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The division algorithm -- Number Theory 3 

Michael Penn
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14 окт 2024

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Комментарии : 58   
@MYlo-lq3wy
@MYlo-lq3wy Год назад
Imagine struggling for 2 semesters to understand a simple proofing division algorithm and this gem explains it in just 27 minutes! Thanks a lot!
@goodplacetostop2973
@goodplacetostop2973 3 года назад
2:20 Good Place To See That 14:47 Michael’s « homework » ? 15:58 0 is already in N in my book 25:22 Michael’s homeworks (great places to start) 27:04 Good Place To Stop Who asked, who cares but I’m going into a 3 week vacation on Wednesday. I should be able to post timestamps still but not everyday.
@malcolms1441
@malcolms1441 3 года назад
Have a wonderful vacation!
@boskayer
@boskayer 3 года назад
9:48 bark bark gone give it to ya rip
@resilientcerebrum
@resilientcerebrum 3 года назад
but according to my book, natural numbers start from 1,2,3,4 .... infinity and 0 U {N} is denoted by {W} that is Whole Numbers
@RobsMiscellania
@RobsMiscellania 3 года назад
I'm currently working on a problem of the distribution of primes in certain intervals. I am astonished time and again at the amount of structure that arises from these basic facts about divisibility. In particular, I'm finishing preparations for a lecture in a couple of weeks on the Sieve of Eratosthenes, which I hope you will enjoy on my channel when I publish it. It is clear to me that there are somewhat more modern and sophisticated perspectives, not far off the beaten path, using this ancient technique coupled with other accessible facts, many of which are presented in this video.
@manucitomx
@manucitomx 3 года назад
I wish I could take the full course. Thank you, professor!
@georgesadler7830
@georgesadler7830 2 года назад
Professor Penn, thank you for a deep learning/explanation and proof of the division algorithm in Number Theory 3.
@spegee5332
@spegee5332 3 года назад
1:34 some googling turns up that the a^n || b is said "a to the n fully divides b", not sure if this is the common usage
@Jack_Callcott_AU
@Jack_Callcott_AU 3 года назад
In class we used to say "a double divides b".
@adhamkassem3058
@adhamkassem3058 2 года назад
Your videos are lovely... Thank you Michael
@davidcroft95
@davidcroft95 3 года назад
Great video! Following the lead of this one, can you make also a video about the tricks to understand if a prime divides a generic integer? Thank you in advance
@TheAlbert18361
@TheAlbert18361 2 года назад
Do we have to prove the uniqueness of r and q for the division algorithm? Isn’t min of a subset of natural number is always unique and q being a one to one mapping of r hence also unique?
@jorgesaxon3781
@jorgesaxon3781 Год назад
I was wondering the same thing! From what I can tell it is enough
@hubertvelasquez1358
@hubertvelasquez1358 2 года назад
Trying to find where to practice these problems more. It's very interesting and not as complicated as one would think but, I like to put in the work. Does anyone know, could suggest, a good book or site to practice this more in depth?
@normanstevens4924
@normanstevens4924 3 года назад
20:45 it seems wrong to me that you're using the Archimedean principle (which is theorem of the real numbers provable only using the completeness of the reals) to prove a basic result in the theory of integers / natural numbers.
@integralboi2900
@integralboi2900 3 года назад
When Michael says "Archimedean principle", it's referring to the fact that any nonempty subset of the natural numbers has a minimum. (BTW if you're referring to the property that says the natural numbers are unbounded, then you can prove it without the completeness of R, because it's also true for the rationals.)
@normanstevens4924
@normanstevens4924 3 года назад
​@@integralboi2900 The Archimedean principle is that given two positive numbers x and y, there is an integer n so that nx > y. It shows that there are no "infinitesimal" numbers. The fact that any nonempty subset of the natural numbers has a minimum is the well ordering principle and can be proven using induction.
@goodplacetostop2973
@goodplacetostop2973 3 года назад
@@normanstevens4924 From what I understand, Michael made the same mistake in the first video of the series, calling the well-ordering principle the Archimedean principle.
@integralboi2900
@integralboi2900 3 года назад
@@normanstevens4924 I think you're referring to the archimedean property, Michael calls the property he used "The Archimedean Principle" in this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-IaLUBNw_We4.html
@normanstevens4924
@normanstevens4924 3 года назад
@@integralboi2900 Then the name is wrong. Archimedes was a geometer. Using his name for a result from set theory is absurd.
@randomlife7935
@randomlife7935 3 года назад
In some books, 0|0 but with a ≠ 0 removed from the definition.
@johnsalkeld1088
@johnsalkeld1088 3 года назад
This obviates the need to define N +0 as N has 0 W does not Z has N and -W and is the equivalence class of pairs of integers defined with a1+b2=a2+b1 means a1,a2 ~ b1,b2 etc.
@Victor_Gonzalez98
@Victor_Gonzalez98 3 года назад
Great video, thank you professor.
@johndeacon3520
@johndeacon3520 3 года назад
Can anyone help me with the 3rd part of the homework?
@johndeacon3520
@johndeacon3520 3 года назад
Any approach to the third part of the homework?
