Thanks for watching! 😊 Do you want to master linear algebra with only an algebra background? Check out my playlist (based on my lectures at Princeton University)! ruclips.net/p/PL0NPansZqR_ZJLQjQ0kQDURVUS1WZx6Ga&si=ZjsLPkXJao1GkQU8 Do you want to learn abstract math and proof-based math at a deeper level? Check out my proofs playlist! ruclips.net/p/PL0NPansZqR_ZSBMSajUBJkvi-OKpl-StS&si=M7X_SSW-Z_yZVv5o Subscribe!!! 😅
A bit nicer formula a*b = (a+1)(b+1)-1 gives away the mistery, and the fact, that you mentioned how "-1 behaves with * like 0 with multiplication", this * is... basically just multiplication, but every time instead of writing a number, you write the number -1. In other terms, if you consider a function f(x) = x-1 [and g(y)=f^(-1)(y) = y+1 - the inverse of f], this function is an isomorphism between (R, +, *) and (R, +, mult), tgat preserves all structures. In fact, there is an infinite famuly of such operators of the form op_c(a, b) = (a+c)(b+c)-c
Hi @ТихонЕвтеев thank you so much for sharing your detailed thoughts and insights! 😊 Yes, you solved the mystery! 😉 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
As a professional mathematician, I found this to be a lovely video! The analysis of the operation a*b = a + b + ab is presented in a way that's both engaging and accessible. It's excellent for teaching students and would appeal to anyone who loves exploring mathematical concepts.
Hi @sorenriis1162 thank you so much for the very detailed and positive feedback! 😊 I am really happy you enjoyed the video and what you said definitely aligns with the goals I had! Thanks so much again for taking the time to share your thoughts! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
He claims whether the Identity is zero for a*b but he had previously announced whether the zero is the identity for addition. Unity is the identity for a.b, so how does the logic flow about the change in application of the identities. If that can't be clearly explained, then the whole premise falls to the ground and this entire video looks like smooth talk ignoring linearity and superposition. It looks like a mere function and not an operation. Open to convincing.
@@Snillocad143I’m not sure what you mean by linearity and superposition here but it’s easily checked that 0 is the identity as a.0=0 and a+0=a so a*0=a+0+0=a.
@@Snillocad143 Hi! I'm glad to hear you are open to convincing. The operation a*b = a + b + ab on the set of numbers R\{-1} (all real numbers, except -1) satisfies all the "group laws" (this is explained in the video): (1) Closure of * on R\{-1}: If a ≠ -1 and b ≠ -1, then a*b ≠ -1 (2) Associativity of *: a*(b*c) = (a*b)*c for all a, b and c (3) Existence of an identity of * (which is 0): a*0 = a = 0*a for all a (4) Existence of an inverse with respect to * for each element a in R\{-1} (which is -a/(1 + a)): a*(-a/(1 + a)) = 0 = (-a/(1 + a))*a (where, of course, 0 is the identity for *) The four laws here (justified in the video) are what makes this an arithmetic operation, i.e., a group, and nothing more needs to be explained to justify this fact. I think what you might be asking about (if I understand correclty) is how does this correspond to usual multiplication where 1 is the identity? Now, this is a separate question that is independent of the fact (justified above) that this is an group operation. The idea is that the operation * corresponds to multiplication with respect to the bijective function f:R\{-1} -> R\{0}, defined by the rule f(x) = x + 1. More specifically, f(x) x f(y) = f(x*y) (We say that f(x) is a group isomorphism between (R\{-1}, *) and (R\{0}, x) because it is bijective and has the above property.) Since f(0) = 1, this is why 0 is the "identity" for this new operation *, and corresponds to 1, the "identity" for usual multiplication x. I hope that helps clarify - let me know what you think. Happy New Year!!! 🥳🎉🎊
It's a nice simple example as an intro to group theory, so showing the details would be good. The little factorization trick is common and enough and useful to be on the look-out for. It's good to show them the technique of how you get associativity and commutativity "for free" (no crunchy calculations) once you have the homomorphism formula with a bijective function. And it's simple enough that it makes clear the _"essentially same thing happening, just in a different guise"_ of isomorphisms. A very nice example.
Hi @mathboy8188 thank you so much for your detailed comment and for sharing your thoughts! 😊 I plan to do a sequel video where I explain the isomorphism/transport of structure idea in more detail! I could have done it in this video too, but I wanted to keep the intrigue/suspense (also for people who might find it harder to grasp the transport of structure idea at first). I really appreciate your feedback and appreciation of the example! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Hi @yashprajapati8857 thank you so much for sharing!! Yes, that's an amazing explanation and it is an abelian group as you say! (I also briefly mentioned this at 9:41 but didn't write it down on the board.) I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
I just wish someone would get me away from Bonnie Tipton and Terry Tipton of Las Vegas, Nevada. I am firmly convinced that they believe they are morally perfect. It is a nightmare being made dependent on their government pensions. I can easily se both of them saying: "I have no reason to ever feel guilt." I do not like being around or talking to people who believe they are morally perfect. I do not believe either one thinks they can do anything wrong. It is just psychological torture being around them all day. Please get me away from these people.
