Really overcomplicated the proof that if p=3 mod 4 then p is not a sum of two squares. It should only take 1 line: squares mod 4 are 0,1, so a sum of two squares is 0,1,2, i.e. never 3.
For the first lemma, there is an alternative proof with quadratic residues. Suppose p does not divide x and does not divide y, which means p does not divide -y^2. Then from x^2 ≡ -y^2 (mod p), -y^2 is a quadratic residue mod p (primes of the form 4k+3 are obviously odd). Hence the Legendre symbol (-y^2/p) = 1, but (-y^2/p) = (-1/p)*(y^2/p) = (-1/p) ≡ (-1)^[(p-1)/2] = -1, a contradiction. Thus either p|x or p|y, and in either cases it's easy to show that p divides the other as well
Why does the theorem proven at 27:13 keep working for even powers of primes p=3(mod 4)? He proved that the product up to (Xk^2 + Yk^2)^Rk is a sum of squares (C^2 + D^2) and then he distributed A^2 in the way (AC)^2 + (AD)^2. But what if that product was 1? In that case who tells us that A^2 can be written as a sum of squares? Additionally, I did not understand why (a-c)/2 and (b-d)/2 would be multiples of X. What if 2^k || a-c and b-d for the same k? In that case wouldn’t 2^k || X also? Please help❤
29:43 little mistake ... Should be (a+c) not (a+b).... I think 34:05 also should be 3 in the bracket not 9....... also...... 65 can be written as sum of two squares in 2 ways. 65 = (1sqd + 8sqd) or (4sqd + 7sqd). 👍😉
But if 65 can be written as a sum of two squares then there is a unique way to do this but you find a two different ways to write it as sum of two squares (1^2+8^2) and (7^2+4^2) i don't know anything 😢😮?
The really cool thing is that a number n can be written as the sum of two squares if and only if it doesn’t have some factor p^k where p is prime and 3 mod 4 and k is odd.
Spoiler alert. For those who want to check their working on the example and the warm-ups. *Example:* 65 = 5*13 = (1² + 2²) * (3² + 2²) Note that the video has this wrong! Now we use m = x² + y² and n = z² + w² to write mn = (xz - yw)² + (xw + yz)² - although the choice of x, y, z, w will sometimes lead to multiple solutions. So 65 = 5*13 = (1² + 2²) * (3² + 2²) = (1*3 - 2*2)² + (1*2 + 2*3)² = 1²+ 8² because (-1)² = (1)² and we're only interested in the solutions in positive integers. But 65 = (1² + 2²) * (3² + 2²) can also be (1*2 - 2*3)² + (1.3 + 2*2)² = 4² + 7² so we have two different representations of 65 as the sum of two squares. *WU1:* 99 = 3² * 11. Now 3² can be written as sum of two squares, but 11 is congruent to 3 (mod 4) so cannot, and therefore, neither can 99. *WU2:* 45 = 3² * 5. Both 3² and 5 (which is congruent to 1 (mod 4)) can be expressed as sum of two squares, so we're good to go. I'll use a short cut, rather than the xyzw formula above as it's easier to just distribute a squared term like this: 45 = 3² * 5 = 3² * (1² + 2²) = (3² * 1²) + (3² * 2²) = (3*1)² + (3*2)² = 3² + 6² That turns out to be the only solution because we cant "break apart" the 3² in the prime factorisation to create another factorisation that will work, since 3 ≡ 3 (mod 4) obviously isn't equal to the sum of two squares. *WU3:* 80 = 2^4 * 5. As 5 ≡ 1 (mod 4), we'll get a solution. 80 = 16 * 5 = 4² * (1² + 2²) = (4*1)² + (4*2)² = 4² + 8² - using the distribution trick from earlier. There are other factorisations of 80, but it turns out that each one leads to the same result: 80 = 8 * 10 = (2² + 2²) * (3² + 1²) = (2*3 - 2*1)² + (2*1 + 2*3)² = 4² + 8² 80 = 4 * 20 = 2² * (2² + 4²) = (2*2)² + (2*4)² = 4² + 8² 80 = 2 * 40 = (1² + 1²) * (2² + 6²) = (1*2 - 1*6)² + (1*2 + 1*6)² = 4² + 8² etc. *WU4:* 490 = 2 * 5 * 7² . The only prime ≡ 3 (mod 4) is 7 and that is to an even power, so we'll get a solution. 490 = 7² * 10 = 7² * (1² + 3²) = (7*1)² + (7*3)² = 7² +21². There are other possible factorisations but it turns out they all lead to the same result: 490 = 98 * 5 = (7² + 7²)(2² + 1²) = (7*2 - 7*1)² + (7*1 + 7*2)² = 7² + 21² 490 = 245 * 2. We'll need to find an expression of 245 as sum of two squares: 245 = 49*5 = 7² * (1² + 2²) = (7*1)² + (7*2)² = 7² + 14² 490 = 245 * 2 = (14² + 7²) * (1² + 1²) = (14*1 - 7*1)² + (14*1 + 7*1)² = 7² + 21²
