Dr. Peyam's clickbait game is on another level. Nice video, I loved seeing a breakdown of what would've been a very boring math paper into an illustrative and exciting video on a concept that was seemingly indeterminate, can't wait to see more. :)
Brilliant video. In the harmonic system in music, rotation by pi/3 is a whole step or going from C to D and pi/2 is a minor third or going from C to Eb. Therefore, both these rotations being roots of unity mean they divide the octave in equal parts thus for a cycle. In this case, because both 3 and 2 divides 6, a tritone is an immediate byproduct of the 6 and 4 cycle all that to say that you can change keys constantly and still end up in the same key you started in without going thorough all 12 possible notes.
@@drpeyam I developed a generator function that takes numbers as inputs and outputs harmonic progressions, chord voicings, Melodie’s or maps of modulation depending on which rhythm you prescribe as the “quantum bit”. Either a measure is the standard unit and your assembling chord progressions i.e. Cmaj7 D-7 G7 or it’s a beat and your assembling chord voicings if all notes are played at once and Melodie’s if all notes are played in temporal succession. Then if a section/phrase is the standard unit then each number represents a key center that you modulate between. You only need 6 angles to consider because of the symmetry of powers of 12. In fact, you can get away with only 3 angles if you allow for negative factors outside and inside the argument. I’ve since written 5 songs with this system.
@@dominicellis1867 then a Major third will become 2π/3 from C to E, or in a higher setting, D to F#, so we will find out whether such an interval will be brighter, like as in A to C#, or darker, like say B to D.
@@byronrobbins8834 exactly, and you can prescribe key center with a delta parameter representing the “new” I chord. It’s essentially a phase shift on an entire musical vector denoting the chords, Melodie’s, or musical movements. Therefore, D maj is equivalent to * d where d in this case = exp(ipi/3).
Love your videos Dr. Peyam! And like another comment said, your clickbait game is on another level! Question though: Do you have a graduate-level textbook or grad-level lecture series that you authored/co-authored or recommend? I just graduated with a B.S. Applied Mathematics degree, and I want to continue sharpening my math skills in areas of Complex analysis, computational linear algebra, etc... I love the way you teach, which - at times in my undergrad studies - has really helped me to solidify my understanding of certain math concepts/principles. So if you happen to see this and have *_ANY_* recommendations or references, I'd love to take as many as you have - doesn't need to be applied mathematics necessarily! Thank you so much, sir!!
Nice little problem. Here's my approach. The problem can be simplified by letting t=x/y, so y/x=1/t (note that, from the given equalion, neither x nor y can be zero, so t≠0, and 1/t is defined), and the equation becomes t+1/t=1. Multiplying by t (which we can do, as long as we bear in mind that t≠0), we get t²+1=t or t²-t+1=0 Let's tweek this slightly more by letting u=-t, so u²+u+1=0, whose solutions we recognise to be the non-real cube roots of 1, ω & ω², where ω=e^(2πi/3)=-1/2+√3/2i. So t=-ω or -ω², and y=-ωx or -ω²x.
@@Francesco_Luligo the cube roots of 1 are the solutions of the equation x³=1, or x³-1=0. Now x³-1=(x-1)(x²+x+1) So x³-1=0 (x-1)(x²+x+1)=0 x=1 or x²+x+1=0. So the roots of x²+x+1=0 are the cube roots of 1 except 1 itself, the non-real cube roots of 1. A similar factorisation argument applies to any positive integer (≥2) order roots of 1, except that if the order is even, we'll also get the real root -1. For example, x⁴-1=(x-1)(x³+x²+x+1), so the 4th roots of 1 other than 1 itself are the roots of x³+x²+x+1=0.
