Perfect Number Proof - Numberphile
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- This video follows on from: • Perfect Numbers and Me...
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Objectivity: / objectivityvideos
Mersenne Primes and Perfect Numbers, featuring Matt Parker.
Matt is the author of Things to Make and Do in the Fourth Dimension. On Amazon US: bit.ly/Matt_4D_US Amazon UK: bit.ly/Matt_4D_UK Signed: bit.ly/Matt_Signed
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Matt is great. I love his sense of humor. He's one of very few people who can take the subject of this video and make in entertaining to non-math nerds.
Is putting the lid on a pen the maths equivalent of dropping a mic, then?
Lol, first at 154 likes
@@rogervanbommel1086 I'm going for 155. : P
or throwing the last piece of chalk to the rubbish trash can
that’s called re-capping
It IS! (Caps pen)
Matt becomes the child he talks about at 3:43, at 13:37
A nice leet reference :)
It's 2am. I've become addicted to watching Numberphile before bed. I'm watching towards the beginning where we are looking at the pattern of the 2, 4, 16, 64... And I think to myself, those are powers of 2. Then I see they are the prime -1. I figure Matt will say "this is obviously just 2 to the power of the prime minus one." When he says he tortures kids with it and it's not obvious at all I feel so happy that I finally understood a non obvious Numberphile concept. I finally feel like I belong. Loved this video!
+Ann Beckman
He tortures KIDS with it, not adults, whom I believe will see the pattern pretty much immediately :P
+Reydriel I'm 12 years old and I saw the pattern immediately... I'm also taking Geometry so I'm familiar with formal proofs already too.
+EpikCloiss37 12 year olds were doing geometry in the late 1800s, too. ☺️
Then what happened to our education system? Now you have to be in super special programs for that... (which are based on IQ of all things... Not a true measure in my opinion...)
I NOTICED THAT TOO WOW! :D
I USED A CALCULATOR TO DETERMINE THAT 8191 is multiplied by 4096 to get 33,550,336!
It blows my mind how similar of a feeling this video gives me to watching my calc 2 professor do proofs for certain series tests...
You should do more of these proof videos, this was really great!
"I use this to torment young people" :)
3:27 Did he just call high school students "young"?
nice
@@quinn7894 secondary (high) schoolers start at around age 10/11 in Australia and UK (where he's from and where he lives respectively), which is young :)
Oh that was beautiful; math truly is the music of logic!
Usually I hear people say the opposite, music is the math of art.
I agree!
Very well said... cool^^
+tggt00 Music is a massles body with a mathematical heart :)
Jasko Z Math is the science of the art of the music of logic.
Matt looks so happy at the end of this video :D
I just pronounce superscripts more quickly when they're together, like parentheses
Yay! For once in my life I did the whole thing myself before watching the video. The only difference with my method was to prove the sum of that particular geometric series by induction, because I already knew what the answer was by inspection, so it seemed like the best proof to use, especially given that I didn't even notice it was a geometric series...
I would have loved to see a proof of the other way around, that is every even perfect number has a Mersenne prime factor.
Matt has got to be the best teacher ever.
If you've ever worked with binary you know that the sum of all the powers of 2 up to n - 1 equals 2^n - 1
I havent even worked with binary I just learned that concept from a Khan Academy video showing how to count to 31 with your fingers xD. I feel like a special snowflake xD
TheRedstoneTaco
Or the binary number with only the nth digit =1 is exactly 2^n, 10000000.... -1 = n-1 ones, which is 2^ (n-1)
technically if you use both hands you could count up to over 1000 lol
and if you can do it with your thumbs they have 2 segments each ( some may say three including the connection to the wrist) and you get up past 1 million then.
Yes! I thought exactly that, the sum of powers up to n-1 is 1111111... with n-1 digits, and if you add 1, it becomes 1000... with a 1 and n-1 zeros, which is 2^n
I saw you on tv! Outrageous acts of science!
