did somebody noticed that he is writing in a sheet of brown paper that is over a white board? ajajajajja i love this guys, they know how to keep the identity of their channel
+Gonzalo Skalari he is left handed. Being left handed his writing on the white board would tend to be rubbed out. Not so much on the paper. It is true. I am a lefty. It is irritating.
I actually first reached when counting down from ∞ and hadn't noticed its alleged significance. I was kinda tired though from being up literally counting forever. That's sounded funnier in my head than it looks on paper. Kind of like mathematical calculations and arithmetic operations.
Ramanujan and Gauss were absolute geniuses. Heegner wasn’t such a slouch either lol. But one of the most amazing parts of this story is that Gauss had the intuition to suspect the end of the list. How??
For those interested, the fact that e^(pi sqrt(163)) is so close to a whole number has to do with properties of the modular function J(tau) as well as the fact that Z[sqrt(-163)] is a unique factorization domain.
@@christopherstoney4154 I don't think you're right about this. The value of Ramanujan's constant is given by a very rapidly converging series, the first two terms of which happen to be integers.
I love that this video starts with explaining how to write a number as a product of a prime, and quickly escalates to the invention of new number systems using unreal numbers.
@@fredyfredo2724 are you aware of the polar form for any complex number a+bi? if so you must know it is r(cosθ+i sinθ). I fail to understand why complex numbers wouldn't work with sine, let alone other trigonometric functions
@grande1899 fair enough... When it comes to the more advanced stuff, it seems we're damned if do and damned if we don't... I hope you like the next one more and appreciate anyone who takes the time to comment constructively.
Wait MISTAKE ALERT.He says square root of -7 gives unique factorization but that's wrong..yiu can write 8 as either 2 times 2 times 2 or as (1-sqr root -7)(1 + sqr root -7) also gives 8! Same reason why sqr root-5 was discarded..sonwhy not discard 7 and 11 and several others for that mater..Didn't anyine else notice this is a mistake??
@@leif1075 It is well-known that -7 yields unique factorization, so my guess is that 2*2*2 and the other factorization you mentioned are what we call "associates". This means that one is a unit multiple of the other, where a "unit" is any element of Z[sqrt(-7)] that has a multiplicative inverse.
@@leif1075 wiki page explains about sqrt(-5): "These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, 1 + sqrt(− 5), 1- sqrt(-5), are associate". I wonder though what are the units for Z[sqrt(-7)]
I must say, as a mathematics major, these videos really keep up my joy for maths. I really enjoy seeing videos on number theory topics and what not. Fascinating, and encourages me to become the best mathematician I can be! Thank you!
@@ludo-ge9fb or by using complex numbers: a + bi Is a number whose distance from the origin is the square root of an integer, so if you square it, it's distance from the origin wil get square and thus, you'll get a complex number whose distance from the origin is an integer. (a + bi)² = (a² - b²) + (2ab)i So that number is at a whole number distance from the origin
What's impressive about this is that it was solved by an amateur mathematician who is as brilliant as all the professional mathematicians combined in number theories
You’d expect a mathematician to be the toughest to break into their suitcase/bank account/etc but it turns out they are the easiest because they use their favorite constant lol
Still my favourite number / numberphile video ! A great example of the delightful surprises that emerge from understanding the most generalised of principles underpinning number 'systems' / Rings / Fields / Groups etc.
Ramanujan was the most talented mathematician to grace the world. He didn't 'proof' what he already knew until they learned him how to. He knew things on his own that the collective mind of math's history took centuries to learn.
In some sense root(-5) isn't really correct. When you take root(a*b) then this is the same as root(a)*root(b). But for -1 root(-1)*root(-1) is equal to i^2=-1, but root(-1*-1) is equal to root(1)=1. Also root(1) does have two solutions, 1 and -1 and we define the root to always give back the positive result (so x^2 does have a bijective inverse function). For root(-1) there are two solutions as well, i and -i, but these are in some sense undistinguishable because there is no notion of comparison in the complex numbers. You can't say i is bigger than -i or vice versa. So it's better to write i*root(5) because that is completely unambiguous and you don't run into problems because it's difficult to define root(z) for a complex number z.
The fact that those three numbers are close to whole numbers means that we still haven’t fully understood the conjecture yet and are a sample in f a second set of numbers, because when working with primes in particular they tend to end up being exponential! Warned
@davidandkaze no I was with you, in fact I think you missed the subtlety of my jokey retort... that I have in fact do have a PIN number... a PINN if you will... a number to protect my number! But I think the moment has passed!
Watching people write left-handed always makes me a bit squeamish, because I naturally imagine myself doing the same, and since I'm right-handed it feels really wrong. XD
Another quibble. The number has nothing to do with Ramanujan. Hermite knew that exp(π√163) is very near an integer. Ramanujan's papers don't mention it.
Oh, stuff and nonsense. He is clearly American -- his American accent is evident every few words -- prahblem, sahlved, liddle, wanna ride it, idennafy, right triangle (instead of right-angled triangle), exhahstive, noo, prahgress etc. etc., and that's just the first couple of minutes, before we get to the Z.
Actually, the Babylonian's used a base 60 system (which is where our time system comes from) because on one hand they would point out only one finger and this would point towards one of their knuckles of the four fingers on the other hand. Each finger has three 'knuckles' if you take a look, hence there are 12 combinations on the one hand, multiplied by the 5 fingers and thumbs of the other hand, to get 60 combinations in total.
Ramanujan himself often didn't understand how exactly he was coming up with his results. And even when he did, often he did not keep the explanation/proof, just the result. He was likely the most talented mathematician ever, but lacked formal faculties and rigor. He started working on those gaps, but died too soon.
