3 = a%3, reflexivity a%3 = b%3 => b%3 = a%3, symmetry if a%3 = b%3 and b%3 = c%3 then c%3 = a%3

[@malikialgeriankabyleswag4200]

these are just words you didnt clear things up for me

[@piglovesasy]

He's too excited and loud, kind of hurts my ear a little. But I guess it's better than my professor's monotone.

[@darkknight3305]

Thank you very much