Haha that's awesome.
Can't wait for objectivity! The onscreen links didn't work though, but the description wasn't far away :)
Rising inflection,,, good work
Is it just me, or does anyone else get a real self-satisfied kick out of people who insist it's not possible to solve infinite sums in the manner described starting at 11:00?
The sum in the video is not even infinite.
Yeah, stuff can get a bit vague when you get to infinite sums. But this one's finite, so there's no real ambiguity to the result. The dots are not necessary, you could as well write the entire sum out and that way it's obvious all of the middle cancels out.
This method only works for infinite sequences whose sum converges. (Unless you're a physicist who doesn't care about rigour).
Wrong, infinity is a concept, not a number.
Well, you shouldn't because they're the ones that are right. That is a perfectly valid method for solving a finite sum, however it is COMPLETELY invalid for an infinite sum other than the small subset that completely converge. Using that method you can get literally any value answer you want. Look up the Riemann series theorem. It is well known that if you manipulate an infinite sum in this way you can arise at any solution you want. For instance 1+2+3+4+... can be shown using this method to equal -50, 2, 17, 99992, 1/6 and absolutely any other value (or also equally be shown not to equal anything).
I saw the pattern, I've never felt so accomplished
The perfect numbers are the triangular numbers of the Mersenne primes, or the factors that you multiply by are half the prime plus 1
You proved each Mersenne prime makes a perfect number of that form. You should prove the converse too: every even perfect number has that specific form.
Is that proven, or have we just not disproven it?
@@Leyrann Euler proved it
An odd number can't be written in that form, and we don't know if there are any odd perfect numbers, therefore this isnt proven
@@coc235 they specified "even perfect number" so yes, it was proven
Isn't the pattern more clear in binary? Aren't we obscuring the pattern by thinking in decimal?
But to prove that about binary, you must still use geometric series, so in the end you get the same result either way.
Yes of course. This is far more easily understood in binary, so some of the algebra could be skipped, but the proof would still be necessary
6 -> 110
28 -> 11100
496 -> 111110000
8128 -> 1111111000000
The amount of 1s is n, the amount of 0s is n-1
A new year, a new Matt Parker video. What a great start to 2015! (Although I'm sure Matt would argue that a year is a meaningless or at least arbitrary measure of time)
There's a mistake at 10:00.
It should be: (1+2+...+2^(n-1)) + (2^n -1) + .........
You wrote: (1+2+...+2^(n-1))*(2^n -1) + (2^n -1) + .........
9+10=21
Agreed, same mistake at 9:19
I think that was originally supposed to be a reminder, that the sum in that line adds up to (2^n)-1 ... but using commentary with round brackets in equations is not a smart thing to do.
***** or 9+4=30
yes, plz correct, i try to follow along but mistakes like these can literally throw the video out of wack
Matt has a book. It's called " Things to Make and Do in the Fourth Dimension Parker Square". Check it out
Fabulous proof. Thank you for a great video.
Another way to see that the geometric sum of 2:s at the end is equal 2^n-1 is to see the sum as a strip of n-1 1:s in binary which is the same 2^n-1
..."negative one plus two to the n"...ambiguity gone
technically that could still mean (-1+2)^n, but i don't think any one normal would actually think that
+stickmandaninacan oh, yah. That hadn't occurred to me.
minus 1 plus the nth power of two is the only case that there's no ambiguity at all, i guess.
+stickmandaninacan No. (-1+2)^n is 1^n, which is 1. The order that you put the base and exponent matter with this operation.
Best IMHO is, "two to the n power minus one" vs "two to the n minus one power."
Completely unambiguous.
"to the" and "power" act like left and right parentheses there.
I have been a fan of both Perfect Numbers and Mersenne Primes since high school (~50+ yrs ago!!), but I have never seen this proof! In the immortal words of Mr. Spock..."Fascinating!"