Is there simple criteria to show how many ways a number can be written as 1) the sum of CONSECUTIVE squares and 2) the sum of consecutive cubes. There is for the sum of consecutive integers. The number of ways that an integer M can be written as the sum of at least two consecutive numbers is the number of divisors of 2M less 1.
Here is a university admission problem from Japan. Prove that for all natural numbers n, x^2+y^2=2009^n always has a solution. Should be fairly easy by applying today's results. 😃
Really interesting! But I'm still waiting for the representation of a given number as a sum of 3 squares? Or may be I missed it? Thank you for the video!
I’ve looked for simple proofs, but I think they require an understanding of 3-variable quadratic forms and some proofs regarding them. Facts about them can get pretty deep, like how the number of representations as a sum of 3 squares depends on the class number of a certain quadratic extension. Sums of two and four squares tend to be nicer.
At least according to his final exam video, there should be 35. He mentioned quadratic forms, and Euler’s Pentagonal Number theorem, so hopefully he’ll post them soon…
For the proof of the lemma with mp, to guarantee a smallest m exists, you should first mention why any exist at all. Obviously, m=p would give you mp=p^2+0^2, so the set of all such m is nonempty.
Could someone explain why the initial lemma was motivated by considering m and n in Z[i] instead of straightforwardly computing mn from their definitions?
Well if you start by saying mn = (x² + y²)(z² + w²), then you have mn = x²z² + x²w² + y²z² + y²w². For most folks, it's not immediately obvious that you can factor that into (xz - yw)² + (xw + yz)². You have this sort of _deus ex machina_ where you add and subtract 2xyzw to obtain the two perfect squares. It's not too jarring if you've seen enough of these, but Michael Penn's method leads progressively through to the desired sum of squares because the imaginary factor keeps separate the appropriate parts is a very satisfactory way.
It is really an adaptation of the multiplicative property of complex modules. If z and t are complex numbers then |zt|^2 = |z|^2|t|^2 which is sufficient to prove the lemma
It's certainly possible to do this by expanding mn, then completing the squares by adding and subtracting 2xyzw from the RHS and recombining to get the required expression as a sum of squares.
As a bit of extension work, I have a small "rabbit hole" that might be worth exploring. As the number gets larger, so the number of different factorisations it has can increase. For example, 50 has the prime factorisation 2 * 5². But since 5 = 1 mod 4, it can be expressed as the sum of two squares, so we can create two different factorisations, each of which is a valid route to expressing it as the sum of two squares: (a) 5² * (1² + 1²) = 5² + 5² (b) 10 * 5 = (3² + 1²)(2² + 1²) = (3*1 - 1*2)² + (3*2 + 1*1)² = 1² + 7² Conjecture: 50 is the smallest number that can be expressed as the sum of two squares in two different, non-trivial ways. Some other numbers that can be expressed as the sum of two squares in two different ways are: 2885, 3385, 3560, 4097, 4625. It is instructive to find the dual solutions to each of these. Question: Is there a way to identify (or even better to _generate_) numbers that can be expressed as the sum of two squares in two different ways? The number 325 can be expressed as the sum of two squares in *three* different, non-trivial ways: 325 = 1² + 18² = 6² + 17² = 10² + 15². Can you find the routes to each of these expressions? Conjecture: 325 is the smallest number that can be expressed as the sum of two squares in three different, non-trivial ways. Some other numbers that can be expressed as the sum of two squares in three different ways are: 425, 725, 1445, 2425. It is instructive to find the triple solutions to each of these. Question: Is there a way to identify (or even better to _generate_) numbers that can be expressed as the sum of two squares in three different ways? Are they all multiples of 5?