Ramanujan number: 1,729 Earth's equatorial radius: 6,378 km. Golden number: 1.61803... • (1,729 x 6,378 x (10^-3)) ^1.61803 x (10^-3) = 3,474.18 Moon's diameter: 3,474 km. Ramanujan number: 1,729 Speed of light: 299,792,458 m/s Earth's Equatorial Diameter: 12,756 km. Earth's Equatorial Radius: 6,378 km. • (1,729 x 299,792,458) / 12,756 / 6,378) = 6,371 Earth's average radius: 6,371 km. The Cubit The cubit = Pi - phi^2 = 0.5236 Lunar distance: 384,400 km. (0.5236 x (10^6) - 384,400) x 10 = 1,392,000 Sun´s diameter: 1,392,000 km. Higgs Boson: 125.35 (GeV) Phi: 1.61803... (125.35 x (10^-1) - 1.61803) x (10^3) = 10,916.97 Circumference of the Moon: 10,916 km. Golden number: 1.618 Golden Angle: 137.5 Earth's equatorial radius: 6,378 Universal Gravitation G = 6.67 x 10^-11 N.m^2/kg^2. (((1.618 ^137.5) / 6,378) / 6.67) x (10^-20) = 12,756.62 Earth’s equatorial diameter: 12,756 km. The Euler Number is approximately: 2.71828... Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2. Golden number: 1.618ɸ (2.71828 ^ 6.67) x 1.618 x 10 = 12,756.23 Earth’s equatorial diameter: 12,756 km. Planck’s constant: 6.63 × 10-34 m2 kg. Circumference of the Moon: 10,916. Gold equation: 1,618 ɸ (((6.63 ^ (10,916 x 10^-4 )) x 1.618 x (10^3)= 12,756.82 Earth’s equatorial diameter: 12,756 km. Planck's temperature: 1.41679 x 10^32 Kelvin. Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2. Speed of Sound: 340.29 m/s (1.41679 ^ 6.67) x 340.29 - 1 = 3,474.81 Moon's diameter:: 3,474 km. Cosmic microwave background radiation 2.725 kelvins ,160.4 GHz, Pi: 3.14 Earth's polar radius: 6,357 km. ((2,725 x 160.4) / 3.14 x (10^4) - (6,357 x 10^-3) = 1,392,000 The diameter of the Sun: 1,392,000 km. Orion: The Connection between Heaven and Earth eBook Kindle
Solved by an "amateur" mathematician? What does that even mean? What makes him an "amateur"? The fact that he didn't have a degree from Oxford? Who came up with that nonsense? You think because you went to university and blew $25 000, that suddenly your a "professional" mathematician"? Mathematics has no degree or level of education. It is a subject that is common to every thinker.
Finlander Ramanujan proved theorems that are applicable in quantum physics and are in use right now, after approximately 100 years of his proofs. Clearly more respect for the man was needed instead of tossing "amateur" out there. Makes it sound like he stumbled upon the theory rather than rigorously and tirelessly worked on it that confounded not only the mathematicians of that era but also the current ones.
This is a video on mathematics, which is a subject based on rigorously defining a system of axioms and proving things using those simple axioms. Do you have a way of rigorously defining the term "amateur" that isn't based on someone not doing an activity as their profesion(ie someone doing something when not being payed to do so)?
Holy wow, chill guys It's a technical term, can't help it that it's a term used for many years and it just so happened that ramanujan fit into this category He is a great mathematician, but he didn't have a degree in math so "technically" under math terminologies, he is a amateur mathematician, that's it Ffs guys in the world...
I have always hated math...since i was kid i never understood it....maybe cause my first teacher used to beat us up if we were wrong...who knows. But it is my biggest regret. I truly wish i could understand it. I love it and i found it fascinating. Great videos even if i got lost once he started talking bout factoring numbers 😅
A eight digit binary sequence with inverse power has a critical wave edge trigger. 101 000 11 next 1. 3×4 is the smallest leap of a right angle for surface symmetry.
10:41 is my favorite moment. I have to agree, I don't think most ordinary people would expect that e^d*pi where d forms a number system with unique factorization, would be very close to, but not quite, a while number.
I had not come across this before and for the briefest of moments I was extremely happy to think that Ramanujan's Constant was an integer. Alas, those thoughts were shattered.
There's mistakes in it sqr root of negative 7 does NOT give you unique factorization because 8 equals (1- sqr root- 7)(1 plus sqr root -7) as well as 2 times 2 times2. So it should be discarded like sqr root of -5....samecscenario.did no one else notice this mistake??
In a straight line y=mx+c, the gradient is m. In a curve the like y=x^2, the gradient has to be worked out differently (it changes as the curve gets steeper). To find the slope you 'differentiate' (you'll learn this later) to find the gradient. The number e is defined to be such that the curve y=e^x differentiates to e^x. Basically the the gradient at any point is equal to the y co-ordinate at any point. 2.718281828 =e (roughly, it's irrational).
What am I doing wrong here? I can see that if we can write numbers as a + b√-5, there aren't unique prime factorizations. In the video, 6 was written as 2 * 3 and (1 + √-5)(1 - √-5). However, it is stated that if √-3 is chosen, there will still be a unique factorization. I don't see how this is the case. Couldn't 4 be written as both 2 * 2 and (1 + √-3)(1 - √-3)?
The only way I could see that sqrt(-3) would be acceptable is that either (1+sqrt(-3)) or (1-sqrt(-3)) aren't "prime numbers". But I have no idea how to check whether they are...
6 = 2×3 = (1+√-1) (1+√-1)× (1+√-2) (1+√-2) Hence, (1+√-5) (1+√-5), can’t be a unique factor. My understanding is that, the Gauss conjecture finds the factors for the prime numbers. These factors are essentially complex and they are formed with a + b, where a can be any real number and b can be any one of √-1, √-2, √-3, √-7, √-11, √-19, √-43, √-67, √-163. Therefore, these numbers form the primer numbers and hence I would call them prime of prime numbers. Similarly, 4 = 2×2 = (1+√-1) (1+√-1)× (1+√-1) (1+√-1) Hence, (1+√-3) (1+√-3), can’t be a unique factor, instead, (2+√-3) (2+√-3) can be a unique factor and it is equal to 7. Correct me if I am wrong.
I messed up with + and - sign in the above reply. Here is the corrected equations. 6 = 2×3 = (1+√-1) (1-√-1)× (1+√-2) (1-√-2) Hence, (1+√-5) (1+√-5), can’t be a unique factor. My understanding is that, the Gauss conjecture finds the factors for the prime numbers. These factors are essentially complex and they are formed with a + b, where a can be any real number and b can be any one of √-1, √-2, √-3, √-7, √-11, √-19, √-43, √-67, √-163. Therefore, these numbers forms the primer numbers and hence I would call them as prime of prime numbers. Similarly, 4 = 2×2 = (1+√-1) (1-√-1)× (1+√-1) (1-√-1) Hence, (1+√-3) (1-√-3), can’t be a unique factor, instead, (2+√-3) (2-√-3) can be a unique factor and it is equal to 7.
