Tsk! Numberphile, why are you eroding the self esteem of real numbers by popularizing the impossible standards of 'perfect' numbers? Do you know what percentage of numbers can actually meet those criteria? You do? Nevermind then.
Matt is the author of Things to Make and Do in the Fourth Dimension. You can support him by checking out his book... On Amazon US: bit.ly/Matt_4D_US Amazon UK: bit.ly/Matt_4D_UK Signed: bit.ly/Matt_Signed
I enjoy watching your videos very much! And I often wonder, to what conclusions good mathematicians like you will come to if you will try to find patterns in the market charts. I'm not talking about finding the "holy grail" but it is still very fascinating. do you really believe that the market is completely random? I would love to see a video on the topic! as a trader and a subscriber : )
baga inferno I think you can find patterns everywhere and use math to understand it. But mysteries about numbers I think are usually only inside mathematics itself.
6 should be a super duper perfect number because not only do you get 6 when you add its proper factors together (3+2+1) but you also get 6 when you multiply its proper factors together (3x2x1) :D
@Amber Berkenveld 1 x 2 x 4 x 7 x 14 is not 28, it is in fact 784. The reason that 6 has this property is that it is small enough that only 1 pair of factors is contained in its proper factors (2 and 3, 1 doesn't count because anything times 1 is itself). What is interesting is that, since 1 and itself do not count, multiplying all of the proper factors of the perfect number gives you a power of the number.
@@matthewstuckenbruck5834 These smallest of numbers have the most interesting properties. For instance, that 30 is the largest number whose smaller co-primes are all prime. All larger numbers have composite co-primes.
thanks. also, thanks for the cadence, cogent flow, and positivity in your methods of teaching this. you're not just teaching, your conveying it with a thought to educational psychology and how people learn. grazie.
Being someone who has previously never learned about Perfect Numbers or Mersenne Primes, I appreciate all the background you provide. I was slightly confused about if n had to be prime or n-1 one had to be prime--so I decided to get out a piece of paper myself and do some exploring. I proved to myself, as you did, that n must be prime. I am continuously satisfied with the patterns that appear in math and I appreciate you exposing them! I wonder though what a practical use for this information would be--or is it just for fun? Even if it is merely amusement, very interesting video!
2,147,483,647 is my favorite number. It's 2^(31)-1 and it's a *double* Mersenne Prime. It's the maximum value for a signed 32-bit integer, and just last month, for the first time ever, a RUclips video surpassed this barrier in views. For a short period of time, Gangnam Style's view count was actually in the negatives until RUclips changed the view count to a signed 64-bit integer. However, this is rumored to have been caused by Google as a joke.
2,147,483,647 is actually not the maximum value of a 32-bit integer! It is the maximum value of a signed 32-bit integer. The maximum of an unsigned 32-bit integer is 2^32-1.
Trevor Whit I don't confidently feel that I understand your attempt at a witty and sarcastic remark. Do you perhaps mean that I searched for information on the topic?
_And so, in fact, that pattern, when people spot these prime values in n, giving these numbers prime, only works in one direction_ **shows a picture of One Direction**
To all those people who read the news of the discovery of 50th Mersenne prime and searched for it on RUclips and landed here to read my comment I say to you, "Have a nice day."
This video helpfully reminded me that 28 is a perfect number. The math teacher in the family recently turned 27, which my father-in-law pointed out is 3^3. Since the next n^n (for whole numbers at least) is far older than we expect she'll ever reach, he was quite proud of himself. I pointed out there were other interesting things for later ages, such as 28 being a perfect number. Of course, all numbers are interesting somehow...
Gonna throw out there that I'm loving Matt's book, and as a maths major in Canada, I feel compelled to thrust it at my fellow students as often as possible. Also, anyone else here watch Doctor Who, then year Matt Parker say 'primey' and immediately think a math-parody of Doctor Who would be perfect?
"The ability to wear a hat is a necessary but not sufficient condition to be a girl" I'm now trying to figure out an haircut strange enough (but possible) such that it would then be impossible to wear a hat (perhaps some kind of mohawk ?). Therefore a girl with such an haircut would not be a girl anymore.
With a, b, n being natural numbers and a>b, we have 1. If (a^n-b^n)/(a-b) is prime, then n is prime. 2. If n is odd and (a^n+b^n)/(a+b) is prime, then n is prime.
