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.
i havent had much of this experience but i generally say "one less than 2 to the n" and "2 to one less than n" rather than having "minus one" in suitable place. others find it annoying for obvious reasons but i like it. :)
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!
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...)
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
Hey Brady, the videos in my subscription feed listed this one before (thus older) the previous one, which made it really hard to watch.. Just something to look out for. Really great videos nonetheless!
Matt Parker, I would have loved to have you as a math professor in school. I've always loved math, and even majored in it as college. It was teachers like you who made it even more interesting.
I love this channel! Matt has told me everything I needed to know about perfect numbers and mersenne primes in this video and his one that came right before it that I can teach it to my classmates that know nothing about it.
@@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 :)
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...
Lovely video, thanks. This link was known to the ancient Greeks... but the converse (that all perfect numbers are of this form) had to wait until Euler. I wish you could dedicate one more video to this other side of the proof.
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.
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)
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
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
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?
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.
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).
To avoid confusion it might help to be a bit more rigorous - and a bit more formal - with the syntax, differentiating the product of 2^n minus one from two to the power of the difference of n-1. It's a litte harder to follow, but if you understand it, it makes it clearer which is which.
An interesting property of even perfect numbers that follows this theorem (although the proof is not as exiting) is that all even perfect numbers end with the digits 6 or 28. Another interesting fact as that the proof in this video was proven in one way by Euclides and by Euler in the other, two of the greatest mathematicians of all time. Euler also did some work on odd perfect numbers.
@@KasabianFan44 I thought "one way by Euclid and by Euler in the other" meant that he was saying they proved the same thing two different ways, which isn't true. Now that you're pointing it out though, I can see how I was probably wrong.
Is it significantly more difficult to prove that every even perfect number fits that pattern? This only proved that everything fitting the pattern is an even perfect number.
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.
I knew that 6 was a perfect number from my childhood, but on a lonely day with nothing to do (and before the internet) I worked out that 28 was the next one and that 496 the third one when I noticed the pattern in the factors and stumbled onto Mersenne primes by accident as I tried to work out more perfect numbers. I was so excited! Alas, that I was not the first (by millennia) - but it was still fun to discover on my own! Awesome video and explanation of why it works out this way. Thanks!
That's always fun. I remember being bored one day and trying to write a proof for the the quadratic equasion, I think it was nearly a decade before I found out what proofs were. So satisfying.
Correct me please if I'm wrong, but he managed to prove that with any Mersenne prime you can get a perfect number multiplying it by 2^(n-1), but he hasn't shown at all that all perfect numbers will have one and only one Mersenne prime factor. This is what he was aiming at.
why equals that in 9:13 ? Let A = 2^(n-1) , B = 2^n -1 sum of the factors of AB = sum of the factors of A + sum of the factors of AB? any sum of the factors of B?
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.
1:30 It is clear to me that an odd perfect number (opn) cannot contain a mersenne prime *to an odd power* (Euler gives that an opn has the form N=q^α p_1^(2e_1) ... p_k^(2e_k) where α and q are 1 (mod 4). A mersenne prime is 3 (mod 4) so it cannot play the role of q here.) But I cannot find anything about *even powers* of mersenne primes, e.g. 3
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.
you dont need geometric series to solve that, just add 1 to the 1+2+4+..., you can see that the 1 you add merge the 1, equal 2, then 2 merge 2 equal 4 and so on until it is 2 to the n, and finally minus 1 which you added earlier.
The power series summation came very naturally to me, because I saw it as all ones written in base b. Most easily in base 10, 10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 10^0 = 111111 (in base ten which is very natural for most of us). But adding powers of 2 aren't any different now. 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = 111111 (in base 2). The challenge in how to get to all ones with a single equation is done by going one power higher, subtracting one, and dividing by the digit repeated from the result. 10^6-1 = 999999, so divide by (10-1) to get all ones. 2^6-1 = 111111 (base 2), already ones but you can divide by (2-1) for giggles 8^6-1 = 777777 (base 8), so divide by (8-1) to get all ones. 5^6-1 = 444444 (in base 5), so divide by (5-1) to get all ones and when you have all ones, you have a sum of a geometric series. Sum b^0 to b^n = (b^(n+1) - 1) / (b - 1) This may seem contrived or odd to many people, but until I saw it this way, it never really clicked. Perhaps it is a way to show it to somebody who doesn't understand it through other ways.
A little-known fact is the converse of the theorem proved here is also true: If an even number is perfect, it must be of the form described here (i.e, 2 ^ (n - 1) * ((2 ^ n) - 1) ). This was proved by either Euler or Fermat, I'm not sure which. The proof is also longer than this one.
