7:50 "That's the worst prime-generating formula ever!" I just discovered a new prime-generating formula: f(n)=2n. So far, I've been able to show that f(1)=2 is prime. I don't have the computing power to prove that f(2)=4 is a prime, but I suspect it is.
After days of computing, I calculated that none of F(2) up to F(32) are prime, and I conjuncture that they are all divisible by 2, but I do not have the capability to prove it.
+Eric Hernandez after many more days of computing I was able to see that F(-1) up to F(-32) all negative and I think none of them are prime but I cannot prove it
"You know what, Professor? I think it would've been easier for him to take the compass and ruler out!" I was trying to think of how to express my dumbfounded reaction to that beast of an equation. That quote doesn't quite reflect it, but it was so good that I decided it would serve well enough. XD
Great video! Interesting to see that this constructing n-gon business came as a result of solving equations with complex numbers and that it had such an application. I spotted a mistake - at 10:10 he says if any of the ith primes are *not* equal to a Fermat prime, then it is not possible to construct the n-gon (which I think is correct) but then he writes down an equal to sign instead of a not equal to sign. I think it's worth pointing out since it might lead to confusion.
Check the construction of a pentagon or try to do it yourself. The ratio between half a side and the radius of the pentagon is the golden ratio (sqrt(5)+1)/2. Sqrt 5 can be constructed as the diagonal of a triangle with one 90-degree angle and sides 1 and 2. For a 17-gon you do the same thing but for a longer time.
I remember something about 2 sqrts of 2, or something. But it was a while ago, and I don't remember it that well. It just would have been nice to have it explained in the video, is all.
That Fermat guy... such a fuzzy mathematician. First this "failed prime-generator" that he belived to work, and then his "fantasic theorem" that he never got time to wright before he died (Fermat's last theorem). What a guy.
Actually you can eliminate one step (and hence introduce one less error) in the construction: no need to bisect the angle 4π/17. Simply mark off by that angle 17 times, getting back to the starting point having gone around the circle twice.
Man, the way that different works of math connect together - it's just like, WOW. Amazing! Feels like there's a whole world of connections to be made that will never finish and you can just keep working on and on and on, and once again, WOW!
I read up on the underlying theory, since a lot of it (n-th roots of unity and finite fields) is related to my communications work. I remember in high school working out exact values for the trig functions of 2 pi / 5. What we did not do was sin(2 pi / 5) has square roots -> sin(2 pi / 5) is constructible -> you can draw a pentagon with a ruler and compass.
It might be worth pointing out that if p = 2^n + 1 is prime, then either n=0 or n is itself a power of two (thus p is 2 or a Fermat prime). That makes it clear why the Fermat primes are fundamental, and how they are conceptually similar to the Mersenne primes.
+StAtheist Yeah, I noticed the same thing, and went to email them at the mistakes email address, but the end of it is chopped off. I hope they fix this.
2^2^k = 2^(2^k) in the standard language, so this gets rid of the ambiguity. This was decided because (2^2)^k = 2^(2k) by the laws of exponents, so there's not as much useful applications in using (2^2)^k as opposed to 2^(2k). 2^(2^k) is more generally useful and cannot be simplified, I believe, so that was decided to be the standard meaning of 2^2^k
THIS IS NOT A JOKE: 8:00 I have found a way to make a N-gon just using ruler and compass. N can be ANY NUMBER! :) I want to know where I should publish this. (if it is new) Plz respond.
They mean using only a compass and straight edge, not a ruler and a regular n-gon such as a regular nonagon or regular hexagon ect. If you want to publish a mathematical discovery you should use a mathematical journal I would assume, but I'm sure you can ask at any university and they can tell you. If that is too difficult try emailing Matt Parker (does some Numberfile videos and has his own maths related youtube channel), he will likely know. I assure you though it is extraordinarily unlikely that you did since it was proven that it can only be done for the ones they discussed in the video (and composite numbers with said primes).
The Z^n bit can be modelled by De'Moivre's theorem, where on the complex plane, each root of the equation z^n -1=0 are of equal angle measure (2pi/n) away from eachother.
DKsilverghost There are four numbers, the cardinal of {0, 1, 2, 3} is four. If you count from zero to three, you will need three seconds, but you will enunciate four numbers. The first three natural numbers are 0, 1 and 2. There is no 3 in the set of the lowest three natural numbers. You can’t get this wrong...
Not to rain on your parade, but all odd numbers in binary end with 1, and all numbers in binary start with 1. However, the number of zeroes is very intresting to think about.
