I discovered your channel a few days ago. What a pity you haven't more subscribers ! I want to learn number theory and go back to the fundamental WITH PROOFS seems an ideal way. Thanks!
This was extemtly useful. Spent 10 minutes trying to understand my textbook explanation of this. I immediately understood what I was missing within 2 minutes of your video. Thank you
What an amazing channel for learning mathematics. I wonder how it has so less subscribers. Simply mind blowing playlist on number theory. Keep growing mate !!
Good algorithm to divide. Can be called "Division by successive aproximation". It's only necesary to generate a seed big enough to reduce the amount of iterations. Good video
Such a great proof. The way you explain things using examples is a lot clear than those literal "bookish" proofs. Thanks a lot. Keeping making things like these :)
Even though it's obvious and doesn't really require it, I want to fill in the gaps of the last step of the uniqueness proof. We have b(q-q')=r'-r, where 0
Does anyone know that this means? Its the first sentence of this cryptography class and I think it means I should drop it: "We say that a nonzero b divides > a if a = mb for some > m, where > a,b, and > m are integers. That is, b divides > a if there is no remainder on division. The notation > b|a is commonly used to mean b divides > a. Also if > b|a , we say that b is a divisor of > a. The positive divisors of 24 are 1,2,3,4,6,8,12 and 24."
Great proof, love your content, I can barely understand while looking at my lecture note, now I feel better when someone explains, keep up the good work, Jesus bless.
@@antoniojg-b8284 what are you studying specifically as an applied math student. I only ask because I will soon be starting this degree as a freshmen in college :D
hi Michael I liked your video about euclidean algorithm but I know an another form to proof, suppose that q e q' are consecutive integers, then exist a rational between them, then we have. q0
Nice one (I'm aware that the length of the videos doesn't allow intuition, but for some, the steps may seem like magic). For the intuition about the set 'S', we can think about all 'b'-steps to the left and right (at the left is how many 'b' can we fit in 'a'). We don't allow it to go less than zero. It becomes apparent that S includes 'a' itself. If 'a' is negative, we simply extend the logic, and go 'b'-steps to the right until we reach something greater than zero, then start adding to the set 'S'. We then want to find 'r'. Of course it's the minimum of 'S', otherwise, we can subtract out another 'b'. For the final part, as we know 'r' and 'r'' are between 0 and 'b', then their difference must be less than 'b'. Now it's obvious that there is only one 'q' to reach between 0 and 'b', as each leap is at least 'b'-length, so anything else would escape.
Great video, but at 6:07 I did not understand how you went from b is greater than zero to 'it is bigger than or equal to one'. By 'it' are you referring to b here or not?
My God :))) I'm not an English speaker and I'm not a high level. But think about it man! How you teach and explain that even I could understand it well 🥺 thank youuuuuu
My godd....just speechless..this proof is just so easy to understand....I wish the same I had in the book. 😔...the way it is given in the book is literally going above my head😂
What would've happened if r >= r' instead of vice versa (10:37)? Can the uniqueness proof be applied also to the generalization of b € Z-{0} instead of b>=0? Thank you
If we considered the other case that r >= r', you'd just rearrange the equation bq+r=bq'+r' to get r-r'=b(q'-q) and the same logic would apply. WLOG" just means that the other cases have the exact same steps and logic but with symbols swapped, and that we don't want to bother writing out something that's essentially identical.
@@salad7389 Since this is from a while ago I forget the details, but in logic, you can assume anything you want at any time. However, whatever consequences you get from an assumption are always chained to that assumption (at first), meaning you need to remember that they were a result of that assumption. If you want to forget about the WLOG business, you can do each case separately. You can first assume r'>=r and work out the consequence, and then assume r>=r' and work out the other consequence, and since both of these are the only possible options, then if their consequences are the same then their consequence has escaped the chains of the assumptions and is true regardless of the assumption. I hope my explanation provides some clarity. You can assume whatever you want at any time basically. We then used this logic theorem: P->Q. R->Q. P or R. Therefore Q.
Oh boy time to binge number theory! I always feel so stupid that I can’t solve the number theory questions on this channel and everyone in the comments are talking about how trivial they are :/ Hopefully this series helps, it’s the only one I can find in number theory on RU-vid, so thank you for making it, Michael Penn!
same!! you probably got better (and I really hope so). I have a test on Thursday and I'm freaking out lol plus: English is not my native language, so it's even harder.
Sometimes I adjust the speed to 1.5 on certain channels, I find it disrespectful, and I am ashamed, even though they don't know. - But with you, Michael, I have to adjust it to 0.75. At that speed you sound more human, I suppose. Sorry. Your math is excellent. ;-)
Not that good actually. The example are contrived and do not explore the realm of possibilities: a being negative, b being larger than a for instance. And because the result is proven over Z this could have been nice. The problem is that the intuition is not that of a problem: you want to measure a with b but the beauty reside in the fact that it is possible even with the case above.