For clarity - The reverse statement doesn't include all primes which is why it works. It is a subset of prime numbers that make (2^n)-1 = a prime number but not the subset of all prime numbers.
1:00 Or you could say that the sum of the number's least n-1 factors is equal to its greatest factor. ie: 6: factors: 1,2,3,6. and 1+2+3=6 28: factors: 1,2,4,7,14,28. and 1+2+4+7+14=28
Wow, this is quite the coincidence, at math class we were talking about this stuff, and our teacher recommender watching this channel (even though I and quite a few others already were subscribed)
I didn't make it far through the video without appreciating the precision SLASH that six received. It makes me want to coin "naught-six" with that symbol just to see it more.
This was a solid warning about casually abusing the modus moron inference rule. People think that logical validity is apriori truth when it may just be persuasively arranged statements that don't have substantial definitions. Grammar -> Logic -> Rhetoric Rhetoric(Grammar, Logic) -> Coherent/Cogent but not necessarily true statements.
It is pretty trivial to prove that for any Mersenne Prime n has to be prime. Just note that 2^n-1 is factorisable for any integer n>1. It is obvious to see that both factors actually are >1. That concludes the proof.
one more, TODAY (06-28) is a perfect day ! Anybody can notice that (1) Every Mersenne prime (excluding Mersenne prime 3) ends ALWAYS with 1 OR 7 ONLY ! (2) The division of every Mersenne prime (excluding Mersenne prime 3) by 6 gives a REMAINDER 1 ONLY ! These above two statements are true for MERSENNE NUMBER too !
06-28 is also τ day, of course. "(1) Every Mersenne prime (excluding Mersenne prime 3) ends ALWAYS with 1 OR 7 ONLY !" That's because every odd power of 2 ends with either 2 or 8. And 2ⁿ-1 can only be prime if n is. 3 is the exceptional Mersenne Prime here, because its seed is the only even prime, 2. This statement, BTW, is not true of all Mersenne numbers; M(4)=15, e.g. "(2) The division of every Mersenne prime (excluding Mersenne prime 3) by 6 gives a REMAINDER 1 ONLY !" That's because every odd power of 2 is congruent to 2 mod 6. And once again, this isn't true of all Mersenne numbers. M(2k) ≡ 3 mod 6. Both these statements can easily be verified using modular arithmetic. Mod 10, the powers of 2, starting with 2¹ go 2 ≡ 2 2·2 ≡ 4 4·2 ≡ 8 8·2 ≡ 6 6·2 ≡ 2 and so the same pattern repeats every 4 steps. And in mod 6, 2 ≡ 2 2·2 ≡ 4 4·2 = 8 ≡ 2 and so that pattern repeats every 2 steps. And the Mersenne numbers are just congruent to 1 less than each of these.
I think doubling is better than using proper factors, because for 1 to be a factor requires that the number itself is a factor, so it seems improper to include 1 if not including the number itself. I'll just translate everything Matt says to make it fit the right way, since he did it wrong 😁.
I said this in the other video, but wouldn't one be a perfect number? It's only factor is itself, so the sum of its factors would be itself. He also said that if there were an odd perfect number, it wouldn't have a Mersenne Prime as a factor, and one isn't prime. Maybe I am mistaken, feel free to correct me if this is the case.
Well... if we're excluding the number itself from the divisors (which we do for the whole rest), the sum of all proper (!) divisors of 1 is 0. Which isn't 1.
Moritz Durtschi Good answer. What I get hung up on is this: the definition of factors are numbers that can be multiplied together to get the original number. How is one considered a proper factor if the number it would be multiplied by isn't considered a proper factor? Wouldn't both lose their "proper factor status" if either of them did? Or to put it another way: given that x times y = z, where x, y and z are all 1: how can x by proper if y is not, when it's just as legit to say that y is proper and x is not? Doesn't that make 1 a lonely number? Oh... wait... uh, what?!? I mean, a factor can't be "lonely" as such because it needs to be paired up to be included in the set. Seriously though, given a formula where x times y = z, it's hard to define one as a factor and not both without seeming to break the meaning of a factor, whether it be proper or not. Even so - with 1 times x = x, it would seem more appropriate to claim that 1 is NOT proper whereas x IS proper, for much the same reason as how primes are considered. Yes - it's all about defining it using their properties, but it's counter intuitive and somewhat contradicting when looked at from the perspective of the definition of factors and the treatment of 1 regarding primes.