Intriguing stuff! Is there a mistake @10:14. The Prof states if any Pi is NOT equal to, however the notation states Pi equal to. I know nothing about Fermat Primes, but this stood out to me :)
Kourii , aer bank is doing (2^2)^3 +1, which is 4^3 +1= 64+1= 65, but they obviously meant 2^(2^3) +1 = 2^8 +1= 256+1= 257. I understand his confusion since I was also taught that if there were no parentheses, then it should be treated as the former and not the latter, so 2^2^3 +1= 65 is true, but obviously in the video they meant for the later interpretation. I always felt that it should be the later interpretation myself, since if someone wanted to do ((a^b)^c)^d they could write it as a^(b*c*d) while if they want to do a^(b^(c^d)) they have no other means of writing it which seems like a ridiculous amount of parentheses (especially as you get more exponential terms). I'd much rather that a^b^c^d= a^(b^(c^d)) instead of ((a^b)^c)^d, but at last I'll just have to live with the standard that is already accepted.
@@olivialuv1 Wolfram Alpha agrees with you and I trust Wolfram Alpha to know standard notation. I don't know why I thought otherwise. I must have been mis-taught.
iabervon aside from Euler not having computers ( :p ), you can use "any" computer, even a Commodore 64, to factor any number. A 64 bit computer can do math with 64bit chunks at a time, but if you want you can calculate much bigger numbers by splitting them in smaller chunks so they can fit the computer architecture. For example, when doing maths with pencil and paper, you basically act as a 1 digit machine, and you just repeat your algorithm until you have calculated all the digits needed. The only limit is how much paper you have, and how much time you are willing to spend. The same goes with computers and very big numbers (you need lot of memory and time for very big numbers).
Antonio Barba Certainly, but the thing about Fermat numbers is that, for any amount of computer storage you're likely to have, there is one of them that just barely doesn't fit. If you've got 4Gbit of RAM, you can store F_32 - 2, but not F_32. It's tempting to think this is the issue if someone can't factor a Fermat number, even if it's before binary computers and clearly nuts.
if you are willing to spend centuries of spare time, you could also store temporary data to an Hard Disk, which addresses data in 64 or even 128 bits! I'm not saying it is practical, but "in theory" the architecture of the computer is not an issue.
James Cuttell Yes, but they both come from the general formula e^ix=(cos x)+i(sin x) anyway. The important point is that 2*pi turns up a lot more often than a lone pi, so it makes more sense to substitue tau for 2*pi.
James Cuttell e^(iτ/2)=-1 has the advantage that the expression has a half in an exponent, suggesting that it's a square root. Of course, e^iτ=1 doesn't tell us which nth root of unity e^(iτ/n) is, but it's at least telling you it is an nth root of unity. e^(iπ/6) being a 12th root of unity is less obvious written than way, and people don't really care about nth roots of -1 as such.
9:56 "any prime" They have to be distinct, of course. (The prime factorization of a given number is unique.) 10:10 p_i are not equal to Fermat primes! He said it, but he didn't write it.
Someone please explain me this: Superior dimensions have different volumes? Like square liter and so on? Or the notion of volume as we know it does not apply to this kind of objects?
The concept of area and volume are, roughly speaking, how much "stuff" an object contains (the mathematical term is "content"). In 1-dimension, this is known as length. In 2-dimensions, it's called area. In 3-dimensions, as you may have guessed, it's called volume. Hopefully, it should be clear that how much "stuff" an object contains is related to how many dimensions you're working with. Mathematicians have generalized this concept to allow for what's called the n-dimensional volume of an object to be calculated. Sadly, there's no direct geometric interpretation of n-dimensional volume, but it's still a very useful tool in applied mathematics.
It would be weird to measure length with ordinary meters and area with ordinary meters as well, because it'd be like saying they've got the same kind of measurable shape or exist in the same dimension (which is false because a length is obviously totally different from an area). Generally, it makes the most sense to describe 1D things with 1D units of measurement and 2D things with 2D units of measurement. You don't have to understand why to do physics or math, but that's an answer to your question of why: 1D (ordinary) meters for 1D measurements (like a length), 2D ("square") meters for 2D measurements (like an area), 3D ("cubic") meters for 3D measurements (like a volume), etc
At about the 10th - 11th minute mark, there is a misprint, not really a mistake. David wrote (paraphrasing), If pi = Fermat primes, then not possible (you cannot construct the n-gon). What he actually SAID was, if some pi ≠ Fermat prime, then not possible, or equivalently if all the pi ARE different Fermat primes then you CAN construct the n-gon.
