Prof Penn, I really appreciate what you have done to advocate higher mathematics education. But I would like to suggest promoting this channel via your main channel so as to reach more people, I personally didn't know the existence of this channel, until your recent vote about the podcast with your students. Glad to see more content of such good quality!
If anybody is psyched about injectivity and surjectivity, the result at 4:00 that a function is invertible iff it is bijective can be broken down into two separate results: a function has a left-inverse iff it is injective, and a function has a right-inverse iff it is surjective. one can then put these together using the fact that if a function has a left-inverse and a right-inverse then these are equal, so it has a two-sided inverse. The implication "if f: A -> B is surjective then it has a right-inverse" is interesting because it requires the axiom of choice (the proof explicitly constructs a right-inverse: we know that for each b in B there is _some_ a in A which maps to b, and we use the axiom of choice to choose such an a for each b). However when you smash the two results together into "f is bijective iff it has a two-sided inverse" then you don't need the axiom of choice (since when we construct the two-sided inverse, we know that there is a _unique_ a in A mapping to each b in B, so there is no choice involved)
Could this be a motivation that the axiom of choice should be considered true? Since you require it to be accepted to show a right inverse's existence, but you already show it by proving an actual inverse's existence.
@@skylardeslypere9909 I only 'already show it' without the axiom of choice for surjective functions which are also injective. Putting "bijective has a two-sided inverse => has a right inverse" and "axiom of choice => (surjective => has a right inverse)" together, we get "axiom of choice => (surjective but not bijective => has a right inverse)", in which the axiom of choice is always needed (since the bijective surjections have been taken out). The axiom of choice has been proved to be independent of the standard set-theoretical foundations for mathematics (ZF), so I don't think it is possible to find any "motivation" for considering it to be true/false (for taking it as an axiom versus not doing so). However, you definitely *don't* want to take Choice as an axiom if: - you are doing constructivist mathematics (in which e.g. the law of excluded middle doesn't hold) - you are doing computable mathematics (e.g. the intersection of math and theoretical computer science) - you find the Banach-Tarski paradox too absurd of a consequence of Choice for the axiom of choice to be reasonable/justifiable
At 8:47, I was trying to wrap my head around this proof. We suppose f bijective. Then we ... define a function g that satisfies all the conditions for invertibility ? But then ... That assumes there exists an inverse function for f, which is what we want to prove ! As far as I can reason it through, I only see a result we're trying to prove being used in the proof itself ! But I see nothing in the comment talking about that, so I am quite surprised.
The introduction of IDa and IDb at about 0:58 is somewhat skimpy, and confused me further on. I had to rewatch that part multiple times and dredge up old memories to realise IDa and IDb are the identity functions! Given the formalism of like videos, I think concepts like IDa and IDb should be treated more formally.
6:16 There is an unspoken thing going on here where you are treating f(x) as a composition of 2 things(f compose x). Then you can apply your composition theorem and you get the identity on A composed with x and y on each side respectively. TLDR: This hinges on the fact that it makes sense to compose a function and an element together.
The proof of the generalized stokes theorem is filmed and with the editor. I think I may eventually re-do those videos on this channel with this type of "course type" structure.
23:07 I think we are assuming the permutations are finite. I'm not sure it generalizes to infinite permutations. Also you say the proof is simple, but I have no idea how to prove it. (which probably means I'm missing something here. I'll move on and occasionally come back and give it a shot.)
buchweiz asked about differential forms, but I want to ask the same question about the most recent number theory course; there’s only one video left, right?
There is a wonderful open source book called “The Book of Proof”. I think if you read that and worked every exercise you would be in extremely good shape for a math major.
I think videos will probably be released in a start and stop fashion. Probably every other day until Linear Algebra is done -- a break and then another subject. My idea is that this channel is less about "new" content and more about creating a lasting reference of material.
12:07 Not quite so sure of the 4th one. I think you're supposed to do an induction proof then in the induction step you do a counting argument. But how you rigorously show that counting argument is something I'm a little weak on.
13:24 I'm dumb you can just skip the Induction entirely with what you just said. The proof is something like there are n elements and they have n choices. Once the first element makes a choice there are now n-1 choices for the second element. Observe this continues until the nth element which has 1 choice. The total number of functions in Sn is n*(n-1)...(1). Therefore the magnitude of Sn is n!.
Great start for this series! How “deep” did you want to go? For example, you never properly defined the identity function nor the equality of functions, yet you used them both in this video?
yeah he also defined function equality, if only in passing. If you didnt know 2 functions are equal if they have the f(a) = g(a) for all a the shared domain
@@gabriel-et3gy thanks. I was not really asking for these definitions for myself, merely asking whether they were left out willingly or not, just out of completeness. For example the equality between function is *slightly* more problematic since the equality between domains and codomains is of both function is often overlooked, yet necessary.