Interestingly enough, one way to tackle the 1 + 2^1 + 2^2 + … 2^(n-2) + 2^(n-1) summation is to write it in binary. What happens when you do that is you get a binary number that’s a series of 1s that’s n-1 digits long, so if you’re familiar with how binary numbers work it becomes immediately obvious what the sum is.
What's à perfekt Numbers?
Lol Swedish spelling correction when typing English :)
What's a perfect number - the perfect question to answer at the start of this video!
I demand a Parker prime!
I immediately saw that pattern as 2^(n-1) because binary, 2^n (because of programming, binary is something I use on the daily), and because it related to the equation (2^n)-1, also related to binary.
It's funny when you think about it, math and programming are so similar yet so different, or at least in my mind they are.
I noticed you published these videos in the reverse order in which they were filmed. You did the same for the geometric constructions/origami videos. Is there a motivation for this?
I find myself discovering I watched them out of order each time.
4:42 x(2x-1) x=2^n
I'm in high school and I got the pattern before you said it. I do feel smug :)
Me too. I'm also in high school.
Pattern at the start seemed obvious to me, but a interesting video none the less
Watching people do math is like watching people dance - I can't do either, but it's fun to watch someone who does it well.
I found the 2,4,16,64 incredibly quickly. I’m not a genius, I just had already read the top comment
Totally above my intelligence! Looking forward to next video
it should be noted, that ALL even perfect numbers are of this form. This means, that even perfect numbers are basically the same thing as Mersenne primes.
"You can choose ANY mersenne prime of your choosing" @ 5:36 WRONG.
he couldn't have stopped at 8 and substituted 7 (instead of 31 for 32) , because 7 isn't a factor of 496.
9/1,3
1+3=4
15/1,3,5
1+3+5=9
P(prime)/P+1
27/1,3,9
1+3+9=13
If there is an odd perfect number, it must have an even amount of odd factors(Excluding itself).
The only number that might be considered to work (That is odd, of course) would be 1.
4:42 I say it as 2 to the power of n minus 1 timesed by 2 to the power of 1 less than the value of n.
that hit me right in the maths
I love these videos!
that's the most depressing video I have ever watched.
about eight years ago, I've discovered that pattern and formula via trail-and-error, while researching prime numbers for fun.
I really thought I had something, until now, when it's obvious anyone with proper math knowledge already known that :/
What I would give to have Matt Parker as my maths teacher...
That's a special case. Since one of the factors is power of 2, the perfect number shall be even.
Wait how do we know that all even perfect numbers can be constructed this way??
I think this was meant to be one way; that all Mersenne primes correspond to a unique perfect number.
could you have done a proof by induction to prove the statement ?
assume P(k)
then P(k) = > P(k+1) or is that what you already did sort of?
I got a pattern like this 1,2,2,4,8,16... haven't tested it to see if it works beyond that it seems to add up tho, 16^2 is 256 so idk yea the pattern goes multiply each successive term by the old one or whatever probably wrong but who cares
myself I like those proofs for the theorem better that do the calculation and logic using binary number representation
Fun proof. Similar to a lot of the proofs I did when studying polygonal numbers.
I miss doing math like this. I need to get back in school. Lousy medical drops, and time off.
love you, matt and brady
Apologies if this has been mentioned before. I haven't read all of the comments.
If this proof is THE equation to find all perfect numbers (as opposed to Mersenne where only some Mersenne numbers are primes), then hasn't he already proved that there cannot be any odd perfect numbers?
All factors of 2^n (including 2^(n-1), where n > 1) are even numbers. Even numbers are defined where 2 is a factor of a number.
Therefore, even number multipled by Mersenne Prime equals even number.
This is irrelevant if there are other ways to work out a perfect number.
As I take it, we haven't proven there isn't some other way to make a perfect number. Just that all the even ones follow this pattern.