Complex numbers? I seem to have missed that particular lecture in *_this_* series where complex numbers were defined/derived. 🙄 Michael started this lecture series with Peano axioms. Back then, I got the idea that that's sufficient-and that additional results would be developed as needed, step-by-step, within this course. But as the lectures proceeded, Michael started using results from other branches of math, like calculus-and today, complex numbers. So now, I'm beginning to think, you need to know quite a lot of advanced math as prerequisites in order to complete this course. I don't have a problem with this. It's just that I would have appreciated a heads-up about the prerequisites.
Number theory is an advanced topic depending on many other high level branches of math. The Peano axioms aren't really number theory per se, but rather part of the logic/set theory which provides the foundation on which number theory sits.
I think it's worth noting that this series of videos is designed to _support_ a course that he's teaching, rather than being a self-contained teaching course in its own right. It may well be that before using complex numbers or calculus to perform a proof, he will check that his students are comfortable with those concepts, and if not, he could present an alternative proof that stayed within the comfort zone of his students. Every time I can recall him using calculus (or this case using complex numbers), there have been well-documented alternative proofs available. But the proof used here is more _elegant_ than the "add and subtract 2xyzw" kludge that would normally be presented, and I for one find it much more interesting when I see a proof I previously was unaware of.
@@RexxSchneider It's not about the students' comfort level or the elegance of proofs. 🤣 Note that I'm not saying that students won't know complex numbers or calculus. In fact, if you want to study Number Theory (MATH-337) at Randolph College, the chain of prerequisites include Elementary Mathematical Modeling (MATH-113), Precalculus (MATH-119), Calculus 1 (MATH-149), and Calculus 2 (MATH-150). So it's quite clear they'll know. My point is that in math, when you're systematically expounding a particular topic, there's one cardinal rule you must follow: In order to prove a result, you're allowed to use only the axioms you've previously selected, along with any results you've previously proved from those axioms. You _cannot_ use other results that haven't been proved yet. Of course I know that the videos are not the full course. Minor continuity gaps are not the problem. For example, Prof. Penn did not include 0 in his version of the Peano axioms. A few lectures later, he was happily using 0, negative numbers, and rational numbers. He didn't mention where they came from. But that's fine; they're not hard to define. But bringing in calculus is a completely different ballgame. At the very least, you have to go from rational to real numbers-using Dedekind cuts or whatever-and that's not trivial. After that, you can say _"Great, now that we have real numbers, let's use calculus, which you've already learnt."_ Until then, you _must_ use the 'kludgy' proof. Otherwise you could end up with circular reasoning. For example, in Euclidean Geometry, you have to prove the Pythagoras Theorem the old-fashioned way. You can't bring in the Distance Formula from Cartesian Geometry, however elegant-trivial, in fact-it might make your proof!