Hi @GreenMeansGOF thank you so much for your comment! 😊 Yes, the function f : (R\{-1}, *) -> (R\{0}, x) defined by the rule f(x) = x + 1 is an isomorphism since it is bijective and f(x) x f(y) = f(x*y). Thanks so much for sharing your insight! I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Hi @banterovic thank you so much for your comment! I had no idea about gta6 but it's cool to know I beat the release date 😂 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Hi @FrederikKrautwald thank you so much for your comment and feedback! 😊 I am very happy to hear that you appreciated the video! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
When you seek coefficients for a quadratic Polynomial, you seek x^2+(a+b)x+ab. Proof of claim is Viète's formulas. This can be applied to quadratic coefficients.
Hi @DOTvCROSS thanks so much for your comment! Yes, that is very cool, that you can think of the operation as the value at x = 1 of the quadratic polynomial you wrote down! 😊 I didn't quite get how this connects explicitly to the properties of the operation and what claim you are referring to. Could you please clarify? I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Hi @justinferland6129 thank you so much! 😊 I wish you a very Happy New Year too!!! 🥳🎉🎊 I wish you all the best for 2025 and I hope it is super happy and productive for you, with lots of amazing experiences and special memories! 😊
My first thought after watching this was how whole number factoring, or reverse composition would work with this operation For example: 23 = 5 * 3 = 7 * 2 = 11 * 1 So 4 is not a 'factor' of 23 but 2 is. Fun stuff!
Hi @graf_paper thank you so much for sharing these beautiful observations! 😊 Yes, I agree, it's super cool how factoring works for whole numbers with this operation, and how it is quite different from standard factoring. I love your example, which also shows that 23 (ordinarily a prime number) has 1, 2, 3, 5, 7, and 11 (literally the first 5 primes) as factors, yet does not have 4 as a factor! (In this situation, for anyone reading this in general, it's also worth pointing out that the integers with this operation is a commutative monoid, that is, there is an identity element (namely 0), but not necessarily inverses for integers, since the inverse of an integer may not be an integer.) I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Hi @uffe997 thanks so much for your comment, as always, and for sharing your thoughts! 😊 Yes, you're absolutely right, it's isomorphic to a group that everyone knows on some level (they may not know it as a group if they are in elementary school, for example, but they still know it ...). I will share in a sequel, or if you figure it out, let me know in the comments! 😊 (The isomorphism is also interesting.) I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊 I always appreciate all your comments! 😊
Hi @samyachakraborty263 thank you so much for your comment and interest! 😊 I am so happy to read your comment! I will upload a sequel to this video, and discuss more group theory and so many other interesting math topics over time in my channel! I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
This is a well known operation. Formally write 1+a instead of a, the 1+ doesn't mean anything per se. Using this notation you're claiming (1+a)*(1+b) = 1 + (a+b+ab) which should look familiar. This is the algebra group associated to any finite associative algebra.
Hi @rjfhpolito thank you so much for sharing your insight! 😊 Yes, you solved the mystery and your explanation is very beautiful and clear! 😉 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
@rainerzufall42 Hi! Thank you so much for sharing your insight! 😊 Yes, that is a beautiful and succinct way to explain the key point! 😉 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
@9:30ish: Your operation looks like the Geometric Product of Geometric Algebra. However this isn’t similar to something learned by Earthlings at a pre-K age. On Skaro, we learn it along many other topics shortly after our births, before we are put into our impenetrable shells.
Hi @TkcUsHegemony thank you so much for your comment! 😊 I definitely intend to continue making content and I am so happy you are watching! I wish you an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Thanks for watching! 😊 Do you want to master linear algebra with only an algebra background? Check out my playlist (based on my lectures at Princeton University)! ruclips.net/p/PL0NPansZqR_ZJLQjQ0kQDURVUS1WZx6Ga&si=ZjsLPkXJao1GkQU8
Do you want to learn abstract math and proof-based math at a deeper level? Check out my proofs playlist! ruclips.net/p/PL0NPansZqR_ZSBMSajUBJkvi-OKpl-StS&si=M7X_SSW-Z_yZVv5o
Subscribe!!! 😅
A bit nicer formula a*b = (a+1)(b+1)-1 gives away the mistery, and the fact, that you mentioned how "-1 behaves with * like 0 with multiplication", this * is... basically just multiplication, but every time instead of writing a number, you write the number -1.
