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 !
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.
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!!!!
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. :-)
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 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.
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?
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?
@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
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
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
Remember that in order to make Well Ordering Principle works, we need to make the set to have only natural numbers (in this case a-nb>=0). So you can see, plugging n=0 and with given condition a>=0, this works.
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!
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!
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 !
Wait! Professor?! I have to do it as a high school 16 yo teenager
@@abrarfahimul6440 Country?
Thank you for this proof. At times it was really hard to follow but I was able to understand the concept behind it.
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.
It helped me a lot thank you 🙏
Thank you 😊😊
thank you !
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.
Nice. That's an interesting proof to study because it ties a lot together.
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!!!!
Thanks!
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. :-)
Thanks! I'm glad I could help.
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
Ok. Figured it out - but he does not explain it well.
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.
Whoever run this account is the goat
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.
you are RIGHT, what's where I got confused too
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
great video man,understood it soo well...thanks alott
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?
This is such a good video! Thank u very much
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
Thanku sir..
Why do you assume b>0? In the Euclid's division a and b can be any number (except 0 for b)
Can you please suggest me that book?
in the existence portion of the proof, where does the 2a come from? Could we have chosen a, 3a, 100a, and so on?
+ssbsnb1 Any of those choices would work just fine. There are no deep reasons behind my choice of 2a. I hope that helps.
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?
@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
Hey, How where you able to get q-(q+1)b>=0 from a-qb-b
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
earlier it was proven that q'b-qb= r'-r, and there was a condition that 0≤r'
Thank you veryyyyy muchhhhh sir! 😊😊😊
What about Mathematical Investigations and Model can you give video lectures...
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 have a stupid question doesnt the well ordering principle work only on positive integers how we use it with zero
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?
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.
It is difficult to understand even with your clear explanation.
u n me brother
Thankyou ❤
what if a and b are both positive and b>a
how (a-(q+1)b)
If a < b, then q will not be an integer. Right? How will the algorithm work in this case?
If 0 < a < b, let q = 0 and hence r = a, which still fulfills the conditions that r
In 10:02 you set n = 0, why?
Remember that in order to make Well Ordering Principle works, we need to make the set to have only natural numbers (in this case a-nb>=0). So you can see, plugging n=0 and with given condition a>=0, this works.
Thank you
thanks
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!
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!
Why can we conclude that ((a-(q+1)b) < r)?
Because r=a-(q)b and a-(q+1)b < a-(q)b
wow this is not easy
this pissed me off
Please, give lectures in hindi
ok bhai