They don't cover the more general definition of a prime in this video, but it's a little different to the case for the integers. A 'prime' is a number p such that if p divides a product xy, then p must divide either x or y. There's a similar definition: an 'irreducible' number z has that if z = ab, either a or b has a reciprocal that also exists in the domain of numbers you're working in (such as 1, -1, i, -i if they exist in the domain - these are called 'units'). They're not identical definitions, although in the integers Z they do turn out to be the same thing, which is the more commonly known definition of a prime. As you say, in Z[sqrt(-3)], 4 = 2x2 = (1+sqrt(-3))(1-sqrt(-3)), but actually 2 isn't prime in Z[sqrt(-3)]! If it were, since it divides (1+sqrt(-3))(1-sqrt(-3)), there would be some number z with 2z = (1+sqrt(-3)) or (1-sqrt(-3)), but e.g. z = 1/2 + 1/2(sqrt(-3)) isn't in Z[sqrt(-3)] because it's written with fraction coefficients. But 2 *is* irreducible. Unfortunately the video is misleading if you want to delve this deeply into the maths, since his example of 6 = 2x3 is also not a prime factorisation in Z[sqrt(-5)]. As you worked out, unique factorisation into *irreducibles* fails more often. It's possible to show that if you have unique irreducible factorisation then you automatically get unique prime factorisation, but not vice versa.
It's interesting...if you take the list of these 9 numbers and line them up in order and subtract the lowest from the second lowest, the 2nd lowest from the 3rd lowest, etc. like you would if you were trying to find the degree of a function, you end up at 164, which is the lowest number (1) added to the highest number (163). Just thought that was interesting.
I'll be honest with you guys... My spidey sense is going crazy. I think there's an intuition hidden in here somewhere which leads me to strongly believe this is one of the most important Numberphile videos of all.
By raising e^( sqrt(-43)pi ) or whatever number you choose from that list, you are walking halfway round a unit circle sqrt(43) times, because the original expression can be rewritten as e^( sqrt(43)*pi*i ) which will give you an point on the unit circle where the y value (sine) is close to 1. Because the x value (cosine) is very irrational, it may be linked to the thing with unique factorisation. When using the formula at the end of the video, e^( sqrt(43)pi ) (notice the number inside the root is now positive) we are essentially taking an i out of the expression and hence moving the number onto the real axis. because the y value was close to a whole number (defined by the sine of sqrt(-43)pi) it rotates to the x axis where the real component is now close to a whole number. This comment is not necessarily the right answer to your question, but it is a guess as to some of the maths involved in the actual proof.
It might be that, if 'e' and 'Pi' is taken to be more accurate, then if the x.9999... could close more in towards the integer. Then, considering limiting value (as we consider more digits after decimal for 'e' and 'Pi') it might be true, i.e. it really could be an integer.
If Ramanujan said that it is an integer then it is. End of story. We will never know how his mind was wired, certainly not like us the mortals. His infinite series to evaluate pi for example is still a wonder to this day.
By the fundamental theorem of arithmetic, in Z there is only one way of factorising any integer larger than 1 into primes up to rearrangement. This is unique factorization. By introducing a subset of C (complex numbers), that is Z[i], you can factorise a^2+b^2, which is irreducible in Z. Factored into a+bi, and a-bi, which can be proven to be squares themselves of the form d(m+ni)^2, for some m, n in Z. You can then solve the real and imaginary parts to find the right m and n to find a triple.
Interesting but we should not write the square root of a negative number. For example we should write sqrt(5) * i instead of sqrt(-5). The number i is not sqrt(-1) but i * i = -1
We do use higher base systems and we do frequently. Oftentimes, when confronted with a 32-bit number, it is easier to express it using 4 hex digits. Therefore [1] * 32 = ffffffff in hex, which is easier than writing 32 ones. In computers, hex numbers are used to represent operations, memory-addresses, bit-fields, etc. Hex is so popular because of how easy it is to go from base 2 to base 16 since both are powers of 2, so 1111 = f, 1010 = a etc. so we can represent alot w/ hex.
I noticed that most of the poeple know who was Ramanujan except many Indians, his own people and they say that there is no great scientist or mathematician here. If you yourself will not appreciate them then how can you expect from the world? Sad but true that there were many but they just died, struggling to print their research and nobody cared about them.
@IamGumbyy If you haven't realized it by now, brown paper witha marker is their trademark image so to speak. It has been in every video and I doubt they are going to use a whiteboard soon.
For people struggling to pronounce his name, here's a little hint. It should be: "Raam" (rhymes with "palm") + "aanooj" (say "AZUL" but replace the Z with n and L with J) + "Un" (as in Un-believable) = Raam aanooj an = Ramanujan Lesson two : Srinivasa.... that's a topic for another time :P
I understand these are factors, but these complex numbers, (at least the imaginary part) are not whole numbers, so I don't understand how you can call them primes. any thoughts?
It's taking the concept of complex numbers (adding root(-1)) and expanding it... you create separate number systems. The normal complex number system works (in generating unique factorizations for all numbers in the system), the one using root(-2) works, root(-3) works etc... the example root(-5) didn't work... up to root(-163), where you are at an end. *I* want to see the Mandelbrot-like set for the complex-like plane with root(-163)!
The more general definition of prime (also called irreducible) is that if a number p is factorized as p = a*b then either a or b is 1 or -1 (in this case). This is equivalent (also, in this case) with the definition of prime you are probably thinking of, only divisible by 1 or itself. - Suppose p is only divisible by 1 and itself, then p = 1*p is the only factorization, therefore p is also prime according to the more general definition. - Suppose p only allows trivial factorizations i.e. p = 1*p or p = -1*-p, then p is only divisible by 1 or itself, because if it was divisible by something else, there would be a non trivial factorization. Therefore the two definitions are equivalent. You can prove 1 + sqrt(-5) and 1-sqrt(-5) are prime in several ways. Hope this helped!
anglo2255 So 1+sqrt(-5) is divisible by 1, -1, itself and -1-sqrt(-5). In the case of regular primes we could limit ourselves to the positive numbers, but since there is no such thing as a positive complex number z (as long Im(z) =/=0), you have to include "minus"-itself and -1 as well
I think it needs some clearing. Z is a ring for it has two operations with a particular structure + and x (times), you should definitely read Wikipedia on what is asked to be a ring. You can do what is called extension of ring, that is a ring that contains Z and uses the same operations. That is the meaning of Z[i]: the smallest ring containing Z and i, using + and x. To define a prime in Z you need to talk about units. Units are the numbers of your ring that end up going to 1 after being multiplied by itself a finite number of time. If I take Z, 1 is already 1, -1 x - 1=1 that's another, and that's it. A prime is then a number p for which any writing p=a x b, implies that a or b is a unit. For Z, it just means that you can only write p = 1.p = - 1.-p, but for Z[i] it's another story since the units are 1,i,-1,-i. In Z[i], a prime can only be written 1.p = i.-ip =-1.-p = -i.ip. Now if we talk about Z[2i], you notice that the units are only 1 and -1, so the definition of prime is essentially the same as in Z except a and b are in Z[2i]. That means, primes before may not be primes anymore. (1+2i)(1-2i)=5, 5 isn't a prime anymore in Z[2i], and in Z[i] either actually. Now the big deal is to check if your ring allows you to do prime decomposition with unicity by the order (and disregarding units, p and -p are said to be the same factor...). What the video tells, and actually what the Stark-Heegner theorem states is that only for the numbers n=1,2,3,7,...,163, Z[ni] allows a unique factorisation. Hope it helps, you might want to check euclidian division, euclidian domain, principal integral domain, etc, on wikipedia it's already nice to start with.
did somebody noticed that he is writing in a sheet of brown paper that is over a white board? ajajajajja i love this guys, they know how to keep the identity of their channel
+Gonzalo Skalari It could be to avoid the glare off the white board.