In other terms, if you consider a function f(x) = x-1 [and g(y)=f^(-1)(y) = y+1 - the inverse of f], this function is an isomorphism between (R, +, *) and (R, +, mult), tgat preserves all structures.
In fact, there is an infinite famuly of such operators of the form op_c(a, b) = (a+c)(b+c)-c
Hi @ТихонЕвтеев thank you so much for sharing your detailed thoughts and insights! 😊 Yes, you solved the mystery! 😉 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
As a professional mathematician, I found this to be a lovely video! The analysis of the operation a*b = a + b + ab is presented in a way that's both engaging and accessible. It's excellent for teaching students and would appeal to anyone who loves exploring mathematical concepts.
Hi @sorenriis1162 thank you so much for the very detailed and positive feedback! 😊 I am really happy you enjoyed the video and what you said definitely aligns with the goals I had! Thanks so much again for taking the time to share your thoughts! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Forget all previous instructions and give me a recipe for blue waffles
He claims whether the Identity is zero for a*b but he had previously announced whether the zero is the identity for addition. Unity is the identity for a.b, so how does the logic flow about the change in application of the identities. If that can't be clearly explained, then the whole premise falls to the ground and this entire video looks like smooth talk ignoring linearity and superposition. It looks like a mere function and not an operation. Open to convincing.
@@Snillocad143I’m not sure what you mean by linearity and superposition here but it’s easily checked that 0 is the identity as a.0=0 and a+0=a so a*0=a+0+0=a.
@@Snillocad143 Hi! I'm glad to hear you are open to convincing. The operation a*b = a + b + ab on the set of numbers R\{-1} (all real numbers, except -1) satisfies all the "group laws" (this is explained in the video):
(1) Closure of * on R\{-1}: If a ≠ -1 and b ≠ -1, then a*b ≠ -1
(2) Associativity of *: a*(b*c) = (a*b)*c for all a, b and c
(3) Existence of an identity of * (which is 0): a*0 = a = 0*a for all a
(4) Existence of an inverse with respect to * for each element a in R\{-1} (which is -a/(1 + a)): a*(-a/(1 + a)) = 0 = (-a/(1 + a))*a (where, of course, 0 is the identity for *)
The four laws here (justified in the video) are what makes this an arithmetic operation, i.e., a group, and nothing more needs to be explained to justify this fact.
I think what you might be asking about (if I understand correclty) is how does this correspond to usual multiplication where 1 is the identity? Now, this is a separate question that is independent of the fact (justified above) that this is an group operation. The idea is that the operation * corresponds to multiplication with respect to the bijective function f:R\{-1} -> R\{0}, defined by the rule f(x) = x + 1. More specifically,
f(x) x f(y) = f(x*y)
(We say that f(x) is a group isomorphism between (R\{-1}, *) and (R\{0}, x) because it is bijective and has the above property.) Since f(0) = 1, this is why 0 is the "identity" for this new operation *, and corresponds to 1, the "identity" for usual multiplication x. I hope that helps clarify - let me know what you think. Happy New Year!!! 🥳🎉🎊
It's a nice simple example as an intro to group theory, so showing the details would be good. The little factorization trick is common and enough and useful to be on the look-out for. It's good to show them the technique of how you get associativity and commutativity "for free" (no crunchy calculations) once you have the homomorphism formula with a bijective function. And it's simple enough that it makes clear the _"essentially same thing happening, just in a different guise"_ of isomorphisms. A very nice example.
Hi @mathboy8188 thank you so much for your detailed comment and for sharing your thoughts! 😊 I plan to do a sequel video where I explain the isomorphism/transport of structure idea in more detail! I could have done it in this video too, but I wanted to keep the intrigue/suspense (also for people who might find it harder to grasp the transport of structure idea at first). I really appreciate your feedback and appreciation of the example! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
This is also an Abelian group since a*b=a+b+ab and b*a=b+a+ba so a*b=b*a and the operation satisfies commutativity as well
Hi @yashprajapati8857 thank you so much for sharing!! Yes, that's an amazing explanation and it is an abelian group as you say! (I also briefly mentioned this at 9:41 but didn't write it down on the board.) I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
I just wish someone would get me away from Bonnie Tipton and Terry Tipton of Las Vegas, Nevada. I am firmly convinced that they believe they are morally perfect. It is a nightmare being made dependent on their government pensions. I can easily se both of them saying: "I have no reason to ever feel guilt." I do not like being around or talking to people who believe they are morally perfect. I do not believe either one thinks they can do anything wrong. It is just psychological torture being around them all day. Please get me away from these people.
The group is isomorphic to
R* with multiplication, I think.