It weirds me out that 1 is included in it, and the fact that we exclude the number itself makes it even weirder to me. That's just feeling though. Out of curiosity, what if we take one out of the equation and simply do all the factors besides one and a number summed, are there numbers where this sum equals the original number itself. Since there appears to be no even perfect number, even though we can't prove it, is there an odd number of the sort I just described? Curious to know
Suppose there were such a number n. Since n is odd, n isn't divisible by 2 and so also isn't divisible by any multiple of 2; hence all divisors are odd. Since we aren't counting 1, suppose all divisors are in pairs m_1, k_1, m_2, k_2, ... such that each m_i < √n and each k_i > √n and m_i * k_i = n. Then since all these divisors must be odd, the sum of all these divisors must be even and so is a contradiction. Instead, n must then have a divisor equal to itself, and so n must be a perfect square. If √n is prime, n's prime factorization is (√n)^2, but √n + √n < n (for n > 4), so √n must be composite. I stopped here because I couldn't think of anything useful beyond this that could narrow down or disprove. To summarize, if there were such a number, it would be an odd perfect square, and its square root is both odd and composite. I can't really tell whether such a number exists, but it would also be larger than the first four perfect numbers, after testing a few small candidates.
Let F(n+1) = 2^F(n) - 1. If we set F(0) to be some non-Mersenne prime p, is it possible for there to be some p for which F(n) is prime for all positive integer values of n? I'm currently researching p = 2, but I'd like some outside input on whether such a series would be finite or not. I'm *hoping* the series is infinite, but I don't want that to skew my ability to form a proof.
If we use the equation (2^n -1)(2^n-1) I noticed that the number of divisors(including the number itself) of a perfect number is always equal 2n. Does someone know why is that?
Maybe that only works with the smaller ones but is it normal that if you go from highest factor to lowest that they always rounded up halfs of each other? like plays that a role in it. So ⌈Factor n⌉ = (Factor at n+1) / 2
A more formal statement for Mersenne Primes would be x⊂M( 2^x -1 ∧ P(x)) where _M_ is the set of Mersenne Primes and P(x) a function which is true if _x_ is a prime.
Since the numbers go off to fast for me to check, but so far can you say that if you start with 2 and work out the Mersenne prime of that you get 3 (Prime!) . Now take 3 and work out the next prime which is 7. Take 7 and get 127. Take 127 and you'll get another prime (170141183460469231731687303715884105727) Now: Does that work indefinitely?
Unlikely we'll ever be able to check that last one, and almost certainly impossible to prove/disprove that the pattern will work forever. Extremely unlikely that it does, however.
I don't understand how the discovery of a new Mersenne prime can lead you to determine a new perfect number. Just because it gives you a factor of the potential perfect number, how does one go about using that factor to determine the corresponding perfect number?
What I like about the argument against prime 1 here is its circularity. 1 is not prime because when you plug it into the formula you get 1... which is not prime. Why? Because when you....
What I like from maths are the patherns we can find. For instance, If the sommes of the composant of n is equal to a multiple of 3. then n is a multiple of 3 Example: 342 = 3+4+2= 9 342/3 = 114 19874989494 = 1+9+8+7+4+9+8+9+4+9+4 = 72 = 7+2 = 9 19874989494/3 = 624996498 and so on :P
Hi i love the way how you make math so simple, i am student of math teacher, and i wonderig if there´s a chance to do a video of group theory plis ? (I am from Chile, so i am apologize if there is something bad wirting)
if the rule that n=any prime number when 2^n-1 is a prime number always applied, there would be no point in finding the biggest primes, because you can do 2^(the largest prime number)-1, which would be a prime if the rule always applied. Close...
Like Mersenne primes could we have another set of primes, P=(3^p -2) where p is prime where p=2 => P=7 , p=5=>P=241 and so on...? Can we call it Arora's prime of its not taken!
I believe there has been some study of primes that are of the form N(p,n) = (pⁿ-1)/(p-1), for primes p. Idk whether these are named. And here, too, n must be prime in order for N to be prime. (This is a necessary, but not sufficient condition.) When p=2, these are the Mersenne primes. When n=2, N = p+1, so that N(p,2) is prime only for p=2.