@@joshuatilley1887 Well, yes and no. Almost everything that Prof. Penn has taught up to this video can be understood by a high-school student. Now, although he used calculus and complex numbers to prove _some_ theorems, it wasn't necessary. Those theorems could have been proved more conventionally, avoiding calculus and complex numbers. However, as @Rex Schneider pointed out, the conventional proofs wouldn't be as 'elegant.' Certainly, Number Theory can be carried forward to very advanced levels, where it requires other advanced branches of math to prove results. For example, Wiles' proof of Fermat's Last Theorem uses Algebraic Geometry, Modular Forms, Elliptic Curves and so on-none of which I even pretend to understand. However, I doubt this course will go that far. As for Peano axioms: There are any number of foundational axioms to choose from-Peano, New Foundations, NBG, ZFC, Morse-Kelley, Tarski-Grothendieck, Category Theory ... For Number Theory, it doesn't matter which one you use. You just need integers and arithmetic operations, and you're off. As far as I'm concerned, you don't even have to start with foundational axioms. Prof. Penn could have said, _"You already know set theory, calculus (which includes arithmetic), and complex numbers (which include natural numbers, integers, rationals, and reals). Let's start from there." But he didn't do that. He took a different approach: _Forget everything you know. We'll start with an element we'll call 'one,' a successor function, and a handful of axioms. We'll build everything we need starting with these._ And so, when he brings in calculus or complex numbers out of nowhere, you can see why there's a disconnect. (BTW, he also used zero, negative numbers, and rational numbers without defining them first. But those slips weren't so egregious-you can easily get to them from natural numbers.)
@@nHans It's not about students at all. This isn't a classroom and the viewers are not students. For _my_ edification, it's all about the elegance or novelty of proofs and you don't get to say who can watch and appreciate these videos and who can't. To date, this video has 6,087 views, we can make a reasonable guess that 6,081 of those are not from students at Randolph College, so their prerequisites are utterly irrelevant. The rest of the views will come from people with a wide range of mathematical abilities and interests. You can't even assume that they will have watched any of the earlier videos in the set. I'm afraid that your assumptions about "one cardinal rule" is nonsense when you are presenting to an unknown, online audience. It's like attempting to teach a topic to a previously unknown group of students without the ability to use feedback to change your teaching as you learn about them. And then next week, you might get a completely different group. It reminds me to some extent of the first time I taught 16 year-olds. I was beginning to deliver a lesson on magnetism to a group of "General Science" students, when one of the many girls in the class politely explained to me that the class was actually half-way through a Human Biology course. At least I was able to use the feedback to switch the lesson. If you've ever tried to deliver a topic to a freshman group of university students, you'll know that their prior experience can vary dramatically, and no matter what you think they know, you'll be wrong. That's magnified a hundred-fold with an online audience. Under the circumstances that apply to these online presentations, you have to admire Michael Penn's bravery in picking his topics and proofs in a way that generally satisfies a very broad audience, without alienating the less and more experienced. The only way that he can get feedback is through the comments, and these play a valuable part in fixing any typos and bridging the gaps in the presentations. I disagree that using the "kludgy" proofs is essential. If you teach to the lowest common ability, you bore the more able. The essence of a good teach is their ability to vary their targets and maintain the interest of as many as possible. If people want to see the "add-and-subtract 2xyzw" proof, I can cheerfully spell that out here in the comments, and so could you, I'm sure. Either of us could pretty probably point questioners to a plethora of other videos that make use of that sort of hack, because a lot of presenters seem to enjoy presenting problems which are easy to solve if you already know the answer. Personally, I've seen the usual geometric construction to prove Pythagoras enough times. I yearn for the occasional elegant proof, even if relies on other branches of maths in its exposition. I wouldn't put the distance formula from Cartesian geometry in that category, as I'm pretty sure that it directly relies on Pythagoras so much that it's merely a restatement of it using coordinates.
In the proof of the first proposition it is not necessary to delve into complex numbers, if one factors out mn, adding and substracting 2xyzw will give mn as a sum of two squares...
Really overcomplicated the proof that if p=3 mod 4 then p is not a sum of two squares. It should only take 1 line: squares mod 4 are 0,1, so a sum of two squares is 0,1,2, i.e. never 3.
One thing i appreciate on these lectures
*colorful chalks*
Guess who found the secrets to time travel
Nice
For the first lemma, there is an alternative proof with quadratic residues. Suppose p does not divide x and does not divide y, which means p does not divide -y^2. Then from x^2 ≡ -y^2 (mod p), -y^2 is a quadratic residue mod p (primes of the form 4k+3 are obviously odd). Hence the Legendre symbol (-y^2/p) = 1, but (-y^2/p) = (-1/p)*(y^2/p) = (-1/p) ≡ (-1)^[(p-1)/2] = -1, a contradiction. Thus either p|x or p|y, and in either cases it's easy to show that p divides the other as well
Why does the theorem proven at 27:13 keep working for even powers of primes p=3(mod 4)?