+Gonzalo Skalari he is left handed. Being left handed his writing on the white board would tend to be rubbed out. Not so much on the paper. It is true. I am a lefty. It is irritating.
+Gonzalo Skalari They write it on the brown papers so that they can donate it to charities that then auction off the papers to people.
+Jonathan Park They didn't do that at the time
+jackcarr45 it is not a case when you write in arabic dude
I first encountered 163 when I moved on from 162.
You made me laugh. Truly. Thx!
You should get more thumbs up
I actually first reached when counting down from ∞ and hadn't noticed its alleged significance. I was kinda tired though from being up literally counting forever. That's sounded funnier in my head than it looks on paper. Kind of like mathematical calculations and arithmetic operations.
😂😂😂
That was very funny. Grounding. Thank you. Ha.
Ramanujan was probably the most original and great mathematician
Itsiwhatitsi
That's true
..well apart from or on par with Euler, Euclid, Fibonacci, gauss
yes eternia
Einstein and Newton and gallelio and Archimedes are the best
Arsh Upadhyaya umm, einstein was not a mathematician.
Ramanujan is great... but he's no Gauss ;)
nothing is more mysterious than the brown paper.
Piyush Kuril THIS COMMENT HAS 163 LIKES LOLLOL
And its artfoolish fringes
Why cover a board specifically designed to write on , cover it with a paper , in order to write on it .
I am digesting moths .
Maybe the camera couldn't see the white Board or something?
Sells the scribbled brown paper on eBay. Can’t do that if the professors write on their board.
"Who knows how he managed to determine this..."
He was Ramanujan, that's how
I was studied my higher secondary in Ramanujan studied school in kumbakonam 😇I really proud of him
He was Ramen Noodles
@@billoddy5637 hey, surprisingly he really sounds like that 😁😁😁😄😄😄
@@billoddy5637 😂😂😂
@@ranjithkumarr9788 really you are very lucky man
Ramanujan and Gauss were absolute geniuses. Heegner wasn’t such a slouch either lol. But one of the most amazing parts of this story is that Gauss had the intuition to suspect the end of the list. How??
Maybe he tried the rest of the primes up to a thousand!
For those interested, the fact that e^(pi sqrt(163)) is so close to a whole number has to do with properties of the modular function J(tau) as well as the fact that Z[sqrt(-163)] is a unique factorization domain.
Now that makes the whole essence of video crystal clear to me.................btw i dont know maths
I'm not sure how the calculation works, but my intuition tells me that the absolute value of (e^(pi sqrt(163)))+i is likely an integer.
@@christopherstoney4154 I don't think you're right about this. The value of Ramanujan's constant is given by a very rapidly converging series, the first two terms of which happen to be integers.
e to the sqrt -1 x pi even closer to a whole number.
I knew that.
I love that this video starts with explaining how to write a number as a product of a prime, and quickly escalates to the invention of new number systems using unreal numbers.
And demonstrate this new number system is false.
This will never work with sine.
@@fredyfredo2724 nothing in mathematics is "wrong" as long as it's logically consistent
@@dielegende9141 undefine is not demonstrate false or wrong and is not true
@@fredyfredo2724 I have no clue what you're trying to say
@@fredyfredo2724 are you aware of the polar form for any complex number a+bi? if so you must know it is r(cosθ+i sinθ). I fail to understand why complex numbers wouldn't work with sine, let alone other trigonometric functions
It's 99 years today (26 April 2019) since he died.
He didn't look that old in the video.
@@kenmolinaro he was 32 when he died.
@@deepaksinghpatwal5755 You need to learn the meaning of "sarcastic humor".
100 years today
100 years today, 26-04-2020
This is one of the most underrated videos on Numberphile. Absolutely fascinating!
Not to a 56 year old man with a 5 year old's math skills . No offence to 5 year old's !
@grande1899 fair enough...
When it comes to the more advanced stuff, it seems we're damned if do and damned if we don't...
I hope you like the next one more and appreciate anyone who takes the time to comment constructively.
Hi Numberphile, I love you
Wait MISTAKE ALERT.He says square root of -7 gives unique factorization but that's wrong..yiu can write 8 as either 2 times 2 times 2 or as (1-sqr root -7)(1 + sqr root -7) also gives 8! Same reason why sqr root-5 was discarded..sonwhy not discard 7 and 11 and several others for that mater..Didn't anyine else notice this is a mistake??
@@leif1075 It is well-known that -7 yields unique factorization, so my guess is that 2*2*2 and the other factorization you mentioned are what we call "associates". This means that one is a unit multiple of the other, where a "unit" is any element of Z[sqrt(-7)] that has a multiplicative inverse.
didnt know grandayy watched numberphile
@@leif1075 wiki page explains about sqrt(-5): "These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, 1 + sqrt(− 5), 1- sqrt(-5), are associate".
I wonder though what are the units for Z[sqrt(-7)]
Hey look, a white-board! We can use it to -
Numberphile: let's stick some brown paper on it!
They didn't use the white board because it would reflect light making it hard to see.
I must say, as a mathematics major, these videos really keep up my joy for maths. I really enjoy seeing videos on number theory topics and what not. Fascinating, and encourages me to become the best mathematician I can be! Thank you!
White boards have glare that shows up strongly on camera and makes writing hard to read. The brown paper is very easy to read on camera.
The camera man for this channel loves zooming in to faces as awkwardly as possible
Jason Palmer he is an amateur, non professional, he must even love the subject of that clip on amateur mathematics
The camera man for this channel is the dude that runs this channel
Obviously this cameramen never review his work; the worse cinematographers of the millennium !
Jason Palmer hahahahahaha heheheeeee!