Would someone be able to clear something up that I'm missing that I'm pretty sure is basic. Its the numbers in the 2n-1 column that are the mersenne primes right? not the n's that it works for. I'm just a little confused and I know only one of them are mersenne primes.
Love the opening: "Today we're gonna do some math." Wow, who would ever guess *that* in a million years? Gee, I came here expecting to get a key lime pie recipe... (That *is* what that logo depicts, isn't it? The one with "π" in it? But wait, that's an awfully odd shade of green . . .)
If we exclude the 1 from the factors we sum up, there has no perfect number been found yet, nor has it been proven that a such number doesn't exist. I wonder why the ancient Greeks didn't exclude the 1.
as someone who hasnt studied math at all, but is a nerd, ive always wonder how an odd perfect number is not considered impossible, but prime factorization is considered fact as to me that would need to break, can someone educate me where im off in thinking watching these sorts of vids.
6:40
Oh, that's cool, who would've guessed that 34,850,340 was a perfect numb-
DIGITS?! 34 MILLION DIGITS?!
It's funny when numbers get bigger than our average ability to reckon them, lol.
computers lol
that's 34/8=4.25MB to store. we've got plenty more memory in any given home tablet pc.
??.
brady must be an honorary mathematician by now
arcadus I seriously doubt it
"You've been in this game for too long!" I love this guy. :D
the "We call those people mathematicians" joke was golden, because it was no nonchalant
"Today we're going to do some maths."
Thanks, Maths Parker, for your influential turn of phrase.
The perfect number doesn't exi-
THIS HAS 42 LIKES! NO ONE RUIN IT
@@hadizahedi5651 Stfu
@@azerefendizade6017 hey I know it's been a year but that was completely uncalled for
@@qk7x Your birth was uncalled for.
@@aljosamrsovic2116 hey you, guess what? Insulting people who are trying to politely give their opinion does not make you cool or funny.
Tsk! Numberphile, why are you eroding the self esteem of real numbers by popularizing the impossible standards of 'perfect' numbers? Do you know what percentage of numbers can actually meet those criteria?
You do? Nevermind then.
Smiled.
underrated comment
Yes, it's 0% (in the limit as n→∞).
To paraphrase Andy Kaufman, as Latka Gravas in the sitcom Taxi, "Boy, math sure is a tough town!"
9
??.
Matt is the author of Things to Make and Do in the Fourth Dimension. You can support him by checking out his book...
On Amazon US: bit.ly/Matt_4D_US Amazon UK: bit.ly/Matt_4D_UK Signed: bit.ly/Matt_Signed
I enjoy watching your videos very much! And I often wonder, to what conclusions good mathematicians like you will come to if you will try to find patterns in the market charts. I'm not talking about finding the "holy grail" but it is still very fascinating. do you really believe that the market is completely random? I would love to see a video on the topic! as a trader and a subscriber : )
baga inferno I mean, if you like to search for patterns, and solve mysteries that involving numbers, the market is the place to be isn't it?
baga inferno I think you can find patterns everywhere and use math to understand it. But mysteries about numbers I think are usually only inside mathematics itself.
But what if you wear a fedora?
AstronomyTheorist At RUclips n is an even number that you plug in to the equation to get the number that the equation equals to
6 should be a super duper perfect number because not only do you get 6 when you add its proper factors together (3+2+1) but you also get 6 when you multiply its proper factors together (3x2x1) :D
And if you get 3 of them you get.....
ILLUMINATI
@Amber Berkenveld 1 x 2 x 4 x 7 x 14 is not 28, it is in fact 784. The reason that 6 has this property is that it is small enough that only 1 pair of factors is contained in its proper factors (2 and 3, 1 doesn't count because anything times 1 is itself). What is interesting is that, since 1 and itself do not count, multiplying all of the proper factors of the perfect number gives you a power of the number.
@@matthewstuckenbruck5834 These smallest of numbers have the most interesting properties.
For instance, that 30 is the largest number whose smaller co-primes are all prime. All larger numbers have composite co-primes.
This is possible only if you have two prime factors(no power) for that perfect no in the form p1 * p2.
Ah, Matt is always awesome, thanks for the videos!
Wow that intro sentence... I never expected they would do maths on this channel ;D
thanks. also, thanks for the cadence, cogent flow, and positivity in your methods of teaching this. you're not just teaching, your conveying it with a thought to educational psychology and how people learn. grazie.