@@l.p.7585 well that in itself is not clear. Some definition s require the domains and codomains to also coincide for a full fledged equality. Some trickster math teachers did play that trick more than once!
The problems with trying to explain absolutely everything are that 1) this obscures the exposition of what you're actually trying to get on with, not to mention sidetracking you and making the videos much longer than they ought to be, and 2) once you get right down to the bottom, things become very complicated: Michael defined a function as an association _f_ between sets _A_ and _B_ such that for all _a_ in _A_ there is a unique _b_ in _B_ such that _f_ associates _a_ with _b_ - in symbols: ∀_a_ϵ_A_ ∃!_b_ϵ_B_ : _f_(_a_) = _b_. But what is a set? What is an association? It is in fact very difficult to rigorously define sets without running into paradoxes, so in set theory we define sets not by what they are, but by what they do: a set is defined as an element of the universe of all sets, which is governed by the axioms of a set theory (most commonly Zermelo-Frankel set theory). Then perhaps you would want to say what these axioms are, but now you're doing a course on axiomatic set theory instead of what you were actually trying to do. Moreover, defining mathematical objects (e.g. relations, functions, ordered pairs, etc.) in terms of sets doesn't actually shed any light on anything; it doesn't give you a better understanding of what an ordered pair really is, you are just showing that it is possible to encode ordered pairs into a set theory - in this way, set-theoretical foundations of mathematics are more like programming than typical mathematics.
i find the injective proof a bit unintuitive, its hard to think contrapositively. so it feels like you are going through the motions, but they are just syntactically and not semantically meaningful
@11:00 Certainly if you take the codomain of exp(x) as only the *positive* reals then it's surjective, but then every function is trivially surjective if you take the codomain equal to the range. So, what makes that fair play?
For continuous functions perhaps, but I don't think that has to be true for all disjointed functions. Continuity let's things like the intermediate value theorem come into play, but as soon as you can jump around I think you lose that surjection property.
@@BridgeBum In an earlier video, OP distinguishes "codomain" and "range". I don't recall the exact words, but he gave the impression that "codomain" is more like "value-type" (e.g. integer or complex-real) whereas "range" is the set of values a given function could ever possibly evaluate to: in other words, precisely that set over which it is, by definition, surjective. In this video, he casually uses the positive subset of the reals as if on an equal footing with the reals complete. In the process, the distinction between "range" and "codomain" becomes a matter of perspective. That's really all I'm on about.
@@IanKjos I appreciate the difference between co-domain and range, although I would say that the set of positive reals is a reasonable co-domain and one which is often used in various contexts. Integers, naturals, etc. all seem like reasonable "targets". In the end for normal everyday functions the distinction is fuzzier than it would be for countable sets as the domain/co-domain where it is far more likely to have a non-injunctive function. I do see where you are coming from, there's a touch of handwaving here.
basically you define "a" to be (f inverse of b). So there exists some element of set A (that you are just calling "a") that is the inverse of b. So for any b, you've found something you call "a" that f will act on to get you b.
Hey im just wondering I'm in my first semester as a math major and im taking Linear Algebra. At most unis I know (at least her in Germany) CS,Physics and Math Majors (among some others) take the same Linear Algebra courses. Ive checked out engineering math courses also and what im seeing in these videos seems to be a regular linear algebra class. What is it about this course that turns it into "abstract linear algebra"
I don’t understand the 4th observation at 13:15 Could someone explain to me how this is possible? I thought the cardinality of a set was equal to the number objects in it. Since the Sn maps n to itself, then why is the size not n?
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)
Prof Penn, I really appreciate what you have done to advocate higher mathematics education. But I would like to suggest promoting this channel via your main channel so as to reach more people, I personally didn't know the existence of this channel, until your recent vote about the podcast with your students. Glad to see more content of such good quality!
Having finished both linear algebra and abstract algebra, this playlist is gonna be a fun one to watch
25:05 Uncut board cleaning… on purpose ?
27:42 Good Place To Stop
Honestly I prefer the uncut board cleaning
"And that's a good pla-"
Turns out, it wasn't such a good place to stop after all
If anybody is psyched about injectivity and surjectivity, the result at 4:00 that a function is invertible iff it is bijective can be broken down into two separate results: a function has a left-inverse iff it is injective, and a function has a right-inverse iff it is surjective. one can then put these together using the fact that if a function has a left-inverse and a right-inverse then these are equal, so it has a two-sided inverse.