It's obvious you bitches have never had to to film in a cramped space.
Fun fact:
(x^2-y^2)^2 + (2xy)^2 = (x^2+y^2)^2
For all x and y. This is bascially just a Pythagorean Triple machine
@@ludo-ge9fb or by using complex numbers:
a + bi Is a number whose distance from the origin is the square root of an integer, so if you square it, it's distance from the origin wil get square and thus, you'll get a complex number whose distance from the origin is an integer.
(a + bi)² = (a² - b²) + (2ab)i
So that number is at a whole number distance from the origin
What's impressive about this is that it was solved by an amateur mathematician who is as brilliant as all the professional mathematicians combined in number theories
I always love these videos where a seemingly ordinary number is shown to be far more interesting than the average person would expect
Haha look at that face in the end... it WAS his PIN heheh
+Entropy3ko Bosco!
Dat Seinfeld ref! hehe
You’d expect a mathematician to be the toughest to break into their suitcase/bank account/etc but it turns out they are the easiest because they use their favorite constant lol
+Sangeet Khatri
Small correction, 5i or 5 times iota is not the root of -5 it is the root of -(5^2) or - 25
I've been watching these videos from newest to oldest and this video is my favorite so far. Great vid!!!
thank you
Still my favourite number / numberphile video ! A great example of the delightful surprises that emerge from understanding the most generalised of principles underpinning number 'systems' / Rings / Fields / Groups etc.
@ParagonProtege Good to hear from people who enjoy being out of their comfort zone (welcome to my word making these videos!!!)
Ramanujan was the most talented mathematician to grace the world. He didn't 'proof' what he already knew until they learned him how to. He knew things on his own that the collective mind of math's history took centuries to learn.
Rooted negative numbers make me uncomfortable.
What should they do? Just use the word "i" behind the number?
you must be feeling complex!😉
it's just another notation for. (also all numbers are imaginary in some sense)
In some sense root(-5) isn't really correct.
When you take root(a*b) then this is the same as root(a)*root(b). But for -1 root(-1)*root(-1) is equal to i^2=-1, but root(-1*-1) is equal to root(1)=1.
Also root(1) does have two solutions, 1 and -1 and we define the root to always give back the positive result (so x^2 does have a bijective inverse function). For root(-1) there are two solutions as well, i and -i, but these are in some sense undistinguishable because there is no notion of comparison in the complex numbers. You can't say i is bigger than -i or vice versa.
So it's better to write i*root(5) because that is completely unambiguous and you don't run into problems because it's difficult to define root(z) for a complex number z.
Why? Do irrational numbers make you feel uncomfortable?
He looks like the mathematician out of 'Good Will Hunting', who takes Will under the wing.
absolutely brilliant. Thank you Alex.
That's MY number.
TheGuardian163
Prove it
That's NumberWang!
gauss a genius. ramanujan an another genius.
Shaantubes an another???? universe just imploded.
No shot.
@@vinaykumarsharma8565 😭😭😂🤣
Gauss just prove it was given by ramanujan
watching left handed writing is like watching a wizard at work 😓
The fact that those three numbers are close to whole numbers means that we still haven’t fully understood the conjecture yet and are a sample in f a second set of numbers, because when working with primes in particular they tend to end up being exponential! Warned
@davidandkaze no I was with you, in fact I think you missed the subtlety of my jokey retort... that I have in fact do have a PIN number... a PINN if you will... a number to protect my number!
But I think the moment has passed!
Watching people write left-handed always makes me a bit squeamish, because I naturally imagine myself doing the same, and since I'm right-handed it feels really wrong. XD
***** …Excuse me?
The opposite for me, I'm left handed and when I see people writing with their right hand I'm like: "magic!" XD LOL
We lefties feel that you're the weirdos xD
I feel the same way, NoriMori.
How sinister...
love these in-depth ones so much more than the "happy number" type ones. MORE!!!
I really enjoyed this video - this feels like the kind of stuff I always wanted them to cover in school!
a quibble: his name is" rama- nujan " not "ramunajan"
I was thinking that too
john landis yep..It should be pronounced just as it is written like Rama.nujan...no extra flavours..I'm an INDIAN.
Another quibble. The number has nothing to do with Ramanujan. Hermite knew that exp(π√163) is very near an integer. Ramanujan's papers don't mention it.
...and Zed instead of Zee. Clearly incorrect.
Oh, stuff and nonsense. He is clearly American -- his American accent is evident every few words -- prahblem, sahlved, liddle, wanna ride it, idennafy, right triangle (instead of right-angled triangle), exhahstive, noo, prahgress etc. etc., and that's just the first couple of minutes, before we get to the Z.
Now "163" is also my favourite number.
“Someone who wasn’t officially a mathematician” - lol, okay…
Brown paper reduces the light reflection and hence comfortable for the eyes
Ramanujan,
the mystery yet unsolved.
Actually, the Babylonian's used a base 60 system (which is where our time system comes from) because on one hand they would point out only one finger and this would point towards one of their knuckles of the four fingers on the other hand. Each finger has three 'knuckles' if you take a look, hence there are 12 combinations on the one hand, multiplied by the 5 fingers and thumbs of the other hand, to get 60 combinations in total.
Ramanujan was beyond any other mathematician....the sheer intuition and imagination was something alien.
How did Ramanujan calculate this? He was really great... Love for Ramanujan from Bangladesh 🇧🇩
Ramanujan himself often didn't understand how exactly he was coming up with his results. And even when he did, often he did not keep the explanation/proof, just the result. He was likely the most talented mathematician ever, but lacked formal faculties and rigor. He started working on those gaps, but died too soon.
@@flashpeter625even proof is not needed,since on higher plane everything look as formulae.Pls dont speculate,easier for you when you are not even him.
Ramanujan number: 1,729
Earth's equatorial radius: 6,378 km.
Golden number: 1.61803...
• (1,729 x 6,378 x (10^-3)) ^1.61803 x (10^-3) = 3,474.18
Moon's diameter: 3,474 km.
Ramanujan number: 1,729
Speed of light: 299,792,458 m/s
Earth's Equatorial Diameter: 12,756 km. Earth's Equatorial Radius: 6,378 km.
• (1,729 x 299,792,458) / 12,756 / 6,378) = 6,371
Earth's average radius: 6,371 km.
The Cubit
The cubit = Pi - phi^2 = 0.5236
Lunar distance: 384,400 km.
(0.5236 x (10^6) - 384,400) x 10 = 1,392,000
Sun´s diameter: 1,392,000 km.
Higgs Boson: 125.35 (GeV)
Phi: 1.61803...