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.
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.
@5:37 you say that you get to a number that is one more than a Mersenne prime then you switch to it and keep doubling... Why didn't you do that for 8? 7 is a Mersenne prime :-) So is 3 :-) Just yanking your chain. I did notice that the Mersenne prime is at the "mid point" of the list of factors.
Well, you can do it for 7, but then you get a different perfect number. Namely 28. But yeah, for a given perfect number you want the right mersenne prime to do the switch.
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!"
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"
GUYS! 31 doubles is 62, 62 doubled is 124 (he put 126), 124 doubled is 248 (he put 268. Now if we add 1+2+4+8+16+31+62+126+268=518. And if we do it the correct way: 1+2+4+8+16+31+62+124+248=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.
Hi Numberphile, thanks for this lovely video. At time 10:10 I look at the workings and can't understand why the first line of the calculation (on top) is multiplied by (2^n) - 1, this is not included on the second line up from the bottom, did I misunderstand somewhere?
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.
Tell me if I've got this wrong, but if the formula for a perfect number is (2^n - 1)(2^(n-1)) then doesn't that mean all perfect numbers must be even. I mean, 2^n-1 must be an odd number because 2^x is always even and if you take away one that will make it odd. But then because 2^x is always even that means 2^(n-1) must be even an even number. So if that's the case you are just multiplying an even number by an odd number - the result will always be even.
The sum of the first n natural numbers is (n×(n+1))/2. What this proof implies is that if we let n equal any Mersenne Prime, the sum of all natural numbers until n is a perfect number.
Unless I missed something, this doesn't just apply to mersenne primes, but mersennes in general. Also, I think he missed proving that it was a perfect number, and all those numbers he added together were factors. That's probably a video in-and-of itself.
This is one of those really neat links in mathematics. You said something about proving even perfect numbers. Does that mean that there is a proof that all even Perfect numbers have a Mersenne prime factor? You've only proved the converse here.
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 :/
congrats i saw it in less than 5 sec and i'm still an undergrad, anyone with a basic understanding of powers can see it stop gloating, in fact if a student can't see the pattern he should be worried
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.
I differentiate between groups of operations not with inflection, but with pauses:
"Two to the... n minus one," versus "two to the n... minus one."
It's not dramatic, or anything, just a quick stop / catch-breath type pause. Half a beat, so everyone knows what I'm doing.
i havent had much of this experience but i generally say "one less than 2 to the n" and "2 to one less than n" rather than having "minus one" in suitable place. others find it annoying for obvious reasons but i like it. :)
2 to the n ·*camera zooms in* minus one
me 2
to the n minus 1
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!
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 got it for Christmas, it's brilliant!
Same, it was one of the best Christmas presents I ever got!
You make me want to go back to university! Why can't all teachers be like Matt?
Hey Brady, the videos in my subscription feed listed this one before (thus older) the previous one, which made it really hard to watch.. Just something to look out for. Really great videos nonetheless!
It's a really good read!
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 Parker, I would have loved to have you as a math professor in school.
I've always loved math, and even majored in it as college. It was teachers like you who made it even more interesting.
Classic Mathematician:
"Let us assume that we know the total"
13:39 Literally the smuggest face ever :')
And admittedly: "...so pleased I'm going to put the caps back on both pens." ! ;-)
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 love this channel! Matt has told me everything I needed to know about perfect numbers and mersenne primes in this video and his one that came right before it that I can teach it to my classmates that know nothing about it.
"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 :)
Matt looks so happy at the end of this video :D
For some reason, I find Brady's incredulity at the beginning to be hilarious."You've already shown a link!"
Haha.. "that's why Australians are so good at math" 4:54
Matt is so funny
"Ethan, count to ten!"
"Yes, ma'am. One alligator, two alligator…"
(Yes, I know there are no alligators in the wild in Australia.)
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 think I would've gotten stuck at the geometric series step, but everything else was explained well and clicked for me. Cool!
I just pronounce superscripts more quickly when they're together, like parentheses
Lovely video, thanks. This link was known to the ancient Greeks... but the converse (that all perfect numbers are of this form) had to wait until Euler. I wish you could dedicate one more video to this other side of the proof.
Love it! I may need to watch it again! Any plans for a Numberphile book?
Omg this would be amazing!
I haven't done math in ages, but I'm proud to say not only did I follow along with the video, but I was a step or two ahead.
