Primes and Composites -- Number Theory 6

[@goodplacetostop2973]

1:59, 3:49, 8:43 This is a « proof by induction » video 14:00 Homework No Good Place To Stop 😢

[@Alex-kp3hd]

how dare you michael

[@tomatrix7525]

Michael, I think the perspective of your viewers migt be something you’d appreciate as feedback and encouragement so I’d just like to say that I really love that your making your classes openly available on your youtube channel. I haven’t ever taken a number theory class but I feel like I’m in yours. You always have the best explanations and never take anything for granted by proving everything. Cheers!

[@Diamond_heaD_0]

Exponent of x in n! = floor(n/x) + floor(n/x^2) + floor(n/x^3) + ... until floor(n/x^k) becomes 0 So, for number of 0's we are looking for n=5; Number of 0's = [n/5]+[n/5^2]+[n/5^3]+[n/5^4] We can see that for n=620 this value of 152 and n=625 this is 156. (Kind of brute force to determine these 2 numbers though. (We can use binary search)

[@WhattheHectogon]

3:47 Let's dive into the proop!

[@blazedinfernape886]

Let's drive into the poop!

[@CM63_France]

Hi, Good sound without micro in the throat, and without single side band effect. For fun: 3:04 : "great", 5:39 : "ok, great", 14:42 : no place to say if it is a good place to stop or not.

[@GhostyOcean]

The last homework exercise is one I remember doing in my textbook. We proved that for k = max{ integers m | p^m | n!} = sum( floor( n / (p^k) ) ) from k=1 to infinity. So to figure out how many zeros are possible, note that the prime factorization of 10 is 2*5. Since 5>2, the exponent of 5 must be smaller than the exponent of 2 in n! . So we only need to check the power of 5 that divides n!. Luckily we have our theorem from the textbook which says that that exponent is sum( floor( n/(5^k))) from k = 1 to infinity. Creating a data table tells us that at n=624, k=152 and at n=625, k=156. This sequence of k is monotonically increasing, so it it is never equal to 154. Therefore n! cannot end in 154 zeros.

[@oida10000]

The trailing zeros of n! are given by f(n)=sum_(k=1)^(floor(log5(n))) floor(n/5^k). This just the power of 5 in the prime factorisation of n!. As 2 is a smaller prime than 5 there are certainly more multiples of 2 from 1 to n, thus the power of 2 is larger so all 5 combine with a 2 giving 10 adding a trailing zero. So we need to show that this sum is strictly more or strictly less than 154. Ok as the sum is greater than floor(n/5) we know for sure that n=155*5=775 satisfies that n! has more zeros than 154 zeros as floor(775/5)=155 alone is larger. So we can assume n

[@m4riel]

This is a sketch for the third homework, I'll take many 'obvious' things for granted without actually proving them. •If N! has 154 zeros 10^154 | N! => 5^154 | N! ^ 2^154 | N! Since N! is defined as the product of all natural numbers up to N, the factors of 2 and 5 must come from all the multiples of 2 and 5 up to N, respectively. But considering how more frequent the multiples of 2 build up compared to those of 5, we'll only need to look for the multiples of 5. If we look how many factors of 5 each multiple of 5 carries we get a simple pattern: 05 | 1 30 | 1 55 | 1 080 | 1 105 | 1 10 | 1 35 | 1 60 | 1 085 | 1 110 | 1 15 | 1 40 | 1 65 | 1 090 | 1 115 | 1 20 | 1 45 | 1 70 | 1 095 | 1 120 | 1 25 | 2 50 | 2 75 | 2 100 | 2 125 | 3 ... (in case you haven't figured out the pattern, we repeat a previous pattern 5 times, adding a 1 to the last number of the fifth one, so the sequence would be 1111211112...1111311112....1111411112....) When these factor get multiplied, the exponents are added, so we're basically looking for a way to determine the sequence represented by this sum of exponents, which isn't hard to prove: a(1) = 1 a(n+1) = 5*a(n)+1 => a(2)=6 ; a(3)=31 ; a(4)=156 What the last number means is that (5^4)! has 156 factors of 5, i.e. 5^156 | 625! . Since 625=5^4, that means that 624! will have 156-4=152 factors of 5. If N! had 154 factors of five, N would have to be a natural number between 624 and 625, which is a contradiction.

[@m4riel]

I know this isn't efficient at all and there certainly are better ways, but I've already done similar problems so most of this sketch was already consolidated in my head.

[@flo_svs]

I was heading to the same direction, but couldnt find a formal way tout write the a(n) series. Thank you

[@megauser8512]

Nice!

[@thomasdemilio6164]

Just notice that the sum of all factors of 5 up to 625 is (31(factors of 5 up to 125) x 5) +1 (because 625 is 5⁴) = 156. This means that in 625!, 156 factors of five will combine with 156 factors of two to give 156 factors of 10, or 10¹⁵⁶, or 156 zeros. So we take a step back and subtract that 4 (factors of 5 in 625), remaining with 152 zeros for 624!. Because we're in the natural numbers, and 625 is the successor of 624, we can say that for n! there could be 152 zeros or 156 zeros, but never 154 zeros. ■

[@stefanoprodigo3443]

Today I decided to study analys math with you !!