It's only that it's proven that all even perfect numbers must have this form. There are various proven restrictions on odd perfect numbers if they happen to exist, but this is not one of them.
Vulcapyro Got any links that discuss them?
ZipplyZane
Wikipedia's a good enough start. en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers
Would this sequence not be called the telescopic series instead of the geometric series? The one Mister Parker mentioned at 12:17.
The geometric series would be SUM (k=0, n)x^k and not SUM(k=0, n)2^n-1. I may have missed something, though I am pretty sure that was a mistake, though it may have been in the haste of the moment.
The series is of the form x^k, where x=2. Each term is twice the last one, so it is geometric
1st few souls to see this one!!
XD don't know if it indicates how responsive my cellphones notifications are.. or how interesting these videos are that it makes me watch them even when i have a test the following day XD
what was that joke, about Australians that are good at math ? :D
So you proved that if 2^n - 1 is prime, then (2^n - 1)(2^(n-1)) is prime. What about the reverse direction you mentioned? How do we know (or do we know?) that every even perfect number is of this form for some n?
In the blue writing starting with "496=", it should be "2(496)=". The way it's written is an inequality.
Can you prove this problem via Induction of the series?
I came up with the right pattern right after looking at it. I wish I had done it in math class so I could be smug about it.
another formula to prove a perfect number, is
2 x 4^n - 2^n
2 x 4^1 - 2^1 = 6
2 x 4^2 - 2^2 = 28
2 x 4^4 - 2^ 4 = 496
2 x 4^6 - 2^6 = 8128
2 x4^12 - 2^12 = 33,550,336
2 x 4^16 - 2^16 = 5,589,869,056
.....
weird that the first part has less views than the second
Noticed this a year ago! Should have said something. But hey, when you're on about the 1 + 2^1 + 2^2 .... You can say the 1 as 2^0!
Matt,
When I go to university I want to be in your class! What university do you teach at? I got your signed book for Christmas with shapes of constant width 2d and 3d, utilities mug, and the heart keyring! (I can't remember what it was called) they were the best presents ever!
what is perfect number of 8191. When he asked that, I paused and figure way to calculate it (not his way), and than continue realizing it that he won't say at the end. My result was
33 554 432. Can any of SMART MATHEMATICIANS check it, I am probably wrong but I wish to know the answer. If it's right I will write here how I found it
8191 x (2^12) =33 550 336
shuld be 8191*(8191+1)/2 = 33 550 336, confirmed here en.wikipedia.org/wiki/Perfect_number
, looks like you accedently double counted 2^12 as a factor...
You were close. I am not smart mathematican but just did some research: en.wikipedia.org/wiki/List_of_perfect_numbers
hmm I was wrong by half of 8192 which I used. But now it doesn't make sense. Will try to do it with number 17. maybe I find something out of it
You were off by 4096 which is 2^12, so you can probably see where is your mistake
I miss a lecture,
¿what is a perfect number again?
thank you guys!!!!!!!
How did Euclid come up with this expression? I mean once you have the expression it is not difficult to prove, but how at first place come up with it?
Did you know that Five Hour Energy('s)tm taste is new? And improved? Another satisfied customer.
Isn't there an easier way to solve that sum with an unknown end ( 10:34 )? Isn't it always so that if you add up all the powers of 2 up until 2^(n-1), you end up with 2^n - 1? Try it, in my experience it always works out...
what do you mean? isn't that exactly the same thing?
sure, it takes him longer to get there, because he proves it, instead of saying "well, I tried it a bunch and it seems to work"
That was beautiful
Why does this not work for any n? Where in the proof does it specify that n needs to be a Mersenne prime?
n does not need to be a mersenne prime, but 2^(n)-1 has to be a mersenne prime, otherwise it must be factorized as well, thus giving you a bigger total sum of all the factors.
if (2^n -1) is not a prime number, then you can write (2^n-1) = a*b where a and b are integers and thus we have at least two more factors in the final sum of coeffents (a and b) and thus it is not a perfect number.
