The only part that tripped me up was on the uniqueness part, where we were able to squeeze (q' - q)b in between 0 and b: 0 r If b is greater than r, and now we're assuming r is greater than r', then surely b is greater than r - r': b > r > r - r' Since r - r' = (q' - q)b, we can say b is greater than all of that: b > r > ( r - r' = (q' - q)b ) Simplifying that down: b > (q' - q)b If I'm wrong then someone please correct me. If not then I hope this helps someone else.
yes, and specifically, r is either zero or a natural number, since it is in range of [0, b). So we know r - r', we know that since r' is not negative, we are not adding, which means that it is still definitely less than b.
Thank you so much for this ! I have spent 2 days trying to understand my professor's proof on this but couldn't understand a thing.Your explanation makes it so clear. I will continue watching your videos throughout the semester to help me pass. Thanks a ton ! Lot of appreciations! Keep these videos coming !
Thanks for this. You make my life easier. In our class lectures I don't understand a thing while in your simple and concise explanation I understand a lot. :-)
very clear explanation, thank you! I was completely lost since my prof gave us a worksheet to prove the theorem with absolutely no direction..... thanks!!!!
I watched some of the videos about the topic you were discussing about. But tbh this video is so simple and easy to understand for beginners like me. Thank you so much sir. Keep the good work. Love from India.
at 3:00 , if you didn't ignore the negatives if would have worked, as the remainder will be -3 instead of 3. Therefore -21 = (-2) * 9 -3 , which is correct. Great video btw. Thank you
beautiful explanation. Could you please explain the idea behind choosing a set for the proof (for example a set was choosen for the proof of division algorithm) i.e. in general how can I see for myself that there underlies a set and work with the set to get a proof?
Do not understand why (q' - q)b < b , you mentioned b is positive integer in the video , so b > 0, but it does not imply b > (q'-q)b , may I have some explanation ? thanks
At 10:30, does n have to be 2a? Wouldn't the properties for membership of our set S still be satisfied if we chose n=a? That way, when a=0 because the least this expression can be is 0 (in the case when b=1). All other possible values of b will evaluate to strictly positive integers. So in either case, S is non-empty.
so hard...finally understood watching over and over again.....which book is this book from/which book are you following? Can you please explain the proof in Gallian's book?
Thank you for great video! Though I have some question in the existence proof namely the part that shows S is nonempty. One question that raised in my mind was if 0 was chosen arbitary in this line "If a >= 0, then a-0*b = a in S". For instance if I choose 2a s.t. a - 2a*b = a(1-2b) and since b > 0 and a >= 0, then a(1-2b) < 0, which is not in S. Does it mean there are some numbers for, "n" in this case, that satisfy the condition a-bn >= 0? A second question is how one can show that a set S is nonempty. Is it enough to somehow show that there exists positive integer values, to say that the S is nonempty? Best regards
I appreciate the video, but you might want to touch up your set notation. You claim S is the "set of remainders," but without specifying what types of values a and b can take, S is very vague and it is not clear which set you are performing division in. In fact, we cannot use well-ordering if this set isn't defined more clearly... Sorry, math makes me extra pedantic, but I do appreciate the video!
@10:27 Why do you also have to consider the case where a < 0 in your proof by cases of the non-emptiness of set S? Isn't it assumed from the definition of the division theorem that a and b are both positive?
b is given to be positive, but a is allowed to be negative, its so the division algorithm looks kinda funny when you do negative numbers, for instance -3 = -2(2) + 1
Prof. Learnifyable: Thanks for the video on the Division Algorithm, a major topic in number theory. As you can see, after 5 days of release, there are 38 views already. You have quite a few followers who are hungry for more basic abstract-algebra videos. Do you plan to release videos on congruences and other topics of modular arithmetic by the end of April 2014? Thank you again -- you are a very talented teacher (abstract math, physics, etc.)! > Benny Lo Calif. 4-21-2014
Thank you for the kind words. The abstract algebra videos do seem to be quite popular. I think I have a few more number theory topics to cover and then I would like to make a video on cyclic groups. There is more to come!