[@AndyGoth111]

It just occurred to me that one way to distinguish (positive) natural numbers from (positive) rational numbers is whether or not their prime factorizations are permitted to have negative exponents

[@cpiantes]

I like how you present your lemmata, in as efficient a manner as possible

[@telnobynoyator_6183]

You could write 1 as a product of no prime numbers ? Like any prime to the zeroth power

[@schweinmachtbree1013]

you could :D

[@georgesadler7830]

Professor Penn, thank you for a massive analysis of Primes and Composites in Number Theory Six.

[@christophniessl9279]

Maybe you should have said at 6:20 that (nx+by) is actually positive, as divisibility is defined only in the natural numbers yet. It is nearly a no-brainer to see it, as if it were negative or even zero, then b as product of this expression with a positive(!) prime p would also be negative or zero => contradiction.

[@victorzoma2749]

Exactly contents as my professor lectures 😂😂😂🔥🔥👍✍️ nice video

[@joshuanugentfitnessjourney3342]

Any chance we can get more infinite products and series?

[@nowshad843]

well, that ended without a good place to stop.

[@elhamidyabderahman5966]

Great job

[@kaan7866]

Hi, is there solutions for the exercises at the end of the video

[@lordvader22]

Excellentia!

[@tostrer2942]

doesn't (10^154)! end with 154 zeros, because it is a multiple of 10^154? edit: if the ending with more than 154 zeros doesn't count than the question is badly stated

[@mistervorex7923]

Is it true that negative numbers aren't primes?

[@schweinmachtbree1013]

The negatives of the prime numbers are what are called "prime elements" in the ring of integers *Z* . A prime element in a ring is an element _p_ such that for any _a_ and _b_, if _p_ | _ab_ then _p_ | _a_ or _p_ | _b_ (which one can show is true for ordinary prime numbers) - it is easy to see that if _p_ is a prime element then so is _-p_ . I like to think of the term "prime number" as being short for "prime natural number", which justifies the restriction that primes are positive - you could call a positive-or-negative prime a "prime integer" if you wanted.

[@sinecurve9999]

What is the smallest prime p such that p = n! + 1 and n > 14? Leave your answer modulo 10^9+7.

[@proninkoystia3829]

can I order your Merch in Russia? Where?

[@ZEPHYRZHANG-mg8zi]

I think I did the questions right

[@laurendoe168]

You totally lost me at 5:03 -> ax+py=1. If a,p,x, and y are all integers.... ax is no less than 1, and py is no less than 1.... so 1+1=1????

[@fransisigos]

x and y are integers, so they can be

[@laurendoe168]

@@fransisigos That possibility occurred to me about 5 minutes after posting, but I left the comment there for confirmation of this possibility.

[@iabervon]

In the "gcd(a,b)=ax+by for some x, y" statement he proved in video 4, if a and b are both positive, exactly one of x and y must be less than or equal to 0, since divisors of a and b are less than or equal to a and b. It's a little weird to write it as addition when you know it's effectively going to be subtraction, but you can't tell trivially which way the subtraction is going to go.

[@SidneySilvaCarnavaleney]

Estimado noble amigo Michael Penn de esta simple Teoría del Número de Canal: Primes y, con mi respeto a los maestros, estudiantes y amigos de este simple canal, reportaré algo muy intrigante acerca de estos números primos, con un simple PA (Progresión Aritmética), Puedo decir con total veracidad, demostrando Científica y Matemáticamente que los números que mencionaré a continuación no son primos, y los primos gemelos no existen: dos; 19; 41; 59; 61; 79; 101; 139; 179; 181; 199; 239; 241; 281; 359; 401; 419; 421; 439; 461; 479; 499; 521; 541; 599; 601; 619; 641; 659; 661; 701; 719; 739; 761; 821; 839; 859; 881; 919; 941; 1019; 1021; 1039; 1061; 1181; 1201; 1259; 1279; 1301; 1319; 1321; 1361; 1381; 1399; 1439; 1459; 1481; 1499; 1559; 1579; 1601; 1619; 1621; 1699; 1721; 1741; 1759; 1801; 1861; 1879; 1901; 1979; ¿Y cómo sería la Hipótesis de Rieman ?, si estos no son primos? Al tratarse de un descubrimiento innovador en el Universo de las Matemáticas, los enunciados de épocas pasadas han perdido fuerza, dice el autor de la obra "Un atrevimiento del π ser racional", Sr. Sidney Silva.

[@laurendoe168]

If 1 is not prime, and if "product" is defined as one number multiplied by at least one other number.... then primes themselves are not a product of primes since they have only one prime number factor (itself).

[@iabervon]

We define a "product" as a list of at least one number, all multiplied together. Actually, we tend to say that the product of an empty list is defined to be 1. It's necessary to count 1 as a prime, because if we did, numbers would have multiple prime factorizations with different numbers of 1s in them.

[@megauser8512]

@@iabervon *not count*

[@rafael7696]

It's nice but I prefer olympiad problems