The map for R+ to R+^(2×2) is x|-> x/2 * J, where J is the all 1s matrix
@Grassmpl5 жыл бұрын
Exercise for the viewer. Repeat this problem in the case where 2×2 matrices are replaced with general n×n matrices. Of course n in Z+ is fixed.
@salvatorezungri69455 жыл бұрын
@@Grassmpl x|------> x/n * J where J is the all 1s matrix of order n?
@MuPrimeMath5 жыл бұрын
I just finished a lecture on isomorphisms in graph theory one hour ago, and now this video shows up in my recommended. Next week I will try to figure out who in my class works at Google.
@EpicMathTime5 жыл бұрын
Hey man, hope you liked it! Congrats on your big integral win. 💪
@mrnogot42514 жыл бұрын
My algebra teacher explained isomorphism as follows “It’s like when a friend of yours shows up to your house wearing a mask and you are kind of scared at first but then you realize: oh it’s just my friend.”
@boonewalker39733 жыл бұрын
We may be the most significant isomorphic sets
@EpicMathTime5 жыл бұрын
"12:12 Hey, your B is not a group!" *yes it is* "but none of those matrices are inve..." *it's a group tho* "but the identity matrix isn't even in th..." *shhhhhh, only dreams now.*
@Grassmpl5 жыл бұрын
Yes I agree. The trick here is the standard identity matrix is NOT the identity of this group. Since none of the matrices have full rank, we are off the hook.
@poutineausyropderable71085 жыл бұрын
+Epic Math Time could you say that The integral of f(x) dx and f(y)dy is isomorphic if the only difference between the two is the symbol?
@EpicMathTime5 жыл бұрын
@@poutineausyropderable7108 I mean, those two things are really one-the-nose equal.
@Grassmpl5 жыл бұрын
The term "isomorphism" is ambiguous unless the intended underlying equivalence relation is well defined
@DrKjoergoe5 жыл бұрын
Maybe I'm not getting the irony but how is this a group?
@quantumgaming91802 жыл бұрын
I'm going to be honest. I just started college and you explained at the same level, or better maybe, as my teachers. Insanely underrated channel
@BurglarBird5 жыл бұрын
Just discovered this channel. This topic is exactly what I should be studying atm! It's amazing to be able to procrastinate, but also learn about my courses at the same time. I Subscribed!
@EpicMathTime5 жыл бұрын
Welcome :)
@lvl3tensorboi9293 жыл бұрын
I think a neat example of isomorphism is the application of the concept of euclidean space to continuuous functions which leads to fourier or laplace transforms. It amazed me at first how functions can be seen as basis vectors given a proper scalar product that fulfills the same properties as the euclidean scalar product.
@chemistro94405 жыл бұрын
0:44 "There is nothing to fear here" [Clock Town music starts] *well played sir*
@keyyyla5 жыл бұрын
I really like the analogy with the chess game! :)
@tengsolomon Жыл бұрын
As a chem student learning about inorganic chemistry, it just dawned on me that point group operations like rotation and reflection are isomorphic functions. Like you perform rotation about a 180 deg angle, and the properties of the molecule after the rotation is still preserved. The only thing that changed are the matrix representations of the atomic orbitals. This is soo mindblowing.
@EpicMathTime Жыл бұрын
Have you used character tables or learned about group representation theory in your studies? These particular fields are strongly linked. The fact that you are able to see a connection like that at such a foundational level is very impressive.
@tengsolomon Жыл бұрын
@@EpicMathTime Yessss! You determine the point group, represent them as matrices, and every time you perform an operation, the molecular structure is still the same (i.e. geometry, properties, etc) but you reduce the irreducible representations of the molecule. It helps us predict the translational, vibrational, and rotational motions of the molecule.
@nadiaarif197 Жыл бұрын
I’m a math major and this video explained better than both of my math classes combined lol
@Quasarbooster5 жыл бұрын
In combinatory calculus, two combinators A and B are isomorphic iff A applied to any x is isomorphic to B applied to x (ie A=B Ax=Bx). For example, the combinator SKK is isomorphic to SKS because when you apply any x, the first reduces to Kx(Kx) and then to x, and the second reduces to Kx(Sx) and then to x. 😊
@mastercilona5 жыл бұрын
Legend of zelda background music and tool as outro?? AND a really helpful explanation??? you have secured my sub.
@kylezs4 жыл бұрын
Lol I came to the comments to see if there was one on the music, I thought it was runescape music
@cristinapasenelli37205 жыл бұрын
Wish my professor had put it in these terms, then I wouldn't have to be scouring the internet for answers. Thanks!
