Відео

@@najibhh8912 You are welcome 🙏

Welcome 😊

Thank you for your thoughtful question! What you see in the video is the final, polished version of the solution, but getting there involved a lot of trial and error-scribbling on scratch paper, failing, reworking, and refining. This process is very common in mathematics, as the final solution often hides the messy, behind-the-scenes efforts. When working on abstract algebra problems, especially during my undergraduate studies in pure mathematics, I learned that intuition is often limited, and solving such problems requires persistence and experience. The decisions to explore specific paths, like working with b^6 and b^9, were motivated by the relationships between the given powers (b^2 and b^3) and how they naturally connect to higher powers like 6 and 9. That said, even in retrospect, I sometimes feel I might have overcomplicated the problem. That’s why I invite viewers to share their insights-it’s always possible there’s a simpler solution I overlooked. Collaboration and different perspectives are invaluable in mathematics.

@@dR-bAbAk Thanks for your reply. I suppose what puzzles me most of all is that, in any situation where I might be trying random things out for a problem like this, regrouping products in different ways to try to make use of the given identities to find some simplification or cancellation, if I ever found myself writing down fourth or sixth powers of an element - let alone the ninth! - (given that in the statement of the problem the highest power was the third), I would automatically assume that the track I was on was very much the wrong one, and I cannot conceive that I would have the obstinacy or self-belief to persist. I also wonder how results like this get discovered in the first place: after all, if one finds such a problem in a textbook, one is generally safe to assume that it is actually true, and I can see that that might give one the confidence required to carry on. But what would motivate someone to pursue such arcane avenues if one didn't already have this assurance? Having said all that, and reviewing your solution and your remarks on the relationship between 2, 3, 6, and 9, I think I can now at least conceive of a hint that might possibly have put me on the right track. Namely to focus on the fact that (writing * for inverse here, for legibility) (ab²a*)³ = (ab³a*)².

@russellsharpe288 I see your point. I have never seriously delved into designing novel problems. Personally, I prefer working on a given problem rather than creating a beautiful one-I don’t think I’m particularly talented in that regard.

Thank you!

You are welcome 🙏

Great! I am happy for you.

You are welcome

Thank you too!

You are most welcome

Thank you so much.

@@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.

Thanks for the instructive comment. I agree with you. I will try write larger.

Thank you. I am glad that you found it useful.

I am delighted that you find my videos helpful. Thanks for the encouraging comment!

Hello! 🙂

You are welcome. I am happy that you liked it. By the way, I didn't know that you are also following my videos on this channel as well.

I am happy you found it helpful. God bless you too!

Thanks 🙏

You are welcome. Thanks for your kind comment.

Hopefully, it will come.

Thanks for the encouragement. I will!

Hi, I am doing well. I hope the same for you. My pleasure!

Don't worry! You just need to practice. It is not as hard as it looks. Good luck with your studies.

I'm thrilled to hear that you find the videos helpful, and I truly appreciate your encouraging comment. While my channel may not have garnered a large audience yet, your support keeps me motivated. It is challenging to stay enthusiastic throughout projects, especially when faced with a smaller audience. As for creating videos on how to read and understand mathematics, it's not in my current plans. However, I can recommend some excellent books that might help you in that regard: 1. A Logical Introduction to Proof, by Daniel W. Cunningham (www.amazon.com/Logical-Introduction-Proof-Daniel-Cunningham/dp/1461436303/ref=tmm_hrd_swatch_0?_encoding=UTF8&qid=1703261511&sr=1-1) 2. Mathematical Writing, by Franco Vivaldi (www.amazon.com/Mathematical-Writing-Springer-Undergraduate-Mathematics/dp/1447165268/ref=sr_1_1?crid=2M2NO9TKQFT32&keywords=mathematical+writing&qid=1703261451&s=books&sprefix=mathematical+writin%2Cstripbooks-intl-ship%2C297&sr=1-1) 3. Transition to Advanced Mathematics, by Danilo R. Diedrichs, and Stephen Lovett (www.amazon.com/Transition-Advanced-Mathematics-Textbooks-ebook/dp/B09YVTHXZH/ref=sr_1_8?crid=1JF881LHOLNLL&keywords=transition+to+advanced+mathematics&qid=1703261005&s=books&sprefix=transitio%2Cstripbooks-intl-ship%2C328&sr=1-8) I hope you find the books helpful. Good Luck!

@@dR-bAbAk Thank you for your feedback. Your videos are really great with positive aura. Wish you all the best. God bless you.

That you are using UA-cam platform is in its own an indication that this has already helped you 🙂. Without these mathematical concepts, these level of technology would be impossible!

Thank you. I am happy that you like them.

Thanks!

Welcome

I am glad you like it. I would appreciate it if you could popularize my videos. This definitely encourages me to work more enthusiastically.

Yes, true. I ran out of time and could not cover that. Now I intend to have one more session on "Abstraction" before switching to the next topic of Mathematical Logic.

You are welcome. Right now, I am busy with other projects, but I will do my best to do so.

Yes, it is. Don’t you think so?

Is that so puzzling? 🙂

@@dR-bAbAk ...no señor...escribí algo que no se transmitió...tarde me di cuenta del error...gracias igual Señor

You are right. I should have been more careful about the choices of the options 😊

​@@dR-bAbAk For a non multiple choice question, maybe extend the range to -1 to 1 instead of 0 to 1, and the insightful can reduce the calculus needed (by symmetries) and reduce the risk of tripping up on a lost - sign.

@@michaeltempsch5282 Of course, for a non multiple choice question, getting the exact answer 17/6 without using calculus, or any other exhaustion method is not possible. So, I don’t think in that case, I need to change anything for a “shorts” video. Actually, 17/6 represents the area of the given region by how we define the notion of the area in calculus. Area is defined by a definite integral.

@@dR-bAbAk Absolutely, you can't eliminate calculus completely, but seeing that you don't need to involve the y= -x part at all, and that the rest is equal to 2x the integral of x²+2 from 0 to 1 does simplify things... (IMHO at least). And I wasn't thinking in terms of shorts.

@@michaeltempsch5282 Thanks for your constructive comments.

Do you think I look like him? 🙂

Thanks for the comment. Do you mean when computing I(y)? If so, I chose that substitution because I prefer to have no surds in my integrand after the substitution. If I understood you right, after your substitution, the integral, up to a factor, becomes $\int_{1}^{y^3}\sqrt{t-y^3} dx$. I agree that the limits are simpler, but the integrand contains a square root. Technically speaking, one must introduce another substitution to solve the new integral. Of course, since $y^3$ is a constant, the primitive function will be 2/3(t-y^3)^(3/2). To appreciate the substitution that I used in the video, try to solve the integral $\int_{0}^{1}x^2\sqrt[5]{1+x} dx$. To compute this, I hope that you agree that introducing 1+x to be t^5 rather than t makes life after substitution easier from an integration point of view. Still, the price I should pay is to work with limits 1 and $\sqrt[5]{2}$ rather than with simpler limits 1 and 2, as in the case of the latter substitution.