The implication "if f: A -> B is surjective then it has a right-inverse" is interesting because it requires the axiom of choice (the proof explicitly constructs a right-inverse: we know that for each b in B there is _some_ a in A which maps to b, and we use the axiom of choice to choose such an a for each b). However when you smash the two results together into "f is bijective iff it has a two-sided inverse" then you don't need the axiom of choice (since when we construct the two-sided inverse, we know that there is a _unique_ a in A mapping to each b in B, so there is no choice involved)
Could this be a motivation that the axiom of choice should be considered true? Since you require it to be accepted to show a right inverse's existence, but you already show it by proving an actual inverse's existence.
@@skylardeslypere9909 I only 'already show it' without the axiom of choice for surjective functions which are also injective. Putting "bijective has a two-sided inverse => has a right inverse" and "axiom of choice => (surjective => has a right inverse)" together, we get "axiom of choice => (surjective but not bijective => has a right inverse)", in which the axiom of choice is always needed (since the bijective surjections have been taken out).
The axiom of choice has been proved to be independent of the standard set-theoretical foundations for mathematics (ZF), so I don't think it is possible to find any "motivation" for considering it to be true/false (for taking it as an axiom versus not doing so). However, you definitely *don't* want to take Choice as an axiom if:
- you are doing constructivist mathematics (in which e.g. the law of excluded middle doesn't hold)
- you are doing computable mathematics (e.g. the intersection of math and theoretical computer science)
- you find the Banach-Tarski paradox too absurd of a consequence of Choice for the axiom of choice to be reasonable/justifiable
رائع.ذكرتني بعام 1994 مع الأستاذ بويخف في كلية العلوم باكادير.
At 8:47, I was trying to wrap my head around this proof. We suppose f bijective. Then we ... define a function g that satisfies all the conditions for invertibility ? But then ... That assumes there exists an inverse function for f, which is what we want to prove ! As far as I can reason it through, I only see a result we're trying to prove being used in the proof itself ! But I see nothing in the comment talking about that, so I am quite surprised.
The introduction of IDa and IDb at about 0:58 is somewhat skimpy, and confused me further on. I had to rewatch that part multiple times and dredge up old memories to realise IDa and IDb are the identity functions!
Given the formalism of like videos, I think concepts like IDa and IDb should be treated more formally.
I always found the term "transposition" verbose - I just call them "swaps"
6:16 There is an unspoken thing going on here where you are treating f(x) as a composition of 2 things(f compose x). Then you can apply your composition theorem and you get the identity on A composed with x and y on each side respectively.
TLDR: This hinges on the fact that it makes sense to compose a function and an element together.
Thank you for this course !
I don't like the term "inverse function" for this concept. I prefer "reciprocal function".
Great video! Any chance you finish your video series on differential forms on this channel? I was really enjoying, but they ended on such cliffhanger!
The proof of the generalized stokes theorem is filmed and with the editor. I think I may eventually re-do those videos on this channel with this type of "course type" structure.
Oh I'm hooked. It takes me back to 35 years ago 😅
Also might want to mention in f: A->B that A and B are non-empty.
Otherwise the results are all true trivially
At 7:10 , I dont understand why we can set a equals to inverse of b, doesnt this immediately implies the goal of the proof, which is f(a)=b ?
23:07 I think we are assuming the permutations are finite. I'm not sure it generalizes to infinite permutations.
Also you say the proof is simple, but I have no idea how to prove it. (which probably means I'm missing something here. I'll move on and occasionally come back and give it a shot.)
buchweiz asked about differential forms, but I want to ask the same question about the most recent number theory course; there’s only one video left, right?
What should I focus on in math as a 3rd year high school student?
Note :I want to major in math after high school
There is a wonderful open source book called “The Book of Proof”. I think if you read that and worked every exercise you would be in extremely good shape for a math major.
@@mathmajor
Thank you very much Professor Penn
Is there a regular schedule for videos on this channel?
I think videos will probably be released in a start and stop fashion. Probably every other day until Linear Algebra is done -- a break and then another subject. My idea is that this channel is less about "new" content and more about creating a lasting reference of material.
what is the identity
It's the function that maps every element in the set to itself. So idA maps each element of A to itself.
12:07 Not quite so sure of the 4th one. I think you're supposed to do an induction proof then in the induction step you do a counting argument. But how you rigorously show that counting argument is something I'm a little weak on.
13:24 I'm dumb you can just skip the Induction entirely with what you just said.
The proof is something like there are n elements and they have n choices. Once the first element makes a choice there are now n-1 choices for the second element. Observe this continues until the nth element which has 1 choice. The total number of functions in Sn is n*(n-1)...(1). Therefore the magnitude of Sn is n!.
How do you actually solve the examples at the end? Does anyone have any idea?