Can’t line segments be split into n equal parts? Wouldn’t this allow any regular n-gon to be constructible iff n has some Fermat prime in its prime factorization?
John OBrien But you can divide a segment into n parts. So couldn’t you start by constructing a pentagon who is inscribed in a circle, then divide each side into 5 pieces, then projecting the points of division onto the circle by drawing a straight line from the centre of the circle through the division point, to the circle edge, then connect all those points? Or would this not work?
Years and most of a math and physics degree later, the answer is that dividing angles is not the same as dividing line segments into equal parts. While we could divide a side of a polygon into, say, 3 equal parts, what we actually want is for each part to cover the same distance along the circle, which is a subtly different problem.
For those curious, here is a 1-line python expression to generate all 31 odd N-gons >>> [ reduce(lambda a,b:a*b, map(lambda x: x[0]**int(x[1]), zip( [ 2**2**m+1 for m in range(4,-1,-1) ], list(bin(n)[2:].zfill(5))))) for n in range(1,32)]
I accidentally found out (using brute force) that if n is a fermat prime, that if you take (n mod p) for any prime p less than n (other than 2) that it's remainder will NOT be 1. Is there any way to formally prove this?
Is there something about Gauss' approach for constructing values from nested square roots require that a value be prime in addition to having the form 2^2^n +1? Or, to put it another way, is there anything about the method that would require that a number's prime factorization contain only Fermat primes and not factors of Fermat "primes"?
I'm really sorry but there is something I don't understand, at the very beginning. He says 360*/17=2pi/17. I hope I'm not being awkward but could someone please explain to me how 360=2pi?? As I am new to this toe of stuff. Thanks! 😃
Okay hold up. My math teacher explained Euler's formula with complicated, complex Taylor polynomial. But right there at the start this man just opened my eyes to a most intuitive way of understanding it. It is a unit circle on the complex plain! Man, who woulda thought?
Wow great. Did you see my Invention of trisecting 180 degrees, Fermat's Last Theorem and General case too, Cubic Root by hand, Collatz Sequence at RUclips ,
Gauss' proof is exactly the sort of thing the ancient geometers were _avoiding_ . And although it allows the construction of a geometric algorithm for making perfect 17-sided polygons, they would have considered the method suspect: possibly it simply happened to be merely _accurate_ , to, say, one part in 10,000. How would they _know_ it worked? They didn't have Gauss' modern epistemology.
From what I gathered, he used an iterative formula for finding the cosines of angles, perhaps studying specifically the cosines related to solutions of z^n - 1 = 0. He might've used the Taylor series for cosine (if it was discovered at the time--maybe he used something similar). The taylor series for cosine involves x^n, so then he could possibly do something with -1 there. Just an educated guess :)
Yeah it'd actually make it super easy to do a 17-gon, you'd just approximate 360/17 degrees and then do the compass thing 17 times, the whole thing would only take like a minute and a half
Every root is the diagonal of a right triangle. Plus is adding distance. Minus is taking distance away. Dividing a line into n equal lengths is an easy construction to do. Multiplying is extending the line you want to multiply, adjust the compasses to the length of the line, draw a circle around the end point of the line. Mark where the circle cuts the extension of the original line. That's 2*x. Put the compasses on the new end point and draw a new circle to get 3*x etc. That's all.
If any of the other fermat numbers contains a factor of a prime that is not a fermat prime could you then devide it by it's fermat prime factors and end up with a polygon that you usually shouldn't be able to construct?
in One of numberphile's videos we saw a proof that one could trisect an arbitrary angle with a pen and folding the paper. Would implementing trisected angled using that method change the n-gons that can be constructed?
I love proofs like these, its a shame that when you are taught these sort of things at school you are told to take what you are being taught as absolutes.
Why were the ancients so obsessed with unmarked straight edges? Even without using a specific unit measure like inches, can't you make an arbitrary length on a straight edge, then use that as a reference to create any regular polygon you want? What is the appeal of that restriction?
Yes here's a mistake - it's hard to draw circles with something that shows the direction of north. If you are talking old style then it's a "pair of compasses".
There are only 5 Fermat primes so there are only 2^5 - 1 = 31 possible combinations of N-gons ever possible if N is an odd positive integer. Since it is small, the author could have published the entire list. Here it is - [3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295]. Every other N has to be even number (some power of 2 multiplied to one of these 31 odd numbers).
Suggestion: You can keep your channel logo as √2 rather than 2(I am not saying that 2 isn't special but it is not as special as √2) because your first channel has a special number π However, I really like your videos. I have really gained a lot of knowledge about mathematical logic from your videos
@1:27 the i was supposed to be on the circle, because if the segments there are cos and sin, the radius is 1. guess you probably drew it without thinking too much, sooo... yeah. whoops
3:44 He just drew world's most perfect square root, freehand
I am absolutely blown away *-* amazing!!