Being someone who has previously never learned about Perfect Numbers or Mersenne Primes, I appreciate all the background you provide. I was slightly confused about if n had to be prime or n-1 one had to be prime--so I decided to get out a piece of paper myself and do some exploring. I proved to myself, as you did, that n must be prime. I am continuously satisfied with the patterns that appear in math and I appreciate you exposing them! I wonder though what a practical use for this information would be--or is it just for fun? Even if it is merely amusement, very interesting video!
2,147,483,647 is my favorite number. It's 2^(31)-1 and it's a *double* Mersenne Prime. It's the maximum value for a signed 32-bit integer, and just last month, for the first time ever, a RUclips video surpassed this barrier in views. For a short period of time, Gangnam Style's view count was actually in the negatives until RUclips changed the view count to a signed 64-bit integer. However, this is rumored to have been caused by Google as a joke.
2,147,483,647 is actually not the maximum value of a 32-bit integer! It is the maximum value of a signed 32-bit integer. The maximum of an unsigned 32-bit integer is 2^32-1.
PassingTimeAlong Oh, oops! I forgot to specify that!
PassingTimeAlong *2^31-1
NO WAY, YOU CAN USE GOOGLE!!!
Trevor Whit I don't confidently feel that I understand your attempt at a witty and sarcastic remark. Do you perhaps mean that I searched for information on the topic?
I'm going to say that a number is primey in front of my Number Systems lecturer and see what their reaction is.
What was their reaction?
511 does look very 'primey' lol
That's because it is
7 * 73
:'(
Also 2047 = 23 * 89 :'(
I loled at this, too
_And so, in fact, that pattern, when people spot these prime values in n, giving these numbers prime, only works in one direction_ **shows a picture of One Direction**
I, too, watched this video.
Moritz Durtschi I didn't watch the video.
Stop everything! Numberphile just uploaded a new video!
"the hat wearing ability" is the peak of the video imo
To all those people who read the news of the discovery of 50th Mersenne prime and searched for it on RUclips and landed here to read my comment I say to you, "Have a nice day."
This video helpfully reminded me that 28 is a perfect number. The math teacher in the family recently turned 27, which my father-in-law pointed out is 3^3. Since the next n^n (for whole numbers at least) is far older than we expect she'll ever reach, he was quite proud of himself. I pointed out there were other interesting things for later ages, such as 28 being a perfect number. Of course, all numbers are interesting somehow...
do you know the proof that all numbers are interesting
@@wyboo2019 I was referencing that, yes
Gonna throw out there that I'm loving Matt's book, and as a maths major in Canada, I feel compelled to thrust it at my fellow students as often as possible.
Also, anyone else here watch Doctor Who, then year Matt Parker say 'primey' and immediately think a math-parody of Doctor Who would be perfect?
I LOVE perfect numbers, I was so excited when I saw this video!!
Cool I was looking for more info on Mersenne numbers but it looks like you in the process summed up perfect numbers(which I already knew)
Love you guys ❤
"The ability to wear a hat is a necessary but not sufficient condition to be a girl"
I'm now trying to figure out an haircut strange enough (but possible) such that it would then be impossible to wear a hat (perhaps some kind of mohawk ?). Therefore a girl with such an haircut would not be a girl anymore.
They could still wear a hat if they squish their mohawk; P
And we can consider a "hat" a infinity of things that could be adapted to virtualy any haircut.
As a manager of a restaurant that requires the wearing of a hat, there are hairstyles that do indeed make wearing a hat impossible.
This video and this comment are offensive to girls with no heads. What a shame.
With a, b, n being natural numbers and a>b, we have
1. If (a^n-b^n)/(a-b) is prime, then n is prime.
2. If n is odd and (a^n+b^n)/(a+b) is prime, then n is prime.
For clarity - The reverse statement doesn't include all primes which is why it works. It is a subset of prime numbers that make (2^n)-1 = a prime number but not the subset of all prime numbers.
4:06, 1023 is obviously not prime. Add its digits together. 1 + 0 + 2 + 3 = 6. 6 is divisible by 3, thus 1023 is divisible by 3.