Holy frick I am blown away, this is one of the coolest things ever
I knew I should have gone into number theory...
I would have loved to see a proof of the other way around, that is every even perfect number has a Mersenne prime factor.
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
I saw the pattern, I've never felt so accomplished
Matt has got to be the best teacher ever.
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)
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
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 you on tv! Outrageous acts of science!
Haha that's awesome.
To avoid confusion it might help to be a bit more rigorous - and a bit more formal - with the syntax, differentiating the product of 2^n minus one from two to the power of the difference of n-1. It's a litte harder to follow, but if you understand it, it makes it clearer which is which.
The perfect numbers are the triangular numbers of the Mersenne primes, or the factors that you multiply by are half the prime plus 1
An interesting property of even perfect numbers that follows this theorem (although the proof is not as exiting) is that all even perfect numbers end with the digits 6 or 28.
Another interesting fact as that the proof in this video was proven in one way by Euclides and by Euler in the other, two of the greatest mathematicians of all time. Euler also did some work on odd perfect numbers.
Actually, Euclid proved this theorem and Euler proved its coverse (that all perfect numbers are of this form.)
@@leadnitrate2194
That’s… literally what he said…
@@KasabianFan44 I thought "one way by Euclid and by Euler in the other" meant that he was saying they proved the same thing two different ways, which isn't true.
Now that you're pointing it out though, I can see how I was probably wrong.
@@leadnitrate2194
Ahhhhh I see, my bad
Can't wait for objectivity! The onscreen links didn't work though, but the description wasn't far away :)
Happy 2016 Matt. I enjoy your videos.
Is it significantly more difficult to prove that every even perfect number fits that pattern? This only proved that everything fitting the pattern is an even perfect number.
At 9:16, I start seeing two sequences multiplied by the Mersenne prime -- instead of just one.
Rising inflection,,, good work
Last two vids were really good. Keep it up Brady.
"I'm so pleased I'm going to put the lids back onto both of the pens" Hahahaha! You're good on camera! Well proof'd :)
And I'm screaming 256 without thinking it through, I guess I subconsciously realized it was powers of two.
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
I don't change my tone when differentiating between 2^(n-1) and 2^n-1. I use pauses. There's 2 to the…n minus one vs 2 to the n…minus 1.
I knew that 6 was a perfect number from my childhood, but on a lonely day with nothing to do (and before the internet) I worked out that 28 was the next one and that 496 the third one when I noticed the pattern in the factors and stumbled onto Mersenne primes by accident as I tried to work out more perfect numbers. I was so excited! Alas, that I was not the first (by millennia) - but it was still fun to discover on my own!
Awesome video and explanation of why it works out this way. Thanks!
Mathematics at its best
That's always fun. I remember being bored one day and trying to write a proof for the the quadratic equasion, I think it was nearly a decade before I found out what proofs were. So satisfying.
Correct me please if I'm wrong, but he managed to prove that with any Mersenne prime you can get a perfect number multiplying it by 2^(n-1), but he hasn't shown at all that all perfect numbers will have one and only one Mersenne prime factor. This is what he was aiming at.
why equals that in 9:13 ?
Let A = 2^(n-1) , B = 2^n -1
sum of the factors of AB = sum of the factors of A + sum of the factors of AB?
any sum of the factors of B?
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.
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
1:30 It is clear to me that an odd perfect number (opn) cannot contain a mersenne prime *to an odd power*
(Euler gives that an opn has the form N=q^α p_1^(2e_1) ... p_k^(2e_k) where α and q are 1 (mod 4). A mersenne prime is 3 (mod 4) so it cannot play the role of q here.)
But I cannot find anything about *even powers* of mersenne primes, e.g. 3
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
So what about the proof that all (even) perfect numbers are of this form?
4:42 x(2x-1) x=2^n
"torment young people" LOL keep that up!
I've seen Matt Parker in countless numbers of these videos and I just realized he reminds me of The Doctor.
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
I like this proof! Helped me to understand what was shown on the previous video.
you dont need geometric series to solve that, just add 1 to the 1+2+4+..., you can see that the 1 you add merge the 1, equal 2, then 2 merge 2 equal 4 and so on until it is 2 to the n, and finally minus 1 which you added earlier.
Fabulous proof. Thank you for a great video.
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.
The power series summation came very naturally to me, because I saw it as all ones written in base b. Most easily in base 10, 10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 10^0 = 111111 (in base ten which is very natural for most of us). But adding powers of 2 aren't any different now. 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = 111111 (in base 2).