Ah, of course. Thanks guys.
Can you tell me why 2 is the only "even" prime number? thx
Well in (2^(n-1))(2^n-1), since n is a prime, and a prime-1=even number (with exception to 2-1), then 2^(n-1)=even number
(2^n)-1, will be odd because 2^n when n=prime will always be even, and even-1=odd.
Even*odd=even which means P.N=even number
P=perfect number, M=Mersenne prime
(2^((M(n)-1)))((2^M(n)-1)=P(n)
Someone please validate my proof.
And in fact, it has been proven (by Euler, I believe) that all even PN's are of this form. It is unknown whether there are any odd PN's.
If there are, they can't be of this form, for the reason you've shown.
Just to reassure Parker _______ connoisseurs, the numbers are perfect, not the proofs.
dude your name...
what about the cealing((2^n-1)/2) it holds
What about the proof in the other direction?
Beautiful.
I think it would have been more useful to start out with the definition of PN's as having the sum of all its factors equalling twice the original number. Then perhaps note that you like the definition with the proper factors more at the end. It's not a big deal but it would've saved you having to switch between definitions. Otherwise two great videos though!
My solution is very simple..
1.IMHO Every Perfect number has a PIVOT. (MEDIAN of Factors in Factorization) ) eg) For 496 the PIVOT is 31
2.This number is indivisible ie) 31
3. 31*16 [Factor before 31]==496.
4. The previous number is 248 as per definition of Perfect Number.
Catch me if you can 😂
No, Brady, *no new channels*. Bad Brady!
(Just kidding, it sounds like a wonderful idea.)
So there are no primes of form 2^n -1 where n is not prime?
Right, because if n>1 has any divisors other than 1 and n, say, p ≥ 2, then n = p·m, where 1 < p,m < n, and 2ⁿ-1 is a difference of powers, which is always factorable:
2ⁿ-1 = 2ᵐᵖ-1 = (2ᵐ)ᵖ-1ᵖ = (2ᵐ-1)(2ᵐᵖ⁻ᵐ + 2ᵐᵖ⁻²ᵐ +...+ 1)
and 2ᵐ-1 > 1, and the other factor is also > 1.
10:13
Doesn't 2^0= 1
Another channel?
Could you please do a video about Bayesian logic?
I worked it out. So stoked ^_^ of course I'm a 30 year old man, but thems the breaks
This only shows the sufficient side. The necessary side isn't much harder and demonstrates these are the only even perfect numbers.
I always say 2 to the power n then minus 1 .
Am I the only one who worked out the 2^n-1 pattern in 15 seconds?
I figured out the pattern, but I just thought it was the even powers of 2 (2^even number), not 2^n-1.
He also did 2^1 and 1 isn't an even number tho
2, 4, 16, 64
When I saw that, I thought it was basically fibonacci except with multiplication instead of addition.
Except that 2·4 ≠ 16.
How?
A shout into the void
Lets call (2^n)-1... F. I paused at 2:13 to try and figure it out and it seems to be ((F+1)/2)F equals a perfect number, if n is a Mersenne prime. Pretty sure I'm right, but not certain.
Well, I watched the part where he answers it and I seem to be right, but the way I did it is unnecessary... like usual.
To note, your n has to be a Mersenne prime, not just prime. Your statement is pretty simple to convert to the usual statement, though: If F = (2^n)-1, then (F+1)/2 = ((2^n)-1+1)/2 = 2^(n-1), so then ((F+1)/2)*F = 2^(n-1) * (2^n)-1.
Vulcapyro Oh, my bad. I meant to say Mersenne prime, I'll fix that. Also, I did try to convert it to the standard version. I'm actually glad I figured it out before watching the whole video, even if I didn't find the simplest answer.
are these numbers useful for anything?
what ist a perfect Number in general?
too much math could not handle