He proved that the product up to (Xk^2 + Yk^2)^Rk is a sum of squares (C^2 + D^2) and then he distributed A^2 in the way (AC)^2 + (AD)^2. But what if that product was 1? In that case who tells us that A^2 can be written as a sum of squares?
Additionally, I did not understand why (a-c)/2 and (b-d)/2 would be multiples of X. What if 2^k || a-c and b-d for the same k? In that case wouldn’t 2^k || X also?
Please help❤
34:00 13=3^2+2^2.
Classic michael...
33:33 Homework
34:43 Good Place To Stop
29:43 little mistake ... Should be (a+c) not (a+b).... I think 34:05 also should be 3 in the bracket not 9....... also...... 65 can be written as sum of two squares in 2 ways. 65 = (1sqd + 8sqd) or (4sqd + 7sqd). 👍😉
But if 65 can be written as a sum of two squares then there is a unique way to do this but you find a two different ways to write it as sum of two squares (1^2+8^2) and (7^2+4^2) i don't know anything 😢😮?
65 is not a prime.
The really cool thing is that a number n can be written as the sum of two squares if and only if it doesn’t have some factor p^k where p is prime and 3 mod 4 and k is odd.
Spoiler alert.
For those who want to check their working on the example and the warm-ups.
*Example:* 65 = 5*13 = (1² + 2²) * (3² + 2²) Note that the video has this wrong!
Now we use m = x² + y² and n = z² + w² to write mn = (xz - yw)² + (xw + yz)² - although the choice of x, y, z, w will sometimes lead to multiple solutions.
So 65 = 5*13 = (1² + 2²) * (3² + 2²) = (1*3 - 2*2)² + (1*2 + 2*3)² = 1²+ 8² because (-1)² = (1)² and we're only interested in the solutions in positive integers.
But 65 = (1² + 2²) * (3² + 2²) can also be (1*2 - 2*3)² + (1.3 + 2*2)² = 4² + 7² so we have two different representations of 65 as the sum of two squares.
*WU1:* 99 = 3² * 11. Now 3² can be written as sum of two squares, but 11 is congruent to 3 (mod 4) so cannot, and therefore, neither can 99.
*WU2:* 45 = 3² * 5. Both 3² and 5 (which is congruent to 1 (mod 4)) can be expressed as sum of two squares, so we're good to go. I'll use a short cut, rather than the xyzw formula above as it's easier to just distribute a squared term like this:
45 = 3² * 5 = 3² * (1² + 2²) = (3² * 1²) + (3² * 2²) = (3*1)² + (3*2)² = 3² + 6²
That turns out to be the only solution because we cant "break apart" the 3² in the prime factorisation to create another factorisation that will work, since 3 ≡ 3 (mod 4) obviously isn't equal to the sum of two squares.
*WU3:* 80 = 2^4 * 5. As 5 ≡ 1 (mod 4), we'll get a solution.
80 = 16 * 5 = 4² * (1² + 2²) = (4*1)² + (4*2)² = 4² + 8² - using the distribution trick from earlier. There are other factorisations of 80, but it turns out that each one leads to the same result:
80 = 8 * 10 = (2² + 2²) * (3² + 1²) = (2*3 - 2*1)² + (2*1 + 2*3)² = 4² + 8²
80 = 4 * 20 = 2² * (2² + 4²) = (2*2)² + (2*4)² = 4² + 8²
80 = 2 * 40 = (1² + 1²) * (2² + 6²) = (1*2 - 1*6)² + (1*2 + 1*6)² = 4² + 8² etc.
*WU4:* 490 = 2 * 5 * 7² . The only prime ≡ 3 (mod 4) is 7 and that is to an even power, so we'll get a solution.
