(Abstract Algebra 1) The Division Algorithm

A proof of the division algorithm using the well-ordering principle.

[@anamtaj]

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 !

[@cm7686]

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.

[@nutzz9990]

It helped me a lot thank you 🙏

[@Shortsss0.1]

Thank you 😊😊

[@azizyosri2058]

thank you !

[@darcash1738]

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.

[@simonherrera9373]

Thank you for this proof. At times it was really hard to follow but I was able to understand the concept behind it.

[@user-fn1zb1wc7f]

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.

[@hayley9546]

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!!!!

[@learnifyable]

Thanks!

[@EarthandHabitants]

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. :-)

[@learnifyable]

Thanks! I'm glad I could help.

[@harshsharma4856]

great video man,understood it soo well...thanks alott

[@debloated9589]

This is such a good video! Thank u very much

[@bibek2599]

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?

[@TuananhNguyen-kl8ud]

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

[@souverain1er]

How do we know that r is the least element of S? It is stated/assumed without proof. All we show is that r>=0 and r

[@souverain1er]

Ok. Figured it out - but he does not explain it well.

[@ankurc]

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?

[@IPear]

Why do you assume b>0? In the Euclid's division a and b can be any number (except 0 for b)

[@tchevskidorvilme7371]

Whoever run this account is the goat

[@EarthandHabitants]

What about Mathematical Investigations and Model can you give video lectures...

[@andy919896]

Hey, How where you able to get q-(q+1)b>=0 from a-qb-b

[@ssbsnb1200]

in the existence portion of the proof, where does the 2a come from? Could we have chosen a, 3a, 100a, and so on?

[@learnifyable]

+ssbsnb1 Any of those choices would work just fine. There are no deep reasons behind my choice of 2a. I hope that helps.

[@nicholascousar4306]

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.

[@junlinli6170]

you are RIGHT, what's where I got confused too

[@nainamat6861]

Thank you veryyyyy muchhhhh sir! 😊😊😊

[@Xardas_]

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

[@abdelrahmaneissa1463]

I have a stupid question doesnt the well ordering principle work only on positive integers how we use it with zero

[@nicholascousar1559]

I thought the Well Ordering Principle only works on sets of positive integers? Is the W.O.P. if a set contains 0 as an element?

[@blownspeakersss]

It works on any subset of the natural numbers. And since the natural numbers are a subset of the integers, it works on *some* subsets of the integers.

[@iqramaqbool8734]

Thanku sir..

[@davidlusagila8939]

Thank you

[@darcash1738]

maybe it is obvious bc a, n, and b are all defined as ints, but should we also say a-nb exists in the nat nums to use the Well ordering principle

[@Anchal-jt9bp]

Thankyou ❤

[@davidjoseph7185]

@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?

[@ghalibsyed3218]

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

[@deepanikarunaratne2075]

Can you please suggest me that book?

[@arnavchauhan3476]

If a < b, then q will not be an integer. Right? How will the algorithm work in this case?

[@houjinpeh7831]

If 0 < a < b, let q = 0 and hence r = a, which still fulfills the conditions that r

[@subashkafle]

what if a and b are both positive and b>a

[@levinkwong3120]

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

[@sunyijin]

earlier it was proven that q'b-qb= r'-r, and there was a condition that 0≤r'

[@shahadatali74]

thanks

[@JKMizzle]

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!

[@janbendrixmalagayo490]

In 10:02 you set n = 0, why?

[@Love_Hope_from_Above]

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

[@learnifyable]

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!

[@robinandrews5613]

It is difficult to understand even with your clear explanation.

[@lemyul]

u n me brother

[@jyo9517]

how (a-(q+1)b)

[@Quintenkonijn]

Why can we conclude that ((a-(q+1)b) < r)?

[@brianbertness9221]

Because r=a-(q)b and a-(q+1)b < a-(q)b

[@lemyul]

wow this is not easy

[@rbin4205]

this pissed me off

[@majidulsk1006]

Please, give lectures in hindi

[@zeeyadkhan2067]

ok bhai