Agreed :D
Me pregunto si también sabe dibujar perfectamente raíces cúbicas
7:50 "That's the worst prime-generating formula ever!"
I just discovered a new prime-generating formula: f(n)=2n. So far, I've been able to show that f(1)=2 is prime. I don't have the computing power to prove that f(2)=4 is a prime, but I suspect it is.
After days of computing, I calculated that none of F(2) up to F(32) are prime, and I conjuncture that they are all divisible by 2, but I do not have the capability to prove it.
+Eric Hernandez after many more days of computing I was able to see that F(-1) up to F(-32) all negative and I think none of them are prime but I cannot prove it
You guys are all the best.
I figured out a truly remarkable proof of your conjecture but this comment section is to small to contain it.
Anthony Khodanian giggles... in such a redundant fashion. 😀
"You know what, Professor? I think it would've been easier for him to take the compass and ruler out!"
I was trying to think of how to express my dumbfounded reaction to that beast of an equation. That quote doesn't quite reflect it, but it was so good that I decided it would serve well enough. XD
Great video! Interesting to see that this constructing n-gon business came as a result of solving equations with complex numbers and that it had such an application.
I spotted a mistake - at 10:10 he says if any of the ith primes are *not* equal to a Fermat prime, then it is not possible to construct the n-gon (which I think is correct) but then he writes down an equal to sign instead of a not equal to sign. I think it's worth pointing out since it might lead to confusion.
Fermat Primes. The Parker Squares of Prime Numbers.
:D
Spotted a mistake? Email us at mistake@...
Nice one ;)
Rohan Skatedude I got a kick out of that too~
i dont get it
3:51
(1:46) 2*pi/2 = tau. I mean pi!
Stuff like this is what keeps me coming back to Numberphile! The coolest bits of math presented in a very accessible way.
I'd love to see how to figure out how to make one of those _n_-gons from the root formula to actually drawing it.
yeah actually that's what i was expecting from this video
Check the construction of a pentagon or try to do it yourself. The ratio between half a side and the radius of the pentagon is the golden ratio (sqrt(5)+1)/2. Sqrt 5 can be constructed as the diagonal of a triangle with one 90-degree angle and sides 1 and 2.
For a 17-gon you do the same thing but for a longer time.
But _why_ can you get sqrt(5) from that triangle? That's what would be nice to have explained in a video, I think.
Thomas Giles pythagorean theorem, should be clear
I remember something about 2 sqrts of 2, or something. But it was a while ago, and I don't remember it that well. It just would have been nice to have it explained in the video, is all.
I barely know my times table, I'm confused af right now and yet this is so appealing to me.
That Fermat guy... such a fuzzy mathematician. First this "failed prime-generator" that he belived to work, and then his "fantasic theorem" that he never got time to wright before he died (Fermat's last theorem). What a guy.
Actually you can eliminate one step (and hence introduce one less error) in the construction: no need to bisect the angle 4π/17. Simply mark off by that angle 17 times, getting back to the starting point having gone around the circle twice.
Actually.... lol
@@yeet3673 "ackchyually"
"So what is F0?"
A Nintendo Sci-fi racing game starring Captain Falcon.
2^(2^0) + 1 = 2^1 + 1 = 3
FurryFlurry F0:3
Yes.
And F(-∞) = 2
Man, the way that different works of math connect together - it's just like, WOW. Amazing! Feels like there's a whole world of connections to be made that will never finish and you can just keep working on and on and on, and once again, WOW!
9:27 - All N below 300 where a regular N-gon can be constructed using ruler and compass, eh?
... I'd like to see a 2-gon.
Here you go:
_____________
the length may vary.
Vasil Ivanich
I only see one side! :P
Persona just flip your monitor :p
Antonio Barba
Not entirely sure that'd help, but thanks for the advice.
Great Scott! You are not thinking in 4 dimensions!
MISTAKE:
At 10:14 he said "not equal" but he wrote equal.
3:51 how ironic. The mistakes@ e-mail address is the mistake.
maybe that's the joke
i thought the joke was we don't care if we commit mistakes
Leo Wattenberg i believe the actual mistake was he wrote a sixteen instead of a seventeen
2^2^3 +1 = 65
I read up on the underlying theory, since a lot of it (n-th roots of unity and finite fields) is related to my communications work.
