Hello, thanks for the video. This could just be a personal issue but I found the board a little difficult to read since the text is small. My only constructive criticism would be to make text relatively larger. Nonetheless, great video!
@@dR-bAbAk I don't think my other comment went through. So please excuse me if the message isn't really clear on what I mean. a*b = a+b-ab can be factored as 1-(a-1)(b-1) this gives a clue on how to prove your result. What I'll do next is essentially provide an isomorphism (of semigroups) but due to its simple nature it could be redone without it being done explicit. Take phi: x -> 1-x (a mapping from R to R). This map is clearly bijective. Notice now that phi(a*b) = phi(a+b-ab) = 1-a-b+ab = (1-a)(1-b) = phi(a)phi(b). Hence phi is a semigroup isomorphism, with the image space having the regular real number multiplication. In particular, we know R to be a group if we remove 0.
@@TC159 Thank you for your instructive comment. I truly enjoyed your elegant idea of using isomorphism to solve the problem, and I appreciate you sharing it with me. To explain my approach, I would like to outline three points: 1. Although I did not explicitly mention it in the video, I was implicitly assuming that I was using only the theorems, definitions, and concepts covered in the first chapter of any standard undergraduate Abstract Algebra textbook. Therefore, I did not expect the audience to be familiar with the notion of group isomorphism at this stage. 2. If you were to explain your idea in a video to make it clear to your audience, I believe it would still require approximately the same amount of time as my video, considering time as a good measure of complexity. You would need to verify the associativity of the binary operation, explain the construction of the isomorphism, and prove that it is indeed an isomorphism. I fully agree that for an experienced person, once your idea comes to mind, the rest can follow almost instantly. This is why I acknowledge that your approach is more powerful than the one I presented. However, from a teaching perspective, I often advocate for solutions that involve following standard steps. This approach ensures that learners can solve many problems, even if their solutions are not the most innovative or elegant. From my experience of many years in mathematics, I understand that the most exciting moments come not when we find standard solutions but when we discover elegant and extraordinary approaches. The ability to do the former relates to the effort we put into our learning, while the latter often depends on innate talent. This seemingly paradoxical contrast between what we enjoy the most and what we can actually achieve is perhaps one of the greatest driving forces in life, and mathematics is no exception. Unfortunately, current scientific research supports the idea that we have limited control over the latter. This brings me to my final point. 3. I simply could not see your solution 😊 My apologies for becoming too philosophical. Thank you again for your thoughtful comment and for contributing to this discussion.
marvelous video
Thank you. I am glad that you found it useful.
Hello, thanks for the video. This could just be a personal issue but I found the board a little difficult to read since the text is small. My only constructive criticism would be to make text relatively larger. Nonetheless, great video!
Thanks for the instructive comment. I agree with you. I will try write larger.
This is overly complicated, you can get the same result in like 2 lines
@@TC159 can you please elaborate on how?
@dR-bAbAk sure, do you have an email, so I can send compiled LaTeX to once I'm home?
@@dR-bAbAk I don't think my other comment went through. So please excuse me if the message isn't really clear on what I mean.
a*b = a+b-ab can be factored as 1-(a-1)(b-1) this gives a clue on how to prove your result.
What I'll do next is essentially provide an isomorphism (of semigroups) but due to its simple nature it could be redone without it being done explicit.
Take phi: x -> 1-x (a mapping from R to R). This map is clearly bijective. Notice now that phi(a*b) = phi(a+b-ab) = 1-a-b+ab = (1-a)(1-b) = phi(a)phi(b).
Hence phi is a semigroup isomorphism, with the image space having the regular real number multiplication. In particular, we know R to be a group if we remove 0.
@@TC159 Thank you for your instructive comment. I truly enjoyed your elegant idea of using isomorphism to solve the problem, and I appreciate you sharing it with me. To explain my approach, I would like to outline three points:
1. Although I did not explicitly mention it in the video, I was implicitly assuming that I was using only the theorems, definitions, and concepts covered in the first chapter of any standard undergraduate Abstract Algebra textbook. Therefore, I did not expect the audience to be familiar with the notion of group isomorphism at this stage.
2. If you were to explain your idea in a video to make it clear to your audience, I believe it would still require approximately the same amount of time as my video, considering time as a good measure of complexity. You would need to verify the associativity of the binary operation, explain the construction of the isomorphism, and prove that it is indeed an isomorphism. I fully agree that for an experienced person, once your idea comes to mind, the rest can follow almost instantly. This is why I acknowledge that your approach is more powerful than the one I presented. However, from a teaching perspective, I often advocate for solutions that involve following standard steps. This approach ensures that learners can solve many problems, even if their solutions are not the most innovative or elegant.
From my experience of many years in mathematics, I understand that the most exciting moments come not when we find standard solutions but when we discover elegant and extraordinary approaches. The ability to do the former relates to the effort we put into our learning, while the latter often depends on innate talent. This seemingly paradoxical contrast between what we enjoy the most and what we can actually achieve is perhaps one of the greatest driving forces in life, and mathematics is no exception. Unfortunately, current scientific research supports the idea that we have limited control over the latter. This brings me to my final point.
3. I simply could not see your solution 😊
My apologies for becoming too philosophical.
Thank you again for your thoughtful comment and for contributing to this discussion.