(125.35 x (10^-1) - 1.61803) x (10^3) = 10,916.97
Circumference of the Moon: 10,916 km.
Golden number: 1.618
Golden Angle: 137.5
Earth's equatorial radius: 6,378
Universal Gravitation G = 6.67 x 10^-11 N.m^2/kg^2.
(((1.618 ^137.5) / 6,378) / 6.67) x (10^-20) = 12,756.62
Earth’s equatorial diameter: 12,756 km.
The Euler Number is approximately: 2.71828...
Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2. Golden number: 1.618ɸ
(2.71828 ^ 6.67) x 1.618 x 10 = 12,756.23
Earth’s equatorial diameter: 12,756 km.
Planck’s constant: 6.63 × 10-34 m2 kg.
Circumference of the Moon: 10,916.
Gold equation: 1,618 ɸ
(((6.63 ^ (10,916 x 10^-4 )) x 1.618 x (10^3)= 12,756.82
Earth’s equatorial diameter: 12,756 km.
Planck's temperature: 1.41679 x 10^32 Kelvin.
Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2.
Speed of Sound: 340.29 m/s
(1.41679 ^ 6.67) x 340.29 - 1 = 3,474.81
Moon's diameter:: 3,474 km.
Cosmic microwave background radiation
2.725 kelvins ,160.4 GHz,
Pi: 3.14
Earth's polar radius: 6,357 km.
((2,725 x 160.4) / 3.14 x (10^4) - (6,357 x 10^-3) = 1,392,000
The diameter of the Sun: 1,392,000 km.
Orion: The Connection between Heaven and Earth eBook Kindle
Great Ramanujan......
They mentioned several mathematicians, but you only noticed Ramanujian just because he happened to be Indian?
+Sanan Guliyev so what...search about him you will understand...and you have problem with indians?
so what problem you have with country tell me ofcourse i also here for math..
+Tanmoy Samanta whatever man try not to be racist/nationalist and appreciate scientists regardless of nationality/ethnicity
+Sanan Guliyev I'm not.....
The last question was hilarious. Whether the 163 is a combination of his safe locker as number 163 was his favorite number.
Solved by an "amateur" mathematician? What does that even mean? What makes him an "amateur"? The fact that he didn't have a degree from Oxford? Who came up with that nonsense? You think because you went to university and blew $25 000, that suddenly your a "professional" mathematician"? Mathematics has no degree or level of education. It is a subject that is common to every thinker.
Finlander Ramanujan proved theorems that are applicable in quantum physics and are in use right now, after approximately 100 years of his proofs. Clearly more respect for the man was needed instead of tossing "amateur" out there. Makes it sound like he stumbled upon the theory rather than rigorously and tirelessly worked on it that confounded not only the mathematicians of that era but also the current ones.
This is a video on mathematics, which is a subject based on rigorously defining a system of axioms and proving things using those simple axioms. Do you have a way of rigorously defining the term "amateur" that isn't based on someone not doing an activity as their profesion(ie someone doing something when not being payed to do so)?
Holy wow, chill guys
It's a technical term, can't help it that it's a term used for many years and it just so happened that ramanujan fit into this category
He is a great mathematician, but he didn't have a degree in math so "technically" under math terminologies, he is a amateur mathematician, that's it
Ffs guys in the world...
John Stuart Just lingo. he also called one guy a recreational mathematician.
Hats off John!
Calculating 3 irrational numbers in power without calculator.....
Me : left the math
At last a normal person on Numberphile.
mathematicians would like to encourage everyone to do maths
Another fantastic number. Great vid!
I have always hated math...since i was kid i never understood it....maybe cause my first teacher used to beat us up if we were wrong...who knows. But it is my biggest regret. I truly wish i could understand it. I love it and i found it fascinating.
Great videos even if i got lost once he started talking bout factoring numbers 😅
I have to say, this is the one numberphile video that i just don't get. But still, this channel keeps you thinking ;) Keep up the good work!
@Mrtheunnameable I refer you to my other reply... this joke has gone down like a lead balloon in pedant's corner!!!!
Ramanujan never dies. Ramanujan lives for infinity.
There are 2 classes of mathematicians..Ordinary mathematicians and Ramanujan
I love dabbling into the complex plane on these videos, keep it up!
"officially a mathematician"
They don't make 'em any more pretentious than that
Even Ramanujan called himself a clerk and not a mathematician. It is a job title, after all (cf. engineer).
A eight digit binary sequence with inverse power has a critical wave edge trigger. 101 000 11 next 1. 3×4 is the smallest leap of a right angle for surface symmetry.
So when are you "officially" a mathematician?
I suspect you'd need a degree? Just a guess.
Note that e^(sqrt(163)*tau) is also really close to a whole number.
Wait isn't Euler's theorem like the new Ram. constant? e^ipi = -1 (whole number)
bassionbean -1 is not a whole number
+bassionbean It is an integer (whole number).
I thought this too! Fascinating.
No, because ramanujan's constant only has real numbers, euler's formula has imaginary exponent
... they refer to different rings
10:41 is my favorite moment. I have to agree, I don't think most ordinary people would expect that e^d*pi where d forms a number system with unique factorization, would be very close to, but not quite, a while number.
He says "right triangles" but his triangles is actually left.
He is left handed
Илья Лагуткин lol tru
He is talking about Right angled triangle
Chutiya
... and he did mark the 90 degr corner.
nice guy this professor is. hes got a good heart. funny how you can tell that about someone.
I had not come across this before and for the briefest of moments I was extremely happy to think that Ramanujan's Constant was an integer. Alas, those thoughts were shattered.
I think the interlocutor guessed his ATM card code at the end.
I didn't understand this video...
Saeed Oshee 😮😐😕
For more info, check out the OEIS sequence: A003173.
Watch again
Not a problem , you are still fit to survive on this planet
There's mistakes in it sqr root of negative 7 does NOT give you unique factorization because 8 equals (1- sqr root- 7)(1 plus sqr root -7) as well as 2 times 2 times2. So it should be discarded like sqr root of -5....samecscenario.did no one else notice this mistake??
In a straight line y=mx+c, the gradient is m. In a curve the like y=x^2, the gradient has to be worked out differently (it changes as the curve gets steeper). To find the slope you 'differentiate' (you'll learn this later) to find the gradient. The number e is defined to be such that the curve y=e^x differentiates to e^x. Basically the the gradient at any point is equal to the y co-ordinate at any point. 2.718281828 =e (roughly, it's irrational).