I remember in high school working out exact values for the trig functions of 2 pi / 5. What we did not do was sin(2 pi / 5) has square roots -> sin(2 pi / 5) is constructible -> you can draw a pentagon with a ruler and compass.
It might be worth pointing out that if p = 2^n + 1 is prime, then either n=0 or n is itself a power of two (thus p is 2 or a Fermat prime). That makes it clear why the Fermat primes are fundamental, and how they are conceptually similar to the Mersenne primes.
I laughed so hard at 3:51... if that was deliberate, well played : )
At 10:14 Eisenbud says if any "...Pi are *not* equal to..." and writes an equals sign, that should be corrected to an inequality sign.
+StAtheist Yeah, I noticed the same thing, and went to email them at the mistakes email address, but the end of it is chopped off. I hope they fix this.
+Daniel Walsh Pretty sure that bit was a joke.
6:22
Is 2^(2^k) equal to (2^2)^k ? Doesn't seem like it. I think it is ambiguous without parentheses. Am i right?
You're right, they aren't equal.
2^2^k = 2^(2^k) in the standard language, so this gets rid of the ambiguity. This was decided because (2^2)^k = 2^(2k) by the laws of exponents, so there's not as much useful applications in using (2^2)^k as opposed to 2^(2k). 2^(2^k) is more generally useful and cannot be simplified, I believe, so that was decided to be the standard meaning of 2^2^k
this is so much more straightforward that I thought it would be
THIS IS NOT A JOKE: 8:00
I have found a way to make a N-gon just using ruler and compass.
N can be ANY NUMBER! :)
I want to know where I should publish this. (if it is new)
Plz respond.
They mean using only a compass and straight edge, not a ruler and a regular n-gon such as a regular nonagon or regular hexagon ect. If you want to publish a mathematical discovery you should use a mathematical journal I would assume, but I'm sure you can ask at any university and they can tell you. If that is too difficult try emailing Matt Parker (does some Numberfile videos and has his own maths related youtube channel), he will likely know. I assure you though it is extraordinarily unlikely that you did since it was proven that it can only be done for the ones they discussed in the video (and composite numbers with said primes).
when I said ruler I mean straightedge. thanks for conforming what I thought he said was proven though.
Fantastic presentation. My very best regards from Cluj Napoca
What's that wierd star with a hole at 4:12 behind?
The Z^n bit can be modelled by De'Moivre's theorem, where on the complex plane, each root of the equation z^n -1=0 are of equal angle measure (2pi/n) away from eachother.
2:27 proof left as an exercise for the reader?
10:12 says "not equal" writes "equal"
10:28 sees four Fermat primes and asks "but only the first three?"
You’re messing with my brain :'( !
He means "not equal".
Johan Särnbratt I know that :-p .
If you start counting from 0 then there's 3. (0,1,2,3).
DKsilverghost There are four numbers, the cardinal of {0, 1, 2, 3} is four. If you count from zero to three, you will need three seconds, but you will enunciate four numbers. The first three natural numbers are 0, 1 and 2. There is no 3 in the set of the lowest three natural numbers. You can’t get this wrong...
DKsilverghost calling a tail a 'leg' doesn't make it a leg.
There's an error in approximately 10:15, prof David forgot the stroke in the equal sign...
Great video, btw
This is what the Fermat Primes look like in binary:
11
101
10001
100000001
10000000000000001
Well that's trivial.
I wasn't claiming it as a discovery. It's just a cool fact!
Not to rain on your parade, but all odd numbers in binary end with 1, and all numbers in binary start with 1. However, the number of zeroes is very intresting to think about.
It's strangely appealing how this follows directly from the definition
It's interesting to see that there are 2^(n-1)-1 zeros (assuming 3 is term n=1) inbetween two 1s.
At 10:15 he says (and obviously meant) "not equal to" but wrote "equal to" instead. Sorry, I couldn't read the email at 3:51 :)
Intriguing stuff!
Is there a mistake @10:14. The Prof states if any Pi is NOT equal to, however the notation states Pi equal to. I know nothing about Fermat Primes, but this stood out to me :)
2^2^3 +1 = 65 not 257. That's definitely a mistake
+aer bank is that some kind of joke I'm missing? 2^2^3 + 1 ≠ 65
Kourii , aer bank is doing (2^2)^3 +1, which is 4^3 +1= 64+1= 65, but they obviously meant 2^(2^3) +1 = 2^8 +1= 256+1= 257. I understand his confusion since I was also taught that if there were no parentheses, then it should be treated as the former and not the latter, so 2^2^3 +1= 65 is true, but obviously in the video they meant for the later interpretation. I always felt that it should be the later interpretation myself, since if someone wanted to do ((a^b)^c)^d they could write it as a^(b*c*d) while if they want to do a^(b^(c^d)) they have no other means of writing it which seems like a ridiculous amount of parentheses (especially as you get more exponential terms). I'd much rather that a^b^c^d= a^(b^(c^d)) instead of ((a^b)^c)^d, but at last I'll just have to live with the standard that is already accepted.