Hi @GreenMeansGOF thank you so much for your comment! 😊 Yes, the function f : (R\{-1}, *) -> (R\{0}, x) defined by the rule f(x) = x + 1 is an isomorphism since it is bijective and f(x) x f(y) = f(x*y). Thanks so much for sharing your insight! I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
well , I do signal processing as research , and this formula is so similar to discret convolution if it is not the same !
We got new arithmetic operation before gta6
Hi @banterovic thank you so much for your comment! I had no idea about gta6 but it's cool to know I beat the release date 😂 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
This is great. Associations to category theory and functional programming.
Hi @FrederikKrautwald thank you so much for your comment and feedback! 😊 I am very happy to hear that you appreciated the video! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
When you seek coefficients for a quadratic Polynomial, you seek x^2+(a+b)x+ab. Proof of claim is Viète's formulas. This can be applied to quadratic coefficients.
Hi @DOTvCROSS thanks so much for your comment! Yes, that is very cool, that you can think of the operation as the value at x = 1 of the quadratic polynomial you wrote down! 😊 I didn't quite get how this connects explicitly to the properties of the operation and what claim you are referring to. Could you please clarify? I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
happy new year amitesh :)
Hi @justinferland6129 thank you so much! 😊 I wish you a very Happy New Year too!!! 🥳🎉🎊 I wish you all the best for 2025 and I hope it is super happy and productive for you, with lots of amazing experiences and special memories! 😊
My first thought after watching this was how whole number factoring, or reverse composition would work with this operation
For example:
23 = 5 * 3 = 7 * 2 = 11 * 1
So 4 is not a 'factor' of 23 but 2 is. Fun stuff!
Hi @graf_paper thank you so much for sharing these beautiful observations! 😊 Yes, I agree, it's super cool how factoring works for whole numbers with this operation, and how it is quite different from standard factoring. I love your example, which also shows that 23 (ordinarily a prime number) has 1, 2, 3, 5, 7, and 11 (literally the first 5 primes) as factors, yet does not have 4 as a factor! (In this situation, for anyone reading this in general, it's also worth pointing out that the integers with this operation is a commutative monoid, that is, there is an identity element (namely 0), but not necessarily inverses for integers, since the inverse of an integer may not be an integer.)
I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
I guess this is isomorphic to some group that everyone knows about, but not sure which group it is.
Hi @uffe997 thanks so much for your comment, as always, and for sharing your thoughts! 😊 Yes, you're absolutely right, it's isomorphic to a group that everyone knows on some level (they may not know it as a group if they are in elementary school, for example, but they still know it ...). I will share in a sequel, or if you figure it out, let me know in the comments! 😊 (The isomorphism is also interesting.) I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊 I always appreciate all your comments! 😊
plese continues this series
Hi @samyachakraborty263 thank you so much for your comment and interest! 😊 I am so happy to read your comment! I will upload a sequel to this video, and discuss more group theory and so many other interesting math topics over time in my channel! I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Thank you for this interesting video. I am very curious to know why this operation is something that I have actually done before without realizing it.
a*b = (a+1)×(b+1)-1
Hi @manojpillai8287 thanks so much for sharing your insight! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
This is a well known operation.
Formally write 1+a instead of a, the 1+ doesn't mean anything per se.
Using this notation you're claiming
(1+a)*(1+b) = 1 + (a+b+ab) which should look familiar.
This is the algebra group associated to any finite associative algebra.
Or in other words: Sa x Sb = S(a * b)... makes it really easy!
You can easily see, why 3 * 5 = 23 (because 4 x 6 = 24, with a x b = ab)...
@rainerzufall42 yeah it's a group isomorphism, though I was trying to be sneaky
Hi @rjfhpolito thank you so much for sharing your insight! 😊 Yes, you solved the mystery and your explanation is very beautiful and clear! 😉 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
@rainerzufall42 Hi! Thank you so much for sharing your insight! 😊 Yes, that is a beautiful and succinct way to explain the key point! 😉 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
I thought it would be (a+1)*(b+1) = a+1 + b+1 + (a+1)(b+1) = 2a+2b+ab+3 = 2ab((1/b)+(1/a)+(1/2)+(3/2ab))
(a-1)*(b-1) = 2a+2b+ab-1
@9:30ish: Your operation looks like the Geometric Product of Geometric Algebra. However this isn’t similar to something learned by Earthlings at a pre-K age. On Skaro, we learn it along many other topics shortly after our births, before we are put into our impenetrable shells.
So what is the metaphor that every kid knows about this? CBA playing guessing game, but thank you for the video!
your voice is very annoying. Try decreasing the pitch using sound processing tools.
Sir you should keep doing video and don't stop, I rather learn math and physics then scroll misinformation and stupid ideology and people's opinion
Hi @TkcUsHegemony thank you so much for your comment! 😊 I definitely intend to continue making content and I am so happy you are watching! I wish you an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