What am I doing wrong here? I can see that if we can write numbers as a + b√-5, there aren't unique prime factorizations. In the video, 6 was written as 2 * 3 and (1 + √-5)(1 - √-5). However, it is stated that if √-3 is chosen, there will still be a unique factorization. I don't see how this is the case. Couldn't 4 be written as both 2 * 2 and (1 + √-3)(1 - √-3)?
The only way I could see that sqrt(-3) would be acceptable is that either (1+sqrt(-3)) or (1-sqrt(-3)) aren't "prime numbers". But I have no idea how to check whether they are...
I have the same doubt...
6 = 2×3 = (1+√-1) (1+√-1)× (1+√-2) (1+√-2)
Hence, (1+√-5) (1+√-5), can’t be a unique factor.
My understanding is that, the Gauss conjecture finds the factors for the prime numbers. These factors are essentially complex and they are formed with a + b, where a can be any real number and b can be any one of √-1, √-2, √-3, √-7, √-11, √-19, √-43, √-67, √-163. Therefore, these numbers form the primer numbers and hence I would call them prime of prime numbers.
Similarly,
4 = 2×2 = (1+√-1) (1+√-1)× (1+√-1) (1+√-1)
Hence, (1+√-3) (1+√-3), can’t be a unique factor, instead, (2+√-3) (2+√-3) can be a unique factor and it is equal to 7.
Correct me if I am wrong.
I messed up with + and - sign in the above reply. Here is the corrected equations.
6 = 2×3 = (1+√-1) (1-√-1)× (1+√-2) (1-√-2)
Hence, (1+√-5) (1+√-5), can’t be a unique factor.
My understanding is that, the Gauss conjecture finds the factors for the prime numbers. These factors are essentially complex and they are formed with a + b, where a can be any real number and b can be any one of √-1, √-2, √-3, √-7, √-11, √-19, √-43, √-67, √-163. Therefore, these numbers forms the primer numbers and hence I would call them as prime of prime numbers.
Similarly,
4 = 2×2 = (1+√-1) (1-√-1)× (1+√-1) (1-√-1)
Hence, (1+√-3) (1-√-3), can’t be a unique factor, instead, (2+√-3) (2-√-3) can be a unique factor and it is equal to 7.
They don't cover the more general definition of a prime in this video, but it's a little different to the case for the integers. A 'prime' is a number p such that if p divides a product xy, then p must divide either x or y. There's a similar definition: an 'irreducible' number z has that if z = ab, either a or b has a reciprocal that also exists in the domain of numbers you're working in (such as 1, -1, i, -i if they exist in the domain - these are called 'units'). They're not identical definitions, although in the integers Z they do turn out to be the same thing, which is the more commonly known definition of a prime. As you say, in Z[sqrt(-3)], 4 = 2x2 = (1+sqrt(-3))(1-sqrt(-3)), but actually 2 isn't prime in Z[sqrt(-3)]! If it were, since it divides (1+sqrt(-3))(1-sqrt(-3)), there would be some number z with 2z = (1+sqrt(-3)) or (1-sqrt(-3)), but e.g. z = 1/2 + 1/2(sqrt(-3)) isn't in Z[sqrt(-3)] because it's written with fraction coefficients. But 2 *is* irreducible. Unfortunately the video is misleading if you want to delve this deeply into the maths, since his example of 6 = 2x3 is also not a prime factorisation in Z[sqrt(-5)]. As you worked out, unique factorisation into *irreducibles* fails more often. It's possible to show that if you have unique irreducible factorisation then you automatically get unique prime factorisation, but not vice versa.
It's interesting...if you take the list of these 9 numbers and line them up in order and subtract the lowest from the second lowest, the 2nd lowest from the 3rd lowest, etc. like you would if you were trying to find the degree of a function, you end up at 164, which is the lowest number (1) added to the highest number (163). Just thought that was interesting.
So I guess 163 is special for something other than it's digits adding up to ten?
Truly Infamous ml
I'll be honest with you guys... My spidey sense is going crazy. I think there's an intuition hidden in here somewhere which leads me to strongly believe this is one of the most important Numberphile videos of all.
Now a Tamil Movie has come in his honour titled 'Ramanujan' -A Budget movie of course.
I understand the proof well enough I was just having fun, because I found some peculiar patterns in the series of numbers. thanks for the comment
I dont get why z[sqrt(-7)] works.
for example, 8 = 2*2*2 = (1+sqrt(-7))(1-sqrt(-7)). Am I missing something
because you don't know what a plus b whole square means
you're a duffer
Is (1+sqrt(-5)) a prime in Z[sqrt(-5)]? Because in the video he says it is.
Danny I Tan Lin
The "square magnitude" (norm?) of 1+sqrt(-5) in Z[sqrt(-6)] is 6, which is not prime.
Danny I Tan Lin just multiply
how do they even come up with these theories and determine the final effing number? this is quite freaking impressive
Is there any significance to those last, almost whole, numbers being similar form to eulers equation
By raising e^( sqrt(-43)pi ) or whatever number you choose from that list, you are walking halfway round a unit circle sqrt(43) times, because the original expression can be rewritten as e^( sqrt(43)*pi*i ) which will give you an point on the unit circle where the y value (sine) is close to 1. Because the x value (cosine) is very irrational, it may be linked to the thing with unique factorisation. When using the formula at the end of the video, e^( sqrt(43)pi ) (notice the number inside the root is now positive) we are essentially taking an i out of the expression and hence moving the number onto the real axis. because the y value was close to a whole number (defined by the sine of sqrt(-43)pi) it rotates to the x axis where the real component is now close to a whole number. This comment is not necessarily the right answer to your question, but it is a guess as to some of the maths involved in the actual proof.
I love this channel :) I'd sure love if you guys and gals could post daily videos though. Keep up the good work!
So.... which specific number is the Ramanujan constant ?
That,s exactly what I was thinking.
e^sqrt(163) pi? although he didn't predict it, I think they just call it after his two other numbers
1729
Binod.
This is probably the best one yet.
Ramanujan was a mathematical wizard♾️
one of the most cliffhanging numberphile videos ever
It might be that, if 'e' and 'Pi' is taken to be more accurate, then if the x.9999... could close more in towards the integer. Then, considering limiting value (as we consider more digits after decimal for 'e' and 'Pi') it might be true, i.e. it really could be an integer.
If Ramanujan said that it is an integer then it is. End of story. We will never know how his mind was wired, certainly not like us the mortals. His infinite series to evaluate pi for example is still a wonder to this day.