1:00 Or you could say that the sum of the number's least n-1 factors is equal to its greatest factor.
ie: 6: factors: 1,2,3,6. and 1+2+3=6
28: factors: 1,2,4,7,14,28. and 1+2+4+7+14=28
Nice ! Thanks for your work Brady !
2:55 Those perfect ticks... I luv it
Wow, this is quite the coincidence, at math class we were talking about this stuff, and our teacher recommender watching this channel (even though I and quite a few others already were subscribed)
Thanks, that was really helpful.
I'm turning 31 this year! So excited!
It's interesting that when 8 is worked into the Mersenne Prime formula the resulting value equals one less than the square of 16!
Half speed at the beginning sounds hilarious :P
So, what is going on at time stamp 4:49?
It's One Direction, the band. They were massively popular at the time of recording this video.
that cut to one direction was crazy, lol
A cheeky 1 direction inserted
5:10 i was wearing a hat... LOL
@Max mersenne primes are the 2^n - 1. For different values of n you will get different mersenne primes
I didn't make it far through the video without appreciating the precision SLASH that six received. It makes me want to coin "naught-six" with that symbol just to see it more.
This makes your last video posted make so much more sense :)
Would 74,207,281 also be a perfect number (as per 6:41) because of the recently discovered Mersenne prime?
This was a solid warning about casually abusing the modus moron inference rule.
People think that logical validity is apriori truth when it may just be persuasively arranged statements that don't have substantial definitions.
Grammar -> Logic -> Rhetoric
Rhetoric(Grammar, Logic) -> Coherent/Cogent but not necessarily true statements.
In binary, 2^n - 1 is written as n ones. So if n is composite, any factor of n ones would also be divisible into n ones.
Nice!
In fact in any base _b_ the number b^n -1 can be written as _n_ (b-1):s.
check my comment on the follow-up video ;)
TheLeftLibertarianAtheist 10^3-1 = 999, not 111. You meant n (b-1)s.
3snoW Correct. Thanks for correcting me. :)
It is pretty trivial to prove that for any Mersenne Prime n has to be prime. Just note that 2^n-1 is factorisable for any integer n>1. It is obvious to see that both factors actually are >1. That concludes the proof.
06-28-2016.
Do we declare this day (28th day of 06th month) as perfect day?
just like π day. (03-14).
if 03-14 (π day) is doubled, 06-28 is a perfect day !.
one more, TODAY (06-28) is a perfect day !
Anybody can notice that
(1) Every Mersenne prime (excluding Mersenne prime 3) ends ALWAYS with 1 OR 7 ONLY !
(2) The division of every Mersenne prime (excluding Mersenne prime 3) by 6 gives a REMAINDER 1 ONLY !
These above two statements are true for MERSENNE NUMBER too !
06-28 is also τ day, of course.
"(1) Every Mersenne prime (excluding Mersenne prime 3) ends ALWAYS with 1 OR 7 ONLY !" That's because every odd power of 2 ends with either 2 or 8. And 2ⁿ-1 can only be prime if n is. 3 is the exceptional Mersenne Prime here, because its seed is the only even prime, 2. This statement, BTW, is not true of all Mersenne numbers; M(4)=15, e.g.
"(2) The division of every Mersenne prime (excluding Mersenne prime 3) by 6 gives a REMAINDER 1 ONLY !" That's because every odd power of 2 is congruent to 2 mod 6. And once again, this isn't true of all Mersenne numbers. M(2k) ≡ 3 mod 6.
Both these statements can easily be verified using modular arithmetic. Mod 10, the powers of 2, starting with 2¹ go
2 ≡ 2
2·2 ≡ 4
4·2 ≡ 8
8·2 ≡ 6
6·2 ≡ 2
and so the same pattern repeats every 4 steps. And in mod 6,
2 ≡ 2
2·2 ≡ 4
4·2 = 8 ≡ 2
and so that pattern repeats every 2 steps.
And the Mersenne numbers are just congruent to 1 less than each of these.
It should be tau day
1:32 I got so triggered when I thought it's always the previous factor *2 and then it went from 16 to 31
I think doubling is better than using proper factors, because for 1 to be a factor requires that the number itself is a factor, so it seems improper to include 1 if not including the number itself.
I'll just translate everything Matt says to make it fit the right way, since he did it wrong 😁.
Thanks sir
Brilliant! Just brilliant!