@@brcoutme 2^2^3 is always 2^8 in standard notation I believe
@@olivialuv1 Wolfram Alpha agrees with you and I trust Wolfram Alpha to know standard notation. I don't know why I thought otherwise. I must have been mis-taught.
The math for this (cyclomatic equations) is found in chapter 3 of "Galois Theory" by Harold M. Edwards (1984), a very nice book
A sure enough testament to the genius of Gauss , truly hats off
I want a T-shirt with that construction :D
"Spotted a mistake? Email us at mistakes@"...I see what you did there.
3:21..... checks out on ti-84
wait wait wait
10:04 - if ai>1, not possible... Why?
why can you do 3*17=51gon, but you can't do 3*3=9gon?
that may be an interesting thing to discuss
I spotted a mistake. Unfortunately I can't send it to the proper email address, because the mistake was that the address was cut off at the @-sign.
I really, really enjoy listening to this gentleman. Nice videos.
I watched your construction vid first and while cool, I lost you pretty quick. This explains it all. Thanks!
Was that mistakes email screwed up on purpose? If so that's brilliant.
Professor Eisenbud has the dexterity of an artist! O_o
Hang on, as a programmer, I know my exponents of 2. It was only when he wrote 65537 that the relation clicked though.
I briefly wondered if they were unable to prove that 4294967297 was not prime until they could get a 64-bit computer.
iabervon You think that Euler had access to a 64bit computer?
iabervon aside from Euler not having computers ( :p ), you can use "any" computer, even a Commodore 64, to factor any number. A 64 bit computer can do math with 64bit chunks at a time, but if you want you can calculate much bigger numbers by splitting them in smaller chunks so they can fit the computer architecture. For example, when doing maths with pencil and paper, you basically act as a 1 digit machine, and you just repeat your algorithm until you have calculated all the digits needed. The only limit is how much paper you have, and how much time you are willing to spend. The same goes with computers and very big numbers (you need lot of memory and time for very big numbers).
Antonio Barba Certainly, but the thing about Fermat numbers is that, for any amount of computer storage you're likely to have, there is one of them that just barely doesn't fit. If you've got 4Gbit of RAM, you can store F_32 - 2, but not F_32. It's tempting to think this is the issue if someone can't factor a Fermat number, even if it's before binary computers and clearly nuts.
if you are willing to spend centuries of spare time, you could also store temporary data to an Hard Disk, which addresses data in 64 or even 128 bits! I'm not saying it is practical, but "in theory" the architecture of the computer is not an issue.
The thing is, he made mistakes that lead to an exact solution of a 20-gon. That is so interesting.
The beauty of Gons with professor Ross
The 17+3√(17)-√(34-2√(17))-2√(34+2√(17)) can be rewritten as 17+3√(17)-√(170+38√(17))
Thanks, Numberphile! Now I can go blow the minds of the people in my math class!
I spy an errant tau at 1:49. Long live tau, down with pi!
And yet at 1:50 tau shows how it pales in comparison with pi.
Yes but e^i*pi already tells us this whereas e^i*tau doesn't tell us the solution to e^i*pi, in fact it would suggest is 1.
James Cuttell Yes, but they both come from the general formula e^ix=(cos x)+i(sin x) anyway. The important point is that 2*pi turns up a lot more often than a lone pi, so it makes more sense to substitue tau for 2*pi.
James Cuttell e^(iτ/2)=-1 has the advantage that the expression has a half in an exponent, suggesting that it's a square root. Of course, e^iτ=1 doesn't tell us which nth root of unity e^(iτ/n) is, but it's at least telling you it is an nth root of unity. e^(iπ/6) being a 12th root of unity is less obvious written than way, and people don't really care about nth roots of -1 as such.
Is it weird that i can't understand what he's saying but i'm enjoying it?
Nah its awesome, its like when a more humanities gf listens to math bf
I spotted a mistake but I don't know where to send it because the domain was covered. Unless that was the mistake.
9:56 "any prime" They have to be distinct, of course. (The prime factorization of a given number is unique.) 10:10 p_i are not equal to Fermat primes! He said it, but he didn't write it.
i feel like, if guys like gauss and euler had a graphing calculator.. man.. it would've been game over for mathematics
Someone please explain me this: Superior dimensions have different volumes? Like square liter and so on?