whenever cancellation of imaginary parts comes up like this, i just think about how someone once pointed out to me... there is no distinction between "positive" and "negative" i. we arbitrarily chose one to be positive, but unlike real numbers, there are no inherent properties you can use to distinguish positive and negative imaginary numbers. you could exchange −i with +i throughout all of mathematics and everything would still work
So when the constant on the RHS is 0, x is a 90 degree rotation away from y. When that constant is 1, x is a 60 degree rotation away from y. When that constant is 2, x is a 0 degree rotation away from y. Next would be if x/y + y/x = k, find the degrees of rotation between complex x and y in terms of k. Well, let's say when abs(k) < 2
setting the RHS to k and using the same method as the first solution yields: x = y/2 ( k ± √(k^2 - 4) ) for k = 0, 1, 2 same results as the video k = 3: x = y/2 ( 3 ± √(9-4)) x = y (3 ± √5) / 2 This can be written as φ+1 and 1-Φ (or 2-φ) where φ is the golden ratio and Φ is the golden ratio conjugate, namely 1/φ or φ-1 For k ≥ 2 all the solutions will be real. k = i x = y/2 (i ± √(-5)) x = y i (1 ± √5)/2 x = φi y or -Φiy Seeing why this works is also fun: x = φi, y = 1 x/y + y/x = φi + (1/i) (1/φ) = φi + (-i)(φ-1) = i (φ + 1 - φ) = i
Once again, Dr. Peyam proves that *_Nothing is Impossible!_* 👍 Reminds me of a phase I went through in high school. Having learnt complex numbers, 3D vectors etc., I set out to prove that all hitherto-unsolvable equations can be solved if you invented the right family of numbers. 👨👩👧👦 After all, all the numbers that we're taught-negative, rational, irrational, real, imaginary, complex etc.-had been progressively invented for that exact reason: to solve equations that couldn't be solved otherwise. I even independently reinvented split-quaternions when a teacher said _"You cannot add scalars to vectors."_ 💡 It's too bad the teacher didn't recognize my genius. 😞 Even now, when I hear _"It has no solution,"_ my response is: _"You mean, using your current toolkit. Get better tools."_ 🧰 In possibly-related news, I'm a very successful engineering manager. 😎 Right now, the only outstanding "impossible" equation is division-by-zero. One of these days, I'll figure out the right family of numbers for that as well! 🔍
I wish there was a solution to x! = 0. I was trying to invent a HyperComplex number system to solve that equation, but I have no idea how algebra would work with that. And you should watch some videos by BritheMathGuy, he talked about division by 0 in the context of stereographic projection. Sadly, even with that system we can't do 0 * Infinity. But at least we could define 0/0 = 1 (under the right context, not in general)
@@Rudxain Allow me to reply to your two points in two separate comments. About BriTheMathGuy's video: Yeah, I've seen it. Using a stereographical projection, he defines +∞ = -∞ = ∞, and all a/0 = ∞. But this doesn't solve the problem. As he himself points out, it leads to 0 = 1 = 2 = ... He goes on to talk about Wheel Algebra, wherein indeterminate forms like 0/0, 0^0, 0.∞, ∞/∞, ∞ - ∞ are all _defined_ to be a new element, the _bottom_ element ⊥. And all operations involving the bottom element result in the bottom element. I'm okay with that. As a computer programmer, I occasionally deal with NaN, which is basically the same thing. The real problem why Wheel Algebra can't be used in practice is because basic rules like a/a = 1 and a(b+c) = ab + ac don't hold. 😨 I've also seen his other videos he defines and/or proves identities like: 👉 0/0 = ∞ (not 1 like you suggested) 👉 0^0 = 1 Unfortunately, those are neither convincing nor useful; I'm pretty sure they were just for fun. I took a different approach in high school: I invented an entirely _new_ family of numbers that I called *_Circled_* numbers. I described it in a different comment, so let me copy-paste myself: All of us know that you shouldn't divide by zero, and/or if you do, you get infinity. And none of us is happy with that. Admit it-we all occasionally wonder if that problem can be solved. 😜 And so-inspired by number families like *hyperreal, infinitesimal, ordinal, dual* etc.-I invented a new family of numbers that are solutions to division by zero. I write them as _circled_ numbers. For example: 1/0 = ①, 2/0 = ②, and in general, a/0 = ⓐ. I even convinced my engineering colleagues at work that such numbers exist. It was easy-I had already blown their minds by teaching them all those other number families I mentioned earlier, plus *surreal, split-complex, quaternion, octonion* etc. So they were ready to believe in anything! 🤣
@@Rudxain Solutions to x!=0, or more generally, Γ(z)=0, is *_exactly_* the kind of problem I'd like to see Dr. Peyam solve using a hitherto-unknown family of numbers!!! 👍 Dr. Peyam's usual trick of using complex numbers to solve seemingly impossible equations _won't work here!_ ☹ After all, it's well-known that there's no solution even among complex numbers. What else ya got, doc? 🤔 I thought a lot about this. I didn't make much progress. So, yeah, I think it's a very interesting problem!