I said this in the other video, but wouldn't one be a perfect number? It's only factor is itself, so the sum of its factors would be itself. He also said that if there were an odd perfect number, it wouldn't have a Mersenne Prime as a factor, and one isn't prime. Maybe I am mistaken, feel free to correct me if this is the case.
Well... if we're excluding the number itself from the divisors (which we do for the whole rest), the sum of all proper (!) divisors of 1 is 0. Which isn't 1.
Oh okay. Thanks.
For the same reason 1 is not considered to be prime.
A prime is number that is only divisible by 1 AND itself. Since 1 is itself, it is not prime.
Moritz Durtschi Good answer.
What I get hung up on is this: the definition of factors are numbers that can be multiplied together to get the original number. How is one considered a proper factor if the number it would be multiplied by isn't considered a proper factor?
Wouldn't both lose their "proper factor status" if either of them did?
Or to put it another way: given that x times y = z, where x, y and z are all 1: how can x by proper if y is not, when it's just as legit to say that y is proper and x is not?
Doesn't that make 1 a lonely number? Oh... wait... uh, what?!? I mean, a factor can't be "lonely" as such because it needs to be paired up to be included in the set.
Seriously though, given a formula where x times y = z, it's hard to define one as a factor and not both without seeming to break the meaning of a factor, whether it be proper or not. Even so - with 1 times x = x, it would seem more appropriate to claim that 1 is NOT proper whereas x IS proper, for much the same reason as how primes are considered.
Yes - it's all about defining it using their properties, but it's counter intuitive and somewhat contradicting when looked at from the perspective of the definition of factors and the treatment of 1 regarding primes.
0:18 *proper divisors
not proper factors
-the proper factors of 6 is just 3, 2
6:23 Did he do the division in his head or did he have the number memorised?
Maker sound gave me chills lol
whats the name for a number where all its proper factors multiply to make the number eg 6=1x2x3
It weirds me out that 1 is included in it, and the fact that we exclude the number itself makes it even weirder to me. That's just feeling though. Out of curiosity, what if we take one out of the equation and simply do all the factors besides one and a number summed, are there numbers where this sum equals the original number itself. Since there appears to be no even perfect number, even though we can't prove it, is there an odd number of the sort I just described? Curious to know
Suppose there were such a number n. Since n is odd, n isn't divisible by 2 and so also isn't divisible by any multiple of 2; hence all divisors are odd. Since we aren't counting 1, suppose all divisors are in pairs m_1, k_1, m_2, k_2, ... such that each m_i < √n and each k_i > √n and m_i * k_i = n. Then since all these divisors must be odd, the sum of all these divisors must be even and so is a contradiction. Instead, n must then have a divisor equal to itself, and so n must be a perfect square. If √n is prime, n's prime factorization is (√n)^2, but √n + √n < n (for n > 4), so √n must be composite.
I stopped here because I couldn't think of anything useful beyond this that could narrow down or disprove. To summarize, if there were such a number, it would be an odd perfect square, and its square root is both odd and composite. I can't really tell whether such a number exists, but it would also be larger than the first four perfect numbers, after testing a few small candidates.
There is a bunch of "SET" games in the background. Will there be a video on this game?
We got 52nd Mersenne prime
you had me at one direction 🤯
Who are the guys at the picture shown at 4:50?
And WHY are they there?!
Let F(n+1) = 2^F(n) - 1. If we set F(0) to be some non-Mersenne prime p, is it possible for there to be some p for which F(n) is prime for all positive integer values of n?
I'm currently researching p = 2, but I'd like some outside input on whether such a series would be finite or not. I'm *hoping* the series is infinite, but I don't want that to skew my ability to form a proof.
Did this in school today. Interesting
The perfect outro doesn’t exi….
over 50 perfect number are found now..
If we use the equation (2^n -1)(2^n-1) I noticed that the number of divisors(including the number itself) of a perfect number is always equal 2n. Does someone know why is that?
Maybe that only works with the smaller ones but is it normal that if you go from highest factor to lowest that they always rounded up halfs of each other? like plays that a role in it. So ⌈Factor n⌉ = (Factor at n+1) / 2
A more formal statement for Mersenne Primes would be x⊂M( 2^x -1 ∧ P(x)) where _M_ is the set of Mersenne Primes and P(x) a function which is true if _x_ is a prime.