Or the notion of volume as we know it does not apply to this kind of objects?
I'm not an expert but my guess would be that they would have an area, A volume, and the 4D version of volume.
The concept of area and volume are, roughly speaking, how much "stuff" an object contains (the mathematical term is "content"). In 1-dimension, this is known as length. In 2-dimensions, it's called area. In 3-dimensions, as you may have guessed, it's called volume. Hopefully, it should be clear that how much "stuff" an object contains is related to how many dimensions you're working with. Mathematicians have generalized this concept to allow for what's called the n-dimensional volume of an object to be calculated. Sadly, there's no direct geometric interpretation of n-dimensional volume, but it's still a very useful tool in applied mathematics.
Thank you both!
It would be weird to measure length with ordinary meters and area with ordinary meters as well, because it'd be like saying they've got the same kind of measurable shape or exist in the same dimension (which is false because a length is obviously totally different from an area). Generally, it makes the most sense to describe 1D things with 1D units of measurement and 2D things with 2D units of measurement. You don't have to understand why to do physics or math, but that's an answer to your question of why: 1D (ordinary) meters for 1D measurements (like a length), 2D ("square") meters for 2D measurements (like an area), 3D ("cubic") meters for 3D measurements (like a volume), etc
At about the 10th - 11th minute mark, there is a misprint, not really a mistake. David wrote (paraphrasing), If pi = Fermat primes, then not possible (you cannot construct the n-gon). What he actually SAID was, if some pi ≠ Fermat prime, then not possible, or equivalently if all the pi ARE different Fermat primes then you CAN construct the n-gon.
Can’t line segments be split into n equal parts? Wouldn’t this allow any regular n-gon to be constructible iff n has some Fermat prime in its prime factorization?
Only the first power of a Fermat prime is allowed. E.g. a 25-gon is not constructible.
John OBrien But you can divide a segment into n parts. So couldn’t you start by constructing a pentagon who is inscribed in a circle, then divide each side into 5 pieces, then projecting the points of division onto the circle by drawing a straight line from the centre of the circle through the division point, to the circle edge, then connect all those points? Or would this not work?
@@grahamrich9956 How do you divide a line segment into, say 11 equal parts?
Years and most of a math and physics degree later, the answer is that dividing angles is not the same as dividing line segments into equal parts. While we could divide a side of a polygon into, say, 3 equal parts, what we actually want is for each part to cover the same distance along the circle, which is a subtly different problem.
Even gauss wasn't above leaving the proof as an exercise. I guess it was trivial to him
I love his voice; It is so relaxing
We need more Tau in this video !
It's a bit ironic that the text telling us where to email for mistakes is a mistake in itself!
Where can I find the prove that the Non-Prime Fermat Number 5 can not be constructed in the Circle? And can it be algebraicaly depicted? And why not?
Try searching up "constructability of regular polygons by Fermat numbers"
spotted the mistake that the full email to report mistakes to is cut out of the video frame and now i just don't know what to do
I've got lost so I'm gonna watch it again :)
This is the same guy who wrote The Geometry of Syzygies. Respect
3:55 yes, I did spot a mistak. The email address wasn't on the screen :p
That was the joke
3:59 @???
Pleasant irony at 3:50 :)
For those curious, here is a 1-line python expression to generate all 31 odd N-gons >>> [ reduce(lambda a,b:a*b, map(lambda x: x[0]**int(x[1]), zip( [ 2**2**m+1 for m in range(4,-1,-1) ], list(bin(n)[2:].zfill(5))))) for n in range(1,32)]
Why does it have to be a fermat prime, though? What does primeness have to do with being able to construct a regular polygon?
I accidentally found out (using brute force) that if n is a fermat prime, that if you take (n mod p) for any prime p less than n (other than 2) that it's remainder will NOT be 1. Is there any way to formally prove this?
Yes. If n is of the form a + 1 where a is a power of 2. So if n = 1 mod p then a + 1 = 1 mod p. So a = 0 mod p. Hence p is 2 since a is a power of 2.
Is there something about Gauss' approach for constructing values from nested square roots require that a value be prime in addition to having the form 2^2^n +1? Or, to put it another way, is there anything about the method that would require that a number's prime factorization contain only Fermat primes and not factors of Fermat "primes"?