The challenge in how to get to all ones with a single equation is done by going one power higher, subtracting one, and dividing by the digit repeated from the result.
10^6-1 = 999999, so divide by (10-1) to get all ones.
2^6-1 = 111111 (base 2), already ones but you can divide by (2-1) for giggles
8^6-1 = 777777 (base 8), so divide by (8-1) to get all ones.
5^6-1 = 444444 (in base 5), so divide by (5-1) to get all ones
and when you have all ones, you have a sum of a geometric series.
Sum b^0 to b^n = (b^(n+1) - 1) / (b - 1)
This may seem contrived or odd to many people, but until I saw it this way, it never really clicked. Perhaps it is a way to show it to somebody who doesn't understand it through other ways.
I demand a Parker prime!
@9:09 a wrong factor (2^n-1) appears in the first line
Yeah
A little-known fact is the converse of the theorem proved here is also true: If an even number is perfect, it must be of the form described here (i.e, 2 ^ (n - 1) * ((2 ^ n) - 1) ). This was proved by either Euler or Fermat, I'm not sure which. The proof is also longer than this one.
Oh my goodness Brady is making a mausoleum channel.
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.
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.
@5:37 you say that you get to a number that is one more than a Mersenne prime then you switch to it and keep doubling... Why didn't you do that for 8? 7 is a Mersenne prime :-) So is 3 :-)
Just yanking your chain. I did notice that the Mersenne prime is at the "mid point" of the list of factors.
Well, you can do it for 7, but then you get a different perfect number. Namely 28. But yeah, for a given perfect number you want the right mersenne prime to do the switch.
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!"
Matt has a book. It's called " Things to Make and Do in the Fourth Dimension Parker Square". Check it out
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"
GUYS! 31 doubles is 62, 62 doubled is 124 (he put 126), 124 doubled is 248 (he put 268.
Now if we add 1+2+4+8+16+31+62+126+268=518.
And if we do it the correct way:
1+2+4+8+16+31+62+124+248=496.
3:35 "professional jerk".
I'd love to see that as your profession on official documents.
5:41 - isn't 8 also 1 above a Mersenne Prime so why didn't he make that 7 and continue doubling?
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.
I love these videos!
Hi Numberphile, thanks for this lovely video. At time 10:10 I look at the workings and can't understand why the first line of the calculation (on top) is multiplied by (2^n) - 1, this is not included on the second line up from the bottom, did I misunderstand somewhere?
..."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.
Pattern at the start seemed obvious to me, but a interesting video none the less
Tell me if I've got this wrong, but if the formula for a perfect number is (2^n - 1)(2^(n-1)) then doesn't that mean all perfect numbers must be even. I mean, 2^n-1 must be an odd number because 2^x is always even and if you take away one that will make it odd. But then because 2^x is always even that means 2^(n-1) must be even an even number. So if that's the case you are just multiplying an even number by an odd number - the result will always be even.
What I would give to have Matt Parker as my maths teacher...
The sum of the first n natural numbers is (n×(n+1))/2. What this proof implies is that if we let n equal any Mersenne Prime, the sum of all natural numbers until n is a perfect number.
I went through another day not having to use these calculations, again.
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.
the comment about pronouncing superscripts made me laugh out loud, mostly because it's actually true and I never even realized it
That's a special case. Since one of the factors is power of 2, the perfect number shall be even.
Just to reassure Parker _______ connoisseurs, the numbers are perfect, not the proofs.
dude your name...
love you, matt and brady
Unless I missed something, this doesn't just apply to mersenne primes, but mersennes in general. Also, I think he missed proving that it was a perfect number, and all those numbers he added together were factors. That's probably a video in-and-of itself.
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 wish to be at one of his classes🤓
Can you prove this problem via Induction of the series?
I am in severe awe of this man's mathematical prowess.
This is one of those really neat links in mathematics. You said something about proving even perfect numbers. Does that mean that there is a proof that all even Perfect numbers have a Mersenne prime factor? You've only proved the converse here.
"... all even Perfect numbers have a Mersenne prime factor?"
Yes, they do. That proof is more involved.
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 :/
As a CS grad, the first thing I saw was the pattern
im sorry, but what's a CS grad?
Matej božič Computer sciences graduate, i guess.
Slithereenn oh yeah.. probably, thanks!
Matej božič You're always welcome :)
congrats i saw it in less than 5 sec and i'm still an undergrad, anyone with a basic understanding of powers can see it stop gloating, in fact if a student can't see the pattern he should be worried
Beautiful ❤️. Congratulations!!!