490 = 7² * 10 = 7² * (1² + 3²) = (7*1)² + (7*3)² = 7² +21². There are other possible factorisations but it turns out they all lead to the same result:
490 = 98 * 5 = (7² + 7²)(2² + 1²) = (7*2 - 7*1)² + (7*1 + 7*2)² = 7² + 21²
490 = 245 * 2. We'll need to find an expression of 245 as sum of two squares: 245 = 49*5 = 7² * (1² + 2²) = (7*1)² + (7*2)² = 7² + 14²
490 = 245 * 2 = (14² + 7²) * (1² + 1²) = (14*1 - 7*1)² + (14*1 + 7*1)² = 7² + 21²
thank you so much for the very good video
29:11 should be a^2-c^2=d^2-b^2, otherwise the equations at 32:27 are wrong
34:07 13=2^2+3^2
Is there simple criteria to show how many ways a number can be written as 1) the sum of CONSECUTIVE squares and 2) the sum of consecutive cubes.
There is for the sum of consecutive integers. The number of ways that an integer M can be written as the sum of at least two consecutive numbers
is the number of divisors of 2M less 1.
Here is a university admission problem from Japan. Prove that for all natural numbers n, x^2+y^2=2009^n always has a solution. Should be fairly easy by applying today's results. 😃
It’s just a matter of factoring 2009, which you would need to check all primes under about 50.
Really interesting! But I'm still waiting for the representation of a given number as a sum of 3 squares? Or may be I missed it?
Thank you for the video!
I’ve looked for simple proofs, but I think they require an understanding of 3-variable quadratic forms and some proofs regarding them. Facts about them can get pretty deep, like how the number of representations as a sum of 3 squares depends on the class number of a certain quadratic extension.
Sums of two and four squares tend to be nicer.
Professor, you killed me with all the letters! 🙂
Hi Michael, is this the final lecture or are there more that haven’t been posted?
At least according to his final exam video, there should be 35. He mentioned quadratic forms, and Euler’s Pentagonal Number theorem, so hopefully he’ll post them soon…
@@noahtaul which "final exam video"?
21:13 Why n/d^2 ∈ N? It is true only if p does not divide d. But there is no proof of that.
For the proof of the lemma with mp, to guarantee a smallest m exists, you should first mention why any exist at all. Obviously, m=p would give you mp=p^2+0^2, so the set of all such m is nonempty.
But gcd(0,p) in that case is not 1
Great work good job !!
Could someone explain why the initial lemma was motivated by considering m and n in Z[i] instead of straightforwardly computing mn from their definitions?
Well if you start by saying mn = (x² + y²)(z² + w²), then you have mn = x²z² + x²w² + y²z² + y²w².
For most folks, it's not immediately obvious that you can factor that into (xz - yw)² + (xw + yz)².
You have this sort of _deus ex machina_ where you add and subtract 2xyzw to obtain the two perfect squares. It's not too jarring if you've seen enough of these, but Michael Penn's method leads progressively through to the desired sum of squares because the imaginary factor keeps separate the appropriate parts is a very satisfactory way.
It is really an adaptation of the multiplicative property of complex modules.
If z and t are complex numbers then |zt|^2 = |z|^2|t|^2 which is sufficient to prove the lemma
@@RexxSchneider Amazingly, that must have been how the identity was first discovered, since it greatly predated complex numbers.
It's certainly possible to do this by expanding mn, then completing the squares by adding and subtracting 2xyzw from the RHS and recombining to get the required expression as a sum of squares.
As a bit of extension work, I have a small "rabbit hole" that might be worth exploring.
As the number gets larger, so the number of different factorisations it has can increase.
For example, 50 has the prime factorisation 2 * 5². But since 5 = 1 mod 4, it can be expressed as the sum of two squares, so we can create two different factorisations, each of which is a valid route to expressing it as the sum of two squares:
(a) 5² * (1² + 1²) = 5² + 5²
(b) 10 * 5 = (3² + 1²)(2² + 1²) = (3*1 - 1*2)² + (3*2 + 1*1)² = 1² + 7²
Conjecture: 50 is the smallest number that can be expressed as the sum of two squares in two different, non-trivial ways.