Great start for this series! How “deep” did you want to go? For example, you never properly defined the identity function nor the equality of functions, yet you used them both in this video?
well, it doesn't seem like it is hard to define the identity function, its just f: A -> A, f(x) = x
yeah he also defined function equality, if only in passing. If you didnt know 2 functions are equal if they have the f(a) = g(a) for all a the shared domain
@@gabriel-et3gy thanks. I was not really asking for these definitions for myself, merely asking whether they were left out willingly or not, just out of completeness.
For example the equality between function is *slightly* more problematic since the equality between domains and codomains is of both function is often overlooked, yet necessary.
@@l.p.7585 well that in itself is not clear. Some definition s require the domains and codomains to also coincide for a full fledged equality. Some trickster math teachers did play that trick more than once!
The problems with trying to explain absolutely everything are that
1) this obscures the exposition of what you're actually trying to get on with, not to mention sidetracking you and making the videos much longer than they ought to be, and
2) once you get right down to the bottom, things become very complicated: Michael defined a function as an association _f_ between sets _A_ and _B_ such that for all _a_ in _A_ there is a unique _b_ in _B_ such that _f_ associates _a_ with _b_ - in symbols: ∀_a_ϵ_A_ ∃!_b_ϵ_B_ : _f_(_a_) = _b_. But what is a set? What is an association? It is in fact very difficult to rigorously define sets without running into paradoxes, so in set theory we define sets not by what they are, but by what they do: a set is defined as an element of the universe of all sets, which is governed by the axioms of a set theory (most commonly Zermelo-Frankel set theory). Then perhaps you would want to say what these axioms are, but now you're doing a course on axiomatic set theory instead of what you were actually trying to do. Moreover, defining mathematical objects (e.g. relations, functions, ordered pairs, etc.) in terms of sets doesn't actually shed any light on anything; it doesn't give you a better understanding of what an ordered pair really is, you are just showing that it is possible to encode ordered pairs into a set theory - in this way, set-theoretical foundations of mathematics are more like programming than typical mathematics.
i find the injective proof a bit unintuitive, its hard to think contrapositively.
so it feels like you are going through the motions, but they are just syntactically and not semantically meaningful
@11:00 Certainly if you take the codomain of exp(x) as only the *positive* reals then it's surjective, but then every function is trivially surjective if you take the codomain equal to the range. So, what makes that fair play?
For continuous functions perhaps, but I don't think that has to be true for all disjointed functions. Continuity let's things like the intermediate value theorem come into play, but as soon as you can jump around I think you lose that surjection property.
@@BridgeBum In an earlier video, OP distinguishes "codomain" and "range". I don't recall the exact words, but he gave the impression that "codomain" is more like "value-type" (e.g. integer or complex-real) whereas "range" is the set of values a given function could ever possibly evaluate to: in other words, precisely that set over which it is, by definition, surjective. In this video, he casually uses the positive subset of the reals as if on an equal footing with the reals complete. In the process, the distinction between "range" and "codomain" becomes a matter of perspective. That's really all I'm on about.
@@IanKjos I appreciate the difference between co-domain and range, although I would say that the set of positive reals is a reasonable co-domain and one which is often used in various contexts. Integers, naturals, etc. all seem like reasonable "targets". In the end for normal everyday functions the distinction is fuzzier than it would be for countable sets as the domain/co-domain where it is far more likely to have a non-injunctive function. I do see where you are coming from, there's a touch of handwaving here.
9:33 so being well-defined is just being bijective?
what does it mean to set a equal to f inverse of b? 07:05
basically you define "a" to be (f inverse of b). So there exists some element of set A (that you are just calling "a") that is the inverse of b. So for any b, you've found something you call "a" that f will act on to get you b.
Hey im just wondering I'm in my first semester as a math major and im taking Linear Algebra.
At most unis I know (at least her in Germany) CS,Physics and Math Majors (among some others) take the same Linear Algebra courses.
Ive checked out engineering math courses also and what im seeing in these videos seems to be a regular linear algebra class.
What is it about this course that turns it into "abstract linear algebra"
Honest question I'm just curious because I'd love to study in the Us one day and you guys seem to do things a bit different.
I don’t understand the 4th observation at 13:15 Could someone explain to me how this is possible? I thought the cardinality of a set was equal to the number objects in it. Since the Sn maps n to itself, then why is the size not n?
Nvm, haven’t worked with combinatorics in a while. Figured it out
I feel this stuff should be simple. However, I have a hard time with this extremely basic stuff (first part of video).
can anybody suggest me any telegram Math channel
Lit
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html
P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)
Is it just me who has poor quality on video? I can't change to higher quality
I think you caught it before hd had been processed.
@@MichaelPennMath right, it works now. Thank you :)
signum function. Don't know that I had heard of it before:
en.wikipedia.org/wiki/Sign_function