@JTan-fq6vy5 ай бұрын
In time 2:03, it seems the notion of sameness is a relation , and so the function mentioned in 4:26 is a subset of the relation. So A and B are isomorphic iff (A,B) is in the relation (that we are interested). Please correct me if I am wrong.
@fermibubbles93755 жыл бұрын
appreciate the sped up writing.. augmented reality for your hand gestures at some point would be wild
@EmilyA-P9 ай бұрын
Brilliant vid. Newbie to this area and not a Maths student. You explained it so well
@FinnBender5 жыл бұрын
I just wanna say, I love your videos! thx bye
@JM-us3fr5 жыл бұрын
Good old Category Theory. Someday I will take a class on it.
@gandalf292 жыл бұрын
This video has made things very easy for me to grasp. Thanks.
@timothymoore21975 жыл бұрын
I really like this medium of showing math - the clear glass board. It makes these math creatures and what we're saying have this property of "floating around" between us (you and the viewers). I want to create some sort of "math glasses" augmented reality thing where if we put them on, we can have this same effect, but anywhere we want - we wouldn't be bounded by a fixed board placement. I also really like how you speed up the timing when you write down things - it makes the flow of the learning experience better.
@JM-us3fr5 жыл бұрын
Yeah I agree. Maybe it would help people understand math concepts better
@davidk72125 жыл бұрын
That hat tells you that, while he may be really into math, he's still hip and sociable.
@rylanbuck13325 жыл бұрын
This is awesome! I just got into my dream college so I can study pure mathematics! I’m currently in calc 2 and it’s very fun! Don’t find it too difficult, just tricky! Of course I’ve started looking ahead at what’s in the future for me (future classes) and I’ve heard this COUNTLESS times! Finally having a clear head on what isomorphic means really makes it easier to understand 😂
@EpicMathTime5 жыл бұрын
Great! I'm glad you were able to get something out of it. That's very motivating. Thanks for the comment!
@masontdoyle5 жыл бұрын
I had to learn about isomorphisms for some undergraduate research I am doing. Your video helped me understand tremendously! Thank you!
@EpicMathTime5 жыл бұрын
Follow me on Instagram to see video previews before they are released: instagram.com/epicmathtime
@halleberry20942 жыл бұрын
Yessssss. the lofi in the background love it.
@morgengabe15 жыл бұрын
That map diagram at 11:13 seems... raw. Why not make it look more like a function whose domain is some set of order pairs S(x,y) and whose codomain is f(S)? What do continuous functions preserve in topological spaces? Would it be fair to say an Isomorphism is a function that preserves differentiable symmetry?
@PhilosophySama2 жыл бұрын
The kokoroki village music is sending me rn!!
@ashishkumarsharma25845 жыл бұрын
Thanks your way of delivering lectures is nice Thanks from India
@SolidSiren4 жыл бұрын
Isomorphism is not only used in math!! Its everywhere- chemistry, biology, physics, etc. Did it begin in math? Wonder when this word first was used.
@ivan.tucakov Жыл бұрын
Thank you! And props for the inverse writing! You must be doing some isomorphic "translating" to pull that off!
@RonaldModesitt4 жыл бұрын
Your presentation was indeed helpful and enlightening!
@siddid76204 жыл бұрын
"we are concerned with their value, not how they look" yeah right
@justinotherpatriot1744 Жыл бұрын
That thumbnail is hilarious AF
@TheAAZSD4 жыл бұрын
Wonderful explanation
@jacques8277 Жыл бұрын
I neveer thought I'd get a maths lesson from Bam Margera... in any case great video!
@SAAARC4 жыл бұрын
Top tier video and explanation
@carlosraventosprieto20652 жыл бұрын
man, thank you!!!! amazing video
@stydras33805 жыл бұрын
Nice :D Gotta love your structure :P Are you planning on doing a category theory video? :)
@perappelgren9483 жыл бұрын
Such great a video! 👍👍
@benjaminbrady23854 жыл бұрын
5:20 🅱️®️⛎🏨, that's a cop right there
@thebiber94014 жыл бұрын
I'm a physicist. Can someone verify that the following terms are correct? The isomorphisms for... - ... geometry are the Euclidean Transformations (e.g. translations and rotations) - ... vector spaces are linear maps - ... groups are homomorphisms (like representations in quantum mechanics) I don't know about topology.