Since the numbers go off to fast for me to check, but so far can you say that if you start with 2 and work out the Mersenne prime of that you get 3 (Prime!) . Now take 3 and work out the next prime which is 7. Take 7 and get 127. Take 127 and you'll get another prime (170141183460469231731687303715884105727)
Now: Does that work indefinitely?
Unlikely we'll ever be able to check that last one, and almost certainly impossible to prove/disprove that the pattern will work forever. Extremely unlikely that it does, however.
I don't understand how the discovery of a new Mersenne prime can lead you to determine a new perfect number. Just because it gives you a factor of the potential perfect number, how does one go about using that factor to determine the corresponding perfect number?
Do you think the day will come when we will have one equation (E=ˠM˳C²)∞ representing one universal process that the laws of physics are based upon?
At 1:00, that doesn't work because of "1".
What I like about the argument against prime 1 here is its circularity. 1 is not prime because when you plug it into the formula you get 1... which is not prime. Why? Because when you....
What a number the sum of whose factors minus 1 and itself equal itself?
What I like from maths are the patherns we can find. For instance, If the sommes of the composant of n is equal to a multiple of 3. then n is a multiple of 3
Example: 342 = 3+4+2= 9 342/3 = 114
19874989494 = 1+9+8+7+4+9+8+9+4+9+4 = 72 = 7+2 = 9
19874989494/3 = 624996498
and so on :P
100000000010001 is also a multiple of 3 You can add as many number to it. as long the somme of it composant remain a multiple of 3
Hi i love the way how you make math so simple, i am student of math teacher, and i wonderig if there´s a chance to do a video of group theory plis ? (I am from Chile, so i am apologize if there is something bad wirting)
why dosnt it work for 11
2047 is divisible by 23 and 89
if the rule that n=any prime number when 2^n-1 is a prime number always applied, there would be no point in finding the biggest primes, because you can do 2^(the largest prime number)-1, which would be a prime if the rule always applied. Close...
what about 4
4 = 2+2
and those are all it factors
their is no known prime with this digit size ?
Like Mersenne primes could we have another set of primes, P=(3^p -2) where p is prime where p=2 => P=7 , p=5=>P=241 and so on...? Can we call it Arora's prime of its not taken!
I believe there has been some study of primes that are of the form N(p,n) = (pⁿ-1)/(p-1), for primes p. Idk whether these are named.
And here, too, n must be prime in order for N to be prime. (This is a necessary, but not sufficient condition.)
When p=2, these are the Mersenne primes. When n=2, N = p+1, so that N(p,2) is prime only for p=2.
Would someone be able to clear something up that I'm missing that I'm pretty sure is basic. Its the numbers in the 2n-1 column that are the mersenne primes right? not the n's that it works for. I'm just a little confused and I know only one of them are mersenne primes.
Any value in the 2^n +1 column that is prime is, by definition, a mersenne prime. However not every value for 2^n +1 is prime at all. Hope that helps
How to publish next Merseene Prime number if I found it? and by not using GIMPS and do I need to submit any thesis?
Did u do it by hand
These people are geniouses.
Hahahhahah your trick is not completely working 😂😂😂😂😂
257 is not 2^n-1, with n integer
6 is also 1*2*3
"You've been in this game for too long" hahah cute.
Are there an infinite number of mersenne primes? Does every perfect number have a factor that is a mersenne prime?
my birthday is june 28th
NAILED IT
Love the opening: "Today we're gonna do some math."
Wow, who would ever guess *that* in a million years? Gee, I came here expecting to get a key lime pie recipe...
(That *is* what that logo depicts, isn't it? The one with "π" in it? But wait, that's an awfully odd shade of green . . .)
I saw you on Discovery Channel :]
If we exclude the 1 from the factors we sum up, there has no perfect number been found yet, nor has it been proven that a such number doesn't exist. I wonder why the ancient Greeks didn't exclude the 1.
Can you do something on the number 23 enigma?
Why isnt 3 a proper factor of 28?
How many people know that the answer to 1 3/8 divided by 1 11/16 has the number 814 recurring?
That equals one... Is it a prime? -NO
Mathematicians can be so silly :D
as someone who hasnt studied math at all, but is a nerd, ive always wonder how an odd perfect number is not considered impossible, but prime factorization is considered fact as to me that would need to break, can someone educate me where im off in thinking watching these sorts of vids.