Number Theory | Divisibility Basics

We present some basics of divisibility from elementary number theory.

[lifeofphyraprun]

Finally a course on Number Theory to help me in self study and Olympiad preparation. Thanks a lot sir!

[sebastianszymczyk]

Should it be that 2 divides 16 because 16 = 2*8?

[yanchujun]

Love the videos :), but isn’t 2 times 4 equal to 8 and not 16?

[NishantKumar-xwlg]

Thanks for your efforts, i really liked this video and honestly one of the reason to like this video was blackboard and chalk .

[manuelfalzoialcantara]

It is clear exposition, and you are making proof, step by step, thank you very much, I like your explanation very much.

[georgesadler]

MR. Penn, thank for a fantastic introduction to Number Theory

[samrubenabraham]

Sir, you must have made a slight mistake.... 2 * 4 = 8 and not 16... did you just mean 2^4 = 16?

[gedapruro]

Great Chanel, the best that I have seen talking about Number Theory. You are helping me a lot. Congratulations from Spain.

[llchan]

I notice that your play list closely follows the chapters in "Elementary Number Theory" by Gareth A Jones and J May Jones by Springer. Is that the textbook you're using?

[victorpaesplinio]

I wonder if my proof of 2 and 3 are correct (I mean, if it respects all the integers axioms, I need to practice my ability to write a proof).
Proof of 2: If a│b then exists an integer k such that a.k=b.
Multiplying both sides by an integer c we get c.(a.k)=c.b
By associative: a.(c.k)=c.b. Denoting c.k=k' we have a.k'=c.b and it implies that a│c.b. #
Proof of 3:
If *a│b* then exists a integer k such that a.k=b.
If *a│b* then exists a integer l such that a.l=c.
By 2, there is a integer k' such that a.k'=b.x for some integer x. (i)
By 2, there is a integer l' such that a.l'=c.y" for some integer y. (ii)
Adding (i) and (ii):
a.k'+a.l'=b.x+c.y
a.(k'+l')=b.x+c.y
If (k'+l')=m, then a.m=b.x+c.y and it follows that a│b.x+c.y #

[GOATSOFFOOTBALL]

thank you very much sir .really helped me a lot .now i can clear my doubts through your videos
.and also i am preparing for isi examination

[PunmasterSTP]

Thank you for making these wonderful videos; you explanations are very clear and help me to understand. Divisibility? More like divis-easy-ability now!

[radheyshyam]

Your videos are realy helping me lot. Thank so much Sir

[Blure]

Your channel is gold, man. Much respect.

[sangraampatwardhan]

Hi, I wanted to ask whether all of the videos in your number theory Playlist (113 videos in total) are properly ordered?

[maxpercer]

equivalently,

a | b iff a has a multiple that is b,
or even briefer, a | b iff a has multiple b.

[maxpercer]

Definition: For integers A, B, A | B if and only if B = A*k for some integer k. That is, A | B iff B is a multiple of A."

As an alternative to the word 'divides', A | B can mean equivalently A is a factor of B .

Note: The notation A | B is actually more nuanced that simply saying A is a factor of B. To be consistent with the definition, we say A | B when B can be factored (split up) in such a way "B = x * y", such that x = A and y is some integer.

So though B | B is true because B is a factor of B since B = B, it is true that b | b using the definition since B = B * 1. Why go through all this trouble, instead of just using the definition? I think saying a is a factor of b is a nice mental shortcut.
Also the word factor has the connotation of being split up. Single factorizations of numbers are done for the sake of making statements more concise.

In any case, when i see a number like B, then i think B = x * y * z * w , then x, y, z, w are factors of B.
Also for any number B, B = 1 * B, so 1 is a factor of B, and B = B * 1 , so B is a factor of B.

The proof that a | 1 then a = {-1, 1} is quite short.
Since 1 = 1 * 1 and 1 = (-1) * (-1) , so a = { 1, -1} .
ok, to prove that there are no other factors might involve some heavy proof methods.

[zyraleighnsotor]

I really thought he would notice the mistakes in 2|16. Anyway, thanks for this video!

[xoppa]

6:10 The absolute value of any integer is greater than equal to 1" that is false, the absolute value of an integer, |a|, can be zero.
however, since we assume a | 1, then we reject the case a = 0 since 0 | 1 is impossible.

[VictorHugo-xnjz]

I personally feel like the absolute value proof to not be satisfying, using more intuition than rigor. My constructive approach :
First realize that a and k must be of same sign. For the positive case, if k > 1, then a < 1. As a is an integer and there exists no integer between two consecutive integers (in this case 0 and 1), then k can't be > 1. As k is positive, and cannot be greater than 1, than k = 1. The negative case is similar. QED.