@EpicMathTime4 жыл бұрын
These are all correct as morphisms. They are isomorphisms with the added stipulation they are also bijective, with inverses that are also morphisms. (that is, an isomorphism is a special kind of morphism). For topology, the morphisms are continuous functions. Sometimes the steps for a morphism to become an isomorphism get satisfied automatically, depending on category. For Euclidean geometry, all morphisms are automatically isomorphisms (every Euclidean transformation is bijective and has an inverse that is also a Euclidean transformation). For most algebraic categories, any bijective morphism is automatically an isomorphism (the inverse of any linear map is a linear map). This isn't true of topology though (there are bijective continuous functions whose inverses are not continuous). So the statement "bijective morphism whose inverse is a morphism" is meant to be an all-encompassing general description.
@supermarc5 жыл бұрын
I have a question: are there also mathematical objects for which an isomorphism is more than just a map f : A -> B on the underlying sets? For example, if the mathematical object has additional properties whose preservation cannot be expressed in terms of the function immediately?
@EpicMathTime5 жыл бұрын
In an abstract categorical sense, certainly, but that isn't enlightening because category theory puts an abstraction on isomorphisms to the point that it may not capture the intuition that I wanted to address here. But in a more concrete mathematical sense, I think a good example is as follows: instead of vector spaces over a field F, with isomorphisms in the usual way (bijective linear transformations), we can look at all vector spaces over any field isomorphic to F, with isomorphisms taken to be bijective semilinear maps. Semilinear maps are more than just a map on the underlying sets because it must not only translate the vectors in the vector space, but also the scalars to the new field. So it effectively contains two maps, translating the vectors, and translating the field elements in that external field acting on the vector space. In the standard study of linear algebra, this is a nonissue and not needed because we lock interacting vector spaces into being over the same field. Once we broaden the class of objects in the way described, one map on the underlying vectors is not enough to preserve what we want.
@supermarc5 жыл бұрын
@@EpicMathTime Nice one! If we cheat a little bit, though, then we could see such an object as an ordered pair (F,V), where F is a field and V is a vector space over F. Then a morphism could still be seen a map f: (F_1,V_1) -> (F_2,V_2).
@EpicMathTime5 жыл бұрын
@@supermarc There's nothing wrong with describing V over F as a single object (which it is) and denoting it (V,F), but the set-wise meaning of this is unclear. If we want to have a map f:(V,F)->(V',F'), we have to know what (V,F) means as a set. Are elements of (V,F) ordered pairs (v,c)? If so, it's hard to make this agree with the notion of our vector space (though I'd have to think about that more). If the elements of (V,F) are just vectors, then we of course have the same issue as before. Now, if we instead are associating (V,F) to (V',F') not by a map between them as sets, but in a way that views them as single mathematical objects in which (V,F) is the "input" and (V',F') is the "output" (as opposed to the domain and target) then we aren't describing a morphism at all, but a functor.
@tasmiafatima6214 жыл бұрын
Really awesome 😍
@nftnick78153 жыл бұрын
Awesome you should do a video on "Hylomorphism" $DAG Constellation network uses this on their Directed Acyclic Graph as they redefine the internet with the HGTP
@Eis461 Жыл бұрын
Why this channel stopped posting
@prestondebetaz43005 жыл бұрын
Hey that LSU hat makes you 10x cooler. Or, in math terms, the limit as x->cool (EpicMathTime)=infinity
@awes0mef4c5 жыл бұрын
oh hell yeah another video
@steveashkarian32015 жыл бұрын
Amazing!
@temurson5 жыл бұрын
Just discovered your channel, it is so cool! And I very like the idea of putting problems at the end. I think I solved this one, but don't wanna spoil it for anyone. I tried to think of a function that could have some of the three properties, but not the others (reflexivity, symmetry and transitivity), but I couldn't. I know it's a lazy question, but could you give a couple of examples of those? Thanks.
@huseyinhalitince44043 жыл бұрын
What does it mean to be the same as itself .is it possible to think of wise versa? Would you give an example of wise versa
@NightmareCourtPictures2 жыл бұрын
Same as itself is supposed to be an obvious fact…like for example say I took you into a cloning machine and made a perfect clone of you…you and this clone are isomorphic because there is a mapping you can do, atom by atom where that mapping is bijective. Say now that I took your clone and I smushed it into just a lumpy pile of goo. This lumpy pile of goo is still isomorphic to you because there is still a bijective mapping you can make with every atom of your smushed clone and you. You can think of a bijective mapping as me taking pieces of your smushed clone, and placing them somewhere new…so a -1 from the clone and a +1 over here…and slowly but surely from the pile of mush I build you up again into a human being. These operations (+1 and -1) is the bijective function in that case…but the function can be any mathematical function, so long as there is an inverse operation to it.