Some other numbers that can be expressed as the sum of two squares in two different ways are: 2885, 3385, 3560, 4097, 4625. It is instructive to find the dual solutions to each of these.
Question: Is there a way to identify (or even better to _generate_) numbers that can be expressed as the sum of two squares in two different ways?
The number 325 can be expressed as the sum of two squares in *three* different, non-trivial ways: 325 = 1² + 18² = 6² + 17² = 10² + 15². Can you find the routes to each of these expressions?
Conjecture: 325 is the smallest number that can be expressed as the sum of two squares in three different, non-trivial ways.
Some other numbers that can be expressed as the sum of two squares in three different ways are: 425, 725, 1445, 2425. It is instructive to find the triple solutions to each of these.
Question: Is there a way to identify (or even better to _generate_) numbers that can be expressed as the sum of two squares in three different ways? Are they all multiples of 5?
Complex numbers? I seem to have missed that particular lecture in *_this_* series where complex numbers were defined/derived. 🙄
Michael started this lecture series with Peano axioms. Back then, I got the idea that that's sufficient-and that additional results would be developed as needed, step-by-step, within this course. But as the lectures proceeded, Michael started using results from other branches of math, like calculus-and today, complex numbers. So now, I'm beginning to think, you need to know quite a lot of advanced math as prerequisites in order to complete this course.
I don't have a problem with this. It's just that I would have appreciated a heads-up about the prerequisites.
Number theory is an advanced topic depending on many other high level branches of math. The Peano axioms aren't really number theory per se, but rather part of the logic/set theory which provides the foundation on which number theory sits.
I think it's worth noting that this series of videos is designed to _support_ a course that he's teaching, rather than being a self-contained teaching course in its own right. It may well be that before using complex numbers or calculus to perform a proof, he will check that his students are comfortable with those concepts, and if not, he could present an alternative proof that stayed within the comfort zone of his students. Every time I can recall him using calculus (or this case using complex numbers), there have been well-documented alternative proofs available.
But the proof used here is more _elegant_ than the "add and subtract 2xyzw" kludge that would normally be presented, and I for one find it much more interesting when I see a proof I previously was unaware of.
@@RexxSchneider It's not about the students' comfort level or the elegance of proofs. 🤣
Note that I'm not saying that students won't know complex numbers or calculus. In fact, if you want to study Number Theory (MATH-337) at Randolph College, the chain of prerequisites include Elementary Mathematical Modeling (MATH-113), Precalculus (MATH-119), Calculus 1 (MATH-149), and Calculus 2 (MATH-150). So it's quite clear they'll know.
My point is that in math, when you're systematically expounding a particular topic, there's one cardinal rule you must follow: In order to prove a result, you're allowed to use only the axioms you've previously selected, along with any results you've previously proved from those axioms. You _cannot_ use other results that haven't been proved yet.
Of course I know that the videos are not the full course. Minor continuity gaps are not the problem. For example, Prof. Penn did not include 0 in his version of the Peano axioms. A few lectures later, he was happily using 0, negative numbers, and rational numbers. He didn't mention where they came from. But that's fine; they're not hard to define.
But bringing in calculus is a completely different ballgame. At the very least, you have to go from rational to real numbers-using Dedekind cuts or whatever-and that's not trivial. After that, you can say _"Great, now that we have real numbers, let's use calculus, which you've already learnt."_
Until then, you _must_ use the 'kludgy' proof. Otherwise you could end up with circular reasoning. For example, in Euclidean Geometry, you have to prove the Pythagoras Theorem the old-fashioned way. You can't bring in the Distance Formula from Cartesian Geometry, however elegant-trivial, in fact-it might make your proof!
@@joshuatilley1887 Well, yes and no.
Almost everything that Prof. Penn has taught up to this video can be understood by a high-school student.
Now, although he used calculus and complex numbers to prove _some_ theorems, it wasn't necessary. Those theorems could have been proved more conventionally, avoiding calculus and complex numbers. However, as @Rex Schneider pointed out, the conventional proofs wouldn't be as 'elegant.'