@debtanaysarkar9744
@debtanaysarkar9744 2 года назад
I don't know if you are still looking for the solution or not but this is how I tried to figure this out:- drive.google.com/file/d/1QV6l9o1_qrW-93uiDUxirzWQgXHuK719/view?usp=drivesdk
@jorgesaxon3781
@jorgesaxon3781 Год назад
@@debtanaysarkar9744 yeah I guess its enough to just change the union to include integers bigger than -b/2 and the rest of the proof holds up, ps you have very nice handwritting!
@debtanaysarkar9744
@debtanaysarkar9744 Год назад
@@jorgesaxon3781 Thanks
@jorgesaxon3781
@jorgesaxon3781 Год назад
I did something very similar to the general divisibility problem note that is min(s) =< b/2, the proof remains identical, I just added that if min(s) > b/2 to use min(k)-b and added one to the quotient (basically using the first number and index that would be negative), you can prove that this has to work on the second case because the difference would have to be greater than b for it not to work, and, given that we are substracting b that would be impossible
@debtanaysarkar9744
@debtanaysarkar9744 Год назад
@@jorgesaxon3781 👌👌✨✨
@Doug_Piranha
@Doug_Piranha 3 года назад
Hej Michael! Have you thought about revising the first video in this series yet? (see comments to that video)
@goodplacetostop2973
@goodplacetostop2973 3 года назад
Indeed, there are several comments but here's the most detailed : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-IaLUBNw_We4.html&lc=UgzHphYr2JcJs4qFsoh4AaABAg
@doctorb9264
@doctorb9264 3 года назад
exactly divides, example 5^2 [[ 50 ( can't find the bar )
@armacham
@armacham 3 года назад
6:46 if a divides b and b divides a, you have that a=b. But couldn't it also be a = -b? 4 divides -4 and -4 divides 4, right?
@starbeta8603
@starbeta8603 3 года назад
The proposition assumes a and b are members of the natural numbers so a=-b is not allowed since numbers such as -4 are integers but not naturals.
@26-dimesional_Cube
@26-dimesional_Cube 3 года назад
(Note: In Vietnamese, we use the "divides notation" oppositely because it makes sense) For example, you can write: 40 | 8 (because 40/8 = 5, 5 is an integer) But you cannot write: 8 | 40 (like the one that has been introduced in the video) because 8/40 = 1/5, the result is NOT an integer.)
@resilientcerebrum
@resilientcerebrum 3 года назад
/ and | are different
@resilientcerebrum
@resilientcerebrum 3 года назад
But yeah it depends how you define and perceive it
@thomasdemilio6164
@thomasdemilio6164 10 месяцев назад
The | notation is different from the operation of division. Notations have a different purpose than the one of operations in general
@timobrien6957
@timobrien6957 3 года назад
Bruh I was just looking up videos on this like 3 days ago, and now Dr. Penn posts a video on it. What are the odds?? Thank you.
@PubicGore
@PubicGore 3 года назад
Dr. p, not Mr.
@timobrien6957
@timobrien6957 3 года назад
@@PubicGore Right. Also, Dr. P or Dr. Penn, not "Dr.p."
@PubicGore
@PubicGore 3 года назад
@@timobrien6957 I pressed shift but my phone had other ideas. But if you're going to be so rudely technical, I too might as well be. "Right." Is not a proper sentence. And you failed to specify what you are speaking in reference to in your second sentence.
@timobrien6957
@timobrien6957 3 года назад
@@PubicGore Unlucky. Also, since this is the context of a dialogue (and I am clearly the one responding to a statement), one word sentences are grammatically correct. And I made no failure to specify anything.
@PubicGore
@PubicGore 3 года назад
@@timobrien6957 Where are you pulling the rule that being in "context of a dialogue" makes it grammatically correct to use a single word sentence? That's not a rule. Also, you did fail to specify the subject precisely.
@johnsalkeld1088
@johnsalkeld1088 3 года назад
I tend to distinguish whole numbers and natural numbers - whole numbers are 1,2,3 natural are those produced from set theory so emptyset - 0 set containing empty set 1 and the sucessor set contains all the priors in sets
@danielfarbowitz671
@danielfarbowitz671 2 года назад
@9:48 -- proof that x is gonna give it to ya.
@yoav613
@yoav613 3 года назад
That's agood place to stuck, sorry to stop
@MultiCarlio
@MultiCarlio 3 года назад
13:39 "[…] we get that m Is equal to a over DWDHJJH " Uberchad
@MultiCarlio
@MultiCarlio 3 года назад
13:42
@MathAdam
@MathAdam 3 года назад
0:18 "Suppose that a and b are integers" Me: But how do you knoooooowwwwwwww? "And a is not equal to zero" Me: Well, if I can just make stuff up. There's always someone like this in the class. Usually it's me.
@ammyvl1
@ammyvl1 2 года назад
just wait until you learn axiomatic set theory. everything is provable from the axioms
@DendrocnideMoroides
@DendrocnideMoroides Год назад
@@ammyvl1 just wait until you learn Gödel’s Incompleteness Theorem. some true statements are not provable from the axioms
@guill3978
@guill3978 3 года назад
Read the comments on youtube more often pls, even I tried to contact to you through your web page but it was impossible.
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