@muckchorris97454 жыл бұрын
Don't miss arrows on coordinate systems.
@euclidselements95223 жыл бұрын
hey i have this monopoly, let's finish our chess game
@Zeegoner5 жыл бұрын
Thanks for this one
@DynestiGTI3 жыл бұрын
Man it's so reassuring to hear other people found the term isomorphism scary, I was just getting to grips with homomorohism and bijection 😅 same thing happened with real analysis, I was so dumbfounded about where these random epsilon/3's were coming from but so was everyone else.
@kalekaleb81485 жыл бұрын
I suddenly love 😍 mathematics.
@valeriobertoncello18095 жыл бұрын
Are you by any chance related to Paul from the LangFocus channel? You two really look alike
@EpicMathTime5 жыл бұрын
Haha nope, although someone else did mention that recently.
@ytseberle3 жыл бұрын
Except Paul speaks with a Canadian accent and Epic sounds southern U.S., probably Louisiana based on his LSU hat :-)
@illiztDesignsHD5 жыл бұрын
Please start a math podcast!
@PolarSky4 жыл бұрын
This is great! Thank you :)
@bobtom12432 жыл бұрын
awesome!!!
@parepidemosproductions47415 жыл бұрын
I have subscribed and the notifications bell is on for when the inverse of a function is not isomorphic. thanks
@lettersfromanihilist90925 жыл бұрын
this is sorta niche, but is anyone else here because they're reading Wittgenstein's tractatus, and was told that one of the ideas is that "language is isomorphic to reality"
@Grassmpl5 жыл бұрын
Exercise. let G be a group consisting of a subset of n×n matrices over a field F wrt the standard multiplication. Prove that the elements of G are either ALL invertible or ALL singular. Furthermore, prove that in the invertible case, the identity of G MUST BE the standard identity matrix.
@EpicMathTime5 жыл бұрын
:) Proof: First, we show if A in G is invertible, then the identity of G is the identity matrix. Let E denote the identity matrix of G, let A* denote the inverse of A in G (let A^(-1) be its standard inverse, and I the standard identity matrix). Then we have that AA* = E and AA^(-1) = I, hence AA*AA^(-1) = EI = E => AEA^(-1) = E => AA^(-1) = E => E = I. Hence E = I. Now, we show if B in G is not invertible, then the identity of G cannot be the identity matrix. B is in G, therefore it has an inverse B* such that BB* = B*B = E. Since B is not invertible, E =/= I. This shows that A and B cannot lie in the same group (as A's membership requires E = I, while B's requires E =/= I).
@Grassmpl5 жыл бұрын
@@EpicMathTime Well done. Your grade is 5/5. Here is another trivial exercise which I know won't stump you. Let G consist of invertible matrices and H consist of singular matrices. Is it possible that G is isomorphic to H? Either give an example or a disproof as appropriate.
@Grassmpl5 жыл бұрын
@@EpicMathTime btw these questions are from my own head which I know the answer to. Ie they are NOT from textbooks, internet, home works, exams, etc.
@EpicMathTime5 жыл бұрын
@@Grassmpl Yes, it is possible, let E be an idempotent matrix, then {I} and {E} are isomorphic. Or do we want to ban that?
@Grassmpl5 жыл бұрын
@@EpicMathTime this is good. As u can see very trivial. Although since I is idempotent u should mention that E is singular as well. Better yet let E=0
@captainsal70745 жыл бұрын
Clocktown Majora mask
@rohitbhosle65214 жыл бұрын
5:54 so funny 😂😂😂😂
@vojislavbelic8965 жыл бұрын
What i really want explained is why some guys wear beanies all the time
@tomkerruish29824 жыл бұрын
Ask Tim Pool.
@NonTwinBrothers3 жыл бұрын
Ooooh here I am thinking he learned how to draw backwards, nice lol
@captainsal70745 жыл бұрын
Lateralus Tool
@nourhanramadan74873 жыл бұрын
you save me thanks
@thehippievan12885 жыл бұрын
You mean my fortnite skins are isomorphic to the standard ones??? Damn. . .
@AR-vb4xy4 жыл бұрын
Bro what's your math qualifications?
@Anteino3 жыл бұрын
Kudos for writing backwards so fast
@Anteino3 жыл бұрын
No wait, you wrote like normally and mirrored the footage. That must be it.
@inigo87404 жыл бұрын
Equally handsome, equally smart.
@FooodConfusion4 жыл бұрын
lovely brother you nailed it dumb like me in maths can even understand this thing