Certainly, Number Theory can be carried forward to very advanced levels, where it requires other advanced branches of math to prove results.
For example, Wiles' proof of Fermat's Last Theorem uses Algebraic Geometry, Modular Forms, Elliptic Curves and so on-none of which I even pretend to understand.
However, I doubt this course will go that far.
As for Peano axioms: There are any number of foundational axioms to choose from-Peano, New Foundations, NBG, ZFC, Morse-Kelley, Tarski-Grothendieck, Category Theory ...
For Number Theory, it doesn't matter which one you use. You just need integers and arithmetic operations, and you're off.
As far as I'm concerned, you don't even have to start with foundational axioms. Prof. Penn could have said, _"You already know set theory, calculus (which includes arithmetic), and complex numbers (which include natural numbers, integers, rationals, and reals). Let's start from there."
But he didn't do that. He took a different approach: _Forget everything you know. We'll start with an element we'll call 'one,' a successor function, and a handful of axioms. We'll build everything we need starting with these._
And so, when he brings in calculus or complex numbers out of nowhere, you can see why there's a disconnect.
(BTW, he also used zero, negative numbers, and rational numbers without defining them first. But those slips weren't so egregious-you can easily get to them from natural numbers.)
@@nHans It's not about students at all. This isn't a classroom and the viewers are not students. For _my_ edification, it's all about the elegance or novelty of proofs and you don't get to say who can watch and appreciate these videos and who can't.
To date, this video has 6,087 views, we can make a reasonable guess that 6,081 of those are not from students at Randolph College, so their prerequisites are utterly irrelevant. The rest of the views will come from people with a wide range of mathematical abilities and interests. You can't even assume that they will have watched any of the earlier videos in the set.
I'm afraid that your assumptions about "one cardinal rule" is nonsense when you are presenting to an unknown, online audience. It's like attempting to teach a topic to a previously unknown group of students without the ability to use feedback to change your teaching as you learn about them. And then next week, you might get a completely different group. It reminds me to some extent of the first time I taught 16 year-olds. I was beginning to deliver a lesson on magnetism to a group of "General Science" students, when one of the many girls in the class politely explained to me that the class was actually half-way through a Human Biology course. At least I was able to use the feedback to switch the lesson.
If you've ever tried to deliver a topic to a freshman group of university students, you'll know that their prior experience can vary dramatically, and no matter what you think they know, you'll be wrong. That's magnified a hundred-fold with an online audience.
Under the circumstances that apply to these online presentations, you have to admire Michael Penn's bravery in picking his topics and proofs in a way that generally satisfies a very broad audience, without alienating the less and more experienced. The only way that he can get feedback is through the comments, and these play a valuable part in fixing any typos and bridging the gaps in the presentations.
I disagree that using the "kludgy" proofs is essential. If you teach to the lowest common ability, you bore the more able. The essence of a good teach is their ability to vary their targets and maintain the interest of as many as possible. If people want to see the "add-and-subtract 2xyzw" proof, I can cheerfully spell that out here in the comments, and so could you, I'm sure. Either of us could pretty probably point questioners to a plethora of other videos that make use of that sort of hack, because a lot of presenters seem to enjoy presenting problems which are easy to solve if you already know the answer.
Personally, I've seen the usual geometric construction to prove Pythagoras enough times. I yearn for the occasional elegant proof, even if relies on other branches of maths in its exposition. I wouldn't put the distance formula from Cartesian geometry in that category, as I'm pretty sure that it directly relies on Pythagoras so much that it's merely a restatement of it using coordinates.
In the proof of the first proposition it is not necessary to delve into complex numbers, if one factors out mn, adding and substracting 2xyzw will give mn as a sum of two squares...
Yo. You really need to be more careful. 32:20 b=yw+xz not yw-xz. Your proof is wrong.
The first one is something called the Brahmagupta's identity.
Imagine clicking the notification instantly and be defeated by viewers who figured out time travel.
This crap is baalin bro