Nope, it can be shown that, with infinitely precise e and pi, it isn't a whole number
Writing looks so tough for left handers
I feel watching this upside down because he is left handed >_>
By the fundamental theorem of arithmetic, in Z there is only one way of factorising any integer larger than 1 into primes up to rearrangement. This is unique factorization.
By introducing a subset of C (complex numbers), that is Z[i], you can factorise a^2+b^2, which is irreducible in Z. Factored into a+bi, and a-bi, which can be proven to be squares themselves of the form d(m+ni)^2, for some m, n in Z. You can then solve the real and imaginary parts to find the right m and n to find a triple.
There are 163 days until christmas
Ramanujan was undoubtedly the greatest math genius
@@I_leave_mean_comments he had no fundamental training in mathematics yet he achieved great things
@@I_leave_mean_comments you got the internet. Read em
@@I_leave_mean_comments why have you deleted your comment fucktard
Interesting but we should not write the square root of a negative number. For example we should write sqrt(5) * i instead of sqrt(-5). The number i is not sqrt(-1) but i * i = -1
We do use higher base systems and we do frequently. Oftentimes, when confronted with a 32-bit number, it is easier to express it using 4 hex digits. Therefore [1] * 32 = ffffffff in hex, which is easier than writing 32 ones. In computers, hex numbers are used to represent operations, memory-addresses, bit-fields, etc. Hex is so popular because of how easy it is to go from base 2 to base 16 since both are powers of 2, so 1111 = f, 1010 = a etc. so we can represent alot w/ hex.
sqrt(163+6)= 13
13+4= 17.... pretty cool ?
no
Makes sense for you? Thats great!
Finally found it! The NP video that isn't sponsored by someone!
I noticed that most of the poeple know who was Ramanujan except many Indians, his own people and they say that there is no great scientist or mathematician here. If you yourself will not appreciate them then how can you expect from the world? Sad but true that there were many but they just died, struggling to print their research and nobody cared about them.
@IamGumbyy If you haven't realized it by now, brown paper witha marker is their trademark image so to speak. It has been in every video and I doubt they are going to use a whiteboard soon.
"Ruh-MOO-ni-John"? Really?
LancesArmorStriking Ramanujan,Indian name because..you know everybody does not live in west have a look there are people outside
Aditya Prakash Did be actually pronounce the name right? Sounds wrong in English because it's written Ramanujan and not Ramunajan. But
For people struggling to pronounce his name, here's a little hint.
It should be: "Raam" (rhymes with "palm") + "aanooj" (say "AZUL" but replace the Z with n and L with J) + "Un" (as in Un-believable) = Raam aanooj an = Ramanujan
Lesson two : Srinivasa.... that's a topic for another time :P
Rama rhymes with bama in Obama, nu rhymes with two, jan rhymes with pun.. Ramanujan
Shiven Mittal yeah that's the typical english pronunciation of his name
the reason that it's so comfortable is probably because you use it hundred times a day
Amazing that my iPhone calculator cannot calculate e^(SQRT(163)*pi)
My android can (;
Mine gives me a really really big number 6725525588.089824502242480889791268597377
Probably goes beyond that XD
Oh and also android and not iphone.
Then you entered something wrong.
e*(sqrt(163)pi)= 262,537,412,640,768,743 . 999 999 999 999 25 (On my Windows calculator)
*****
Indeed. Seems like i didnt put something in correctly. Your result is the correct one.
Get's 2.62537412641E+17 on my Iphone
If ramanunjan would have lived longer, he would have solved math.
So Al Gore has left Global Warming and moved into Math...
The connection between those numbers being close to whole numbers and the class number being 1 is as eerie as I have ever heard.
I understand these are factors, but these complex numbers, (at least the imaginary part) are not whole numbers, so I don't understand how you can call them primes. any thoughts?
It's taking the concept of complex numbers (adding root(-1)) and expanding it... you create separate number systems. The normal complex number system works (in generating unique factorizations for all numbers in the system), the one using root(-2) works, root(-3) works etc... the example root(-5) didn't work... up to root(-163), where you are at an end. *I* want to see the Mandelbrot-like set for the complex-like plane with root(-163)!
The more general definition of prime (also called irreducible) is that if a number p is factorized as p = a*b then either a or b is 1 or -1 (in this case). This is equivalent (also, in this case) with the definition of prime you are probably thinking of, only divisible by 1 or itself.
- Suppose p is only divisible by 1 and itself, then p = 1*p is the only factorization, therefore p is also prime according to the more general definition.
- Suppose p only allows trivial factorizations i.e. p = 1*p or p = -1*-p, then p is only divisible by 1 or itself, because if it was divisible by something else, there would be a non trivial factorization.
Therefore the two definitions are equivalent.
You can prove 1 + sqrt(-5) and 1-sqrt(-5) are prime in several ways.
Hope this helped!
so, instead of 1 and itself (or P), (1+sqrt(-5) and itself (or P)?
anglo2255 So 1+sqrt(-5) is divisible by 1, -1, itself and -1-sqrt(-5). In the case of regular primes we could limit ourselves to the positive numbers, but since there is no such thing as a positive complex number z (as long Im(z) =/=0), you have to include "minus"-itself and -1 as well
I think it needs some clearing. Z is a ring for it has two operations with a particular structure + and x (times), you should definitely read Wikipedia on what is asked to be a ring. You can do what is called extension of ring, that is a ring that contains Z and uses the same operations. That is the meaning of Z[i]: the smallest ring containing Z and i, using + and x.
To define a prime in Z you need to talk about units. Units are the numbers of your ring that end up going to 1 after being multiplied by itself a finite number of time. If I take Z, 1 is already 1, -1 x - 1=1 that's another, and that's it. A prime is then a number p for which any writing p=a x b, implies that a or b is a unit. For Z, it just means that you can only write p = 1.p = - 1.-p, but for Z[i] it's another story since the units are 1,i,-1,-i. In Z[i], a prime can only be written 1.p = i.-ip =-1.-p = -i.ip.
Now if we talk about Z[2i], you notice that the units are only 1 and -1, so the definition of prime is essentially the same as in Z except a and b are in Z[2i]. That means, primes before may not be primes anymore. (1+2i)(1-2i)=5, 5 isn't a prime anymore in Z[2i], and in Z[i] either actually.
Now the big deal is to check if your ring allows you to do prime decomposition with unicity by the order (and disregarding units, p and -p are said to be the same factor...). What the video tells, and actually what the Stark-Heegner theorem states is that only for the numbers n=1,2,3,7,...,163, Z[ni] allows a unique factorisation. Hope it helps, you might want to check euclidian division, euclidian domain, principal integral domain, etc, on wikipedia it's already nice to start with.