I'm really sorry but there is something I don't understand, at the very beginning. He says 360*/17=2pi/17. I hope I'm not being awkward but could someone please explain to me how 360=2pi?? As I am new to this toe of stuff. Thanks! 😃
en.wikipedia.org/wiki/Radian
Okay hold up. My math teacher explained Euler's formula with complicated, complex Taylor polynomial. But right there at the start this man just opened my eyes to a most intuitive way of understanding it. It is a unit circle on the complex plain! Man, who woulda thought?
Wow great. Did you see my Invention of trisecting 180 degrees, Fermat's Last Theorem and General case too, Cubic Root by hand, Collatz Sequence at RUclips ,
Gauss' proof is exactly the sort of thing the ancient geometers were _avoiding_ . And although it allows the construction of a geometric algorithm for making perfect 17-sided polygons, they would have considered the method suspect: possibly it simply happened to be merely _accurate_ , to, say, one part in 10,000. How would they _know_ it worked? They didn't have Gauss' modern epistemology.
But how exactly did Gauss work out that formula?
From what I gathered, he used an iterative formula for finding the cosines of angles, perhaps studying specifically the cosines related to solutions of z^n - 1 = 0. He might've used the Taylor series for cosine (if it was discovered at the time--maybe he used something similar). The taylor series for cosine involves x^n, so then he could possibly do something with -1 there. Just an educated guess :)
Nice video edit at 0:50 brady.
i loved the little story about the gauss way on the original video. =) you should keep such small "lessons" in all of the numberphile videos.
Would these constructions be easier if a protractor was allowed? (Sorry if this is a stupid question, I'm a humanities scholar.)
Yeah it'd actually make it super easy to do a 17-gon, you'd just approximate 360/17 degrees and then do the compass thing 17 times, the whole thing would only take like a minute and a half
But how do you get from the formula to the construction steps??
Every root is the diagonal of a right triangle. Plus is adding distance. Minus is taking distance away. Dividing a line into n equal lengths is an easy construction to do. Multiplying is extending the line you want to multiply, adjust the compasses to the length of the line, draw a circle around the end point of the line. Mark where the circle cuts the extension of the original line. That's 2*x. Put the compasses on the new end point and draw a new circle to get 3*x etc. That's all.
I need to get a compass and try this out!!
So there are 31 known constructible n-gons with n being odd. The largest one is a 4,294,967,295-gon.
¿Y a qué equivale el seno (sin) de 360°/17?
If any of the other fermat numbers contains a factor of a prime that is not a fermat prime could you then devide it by it's fermat prime factors and end up with a polygon that you usually shouldn't be able to construct?
in One of numberphile's videos we saw a proof that one could trisect an arbitrary angle with a pen and folding the paper. Would implementing trisected angled using that method change the n-gons that can be constructed?
Brady, can you make a video or two revisiting the concept of complex numbers? Maybe show it in a context that we haven't seen before?
I love proofs like these, its a shame that when you are taught these sort of things at school you are told to take what you are being taught as absolutes.
i didnt know u could use square roots inside square roots
There is more Euler than any other mathematician in these videos.
Why were the ancients so obsessed with unmarked straight edges? Even without using a specific unit measure like inches, can't you make an arbitrary length on a straight edge, then use that as a reference to create any regular polygon you want? What is the appeal of that restriction?
All these constructions start out with a few "arbitrary" lengths. There just isn't any point in putting it on the ruler since there was no standard.
The markings would be biased and you'd have to base your entire graph off them
I want to know what that chart is on the chalkboard behind him
Yes here's a mistake - it's hard to draw circles with something that shows the direction of north. If you are talking old style then it's a "pair of compasses".
You know it's going to be heavy maths if it takes a 12 minute video to explain it.
Brady, this was better than the ruler and compass construction video, why was it relegated to the second channel?
There are only 5 Fermat primes so there are only 2^5 - 1 = 31 possible combinations of N-gons ever possible if N is an odd positive integer. Since it is small, the author could have published the entire list. Here it is - [3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295]. Every other N has to be even number (some power of 2 multiplied to one of these 31 odd numbers).
@ 10:10 David forgot to make the "=" into an "≠" although he says it.
F₆ = 18446744073709551617 = 274177*67280421310721
found with Big Integer calculator from link in my last comment.
How did he find the iterative square root stuff for 2pi over 17?
Suggestion: You can keep your channel logo as √2 rather than 2(I am not saying that 2 isn't special but it is not as special as √2) because your first channel has a special number π
However, I really like your videos.
I have really gained a lot of knowledge about mathematical logic from your videos
@1:27 the i was supposed to be on the circle, because if the segments there are cos and sin, the radius is 1.
guess you probably drew it without thinking too much, sooo... yeah. whoops