@@nHans All of that is true. The circled numbers seem like a good idea to preserve numerical data. I also tought of something similar for Infinity, like ♾ = ♾ but ♾+1 != ♾, this would preserve a lot of algebraic properties at the cost of making each Infinity unique. This also meant that sqrt(♾)^2 = ♾ and sqrt(♾^2) = +-♾ (because sqrt has 2 solutions) I also thought about how negative numbers are actually positive numbers multiplied by -1, and we could call -1 the "inverted unit" and give it a special symbol like "@", so -5 in this algebraic system would be written as 5@ and read as "5 inverse units". @ is the solution to the equations x + 1 = 0, x^2 = 1 && x != 1 (this one is similar to epsilon, the 2nd solution to sqrt(0), which is x^2 = 0 && x != 0) This algebraic way of representing negatives seems useless, but it actually makes it easier to understand some of the properties of negative numbers in a different perspective. It also made me realize that @@ = 1 is *actually an axiom,* not a theorem, so we could have a different kind of arithmetic where "negative times negative is negative" instead of positive, but nobody uses that axiom because it's more useful for @ to act as a sign flipper instead of a black hole
@@nHans (replying to the factorial thing) exactly! But it really depends on which factorial extension we're using as "base". We could try using the Pi function, spline interpolation, analytic continuation, etc. But since 0 is an integer, we could use a definition that only works in the domain of ints. There's a factorial extension that returns (-1)^n * n! for negative int values of n. And there's another definition that's based in terms of divison, so n! = -1/n! (again, for negative n). Under that definition we can calculate the limit, and the solution to x! = 0 becomes -♾
can you take x/y + y/x = 1 as x² + y² = xy, then take x² - xy + y² = 0, then under the condition that x ≠ -y multiply both sides by x + y to get (x + y)(x² - xy + y²) = 0 or x³ + y³ = 0? thus you'd get, as x ≠ -y, x = ³√y³ with either one of the two complex branches of the cube root function?
@@drpeyam I did, but you went in a different direction and I'm not profecient enough with the complex roots of a number to tell whether my solution set is equivalent to yours.
@@drpeyam hmm I went back and referred my old high-school mah textbook, and I think my answer set is equivalent, if you take x = (ω/ω²)y where ω and ω² are the complex cube roots of unity
Just one small comment, the operation on irreducible fractions where you add the numerators and the denominators to find a new fraction actually has some interesting mathematical properties, and it's called the mediant.
Nice! I had been expecting a video about Farey addition (and, possibly, Ford circles), which obviously gives x/y ⊕ y/x = (x+y)/(y+x) = 1. It's always fun to be surprised.
Perhaps we should teach kids complex mathematics from and early age. Up to the end of secondary school 15-16 years old, we were being taught that the solution for a quadratic doesn't exist if (bb-4ac)
I totally agree with this! Gosh, I remember in elementary school (I think it was 1st grade) I asked my teacher about a simple addition/subtraction problem in which the answers were negative, and I kid you not she told me that we can't have negative numbers and that we can't solve equations like this because "we can't have negative numbers..." I responded by pointing to the thermometer on the wall which clearly showed it was 29-30˚F today (i.e. -1˚C), which she justified as an exception and that I didn't need to worry about it. Lol
I don’t know if you need to teach complex mathematics, but you can hint at it. For example, they are going to know about whole numbers and fractions. So you can do something like, can you divide 5 by 3? No, if the answer has to be a whole number, but yes if it can be a fraction. Can you find the square root of -3? No, if it has to be a real number, but there are other numbers where the answer is yes. In fact, there are lots of other types of interesting numbers that the students would want to learn about in the future. If you try to teach complex numbers really early, then where do you stop? Complex, quaternions, matrixes, infinities, etc. But I think it is bad to say there is no solution. Better to say, there is no solution with the numbers that we have studied so far. Sad thing is, some that teach ages up to 15 or so, may not even know about complex numbers themselves.
No, I'm afraid I disagree about teaching complex numbers to kids. In fact, a lot of well-intentioned people-like yourself-often suggest that some particular topic or another should be taught to all schoolchildren at an early age, because they found it very interesting or useful for themselves. Not just math topics, but also subjects like philosophy, cursive writing, farming, quantum mechanics, foreign languages, touch-typing, ethics, music etc. There are several reasons why I disagree. Let me discuss the two most important ones: Time and Relevance. (The others are Pedagogy and Teaching Resources.) It's not like schoolchildren are idling on standby with nothing to do. Rather, their curricula and schedules are already overcrowded. They don't have enough time even to absorb what they're being taught, let alone play and just enjoy being children. Every new topic that you want to add has to take time away from something else-eating, playing, sleeping, or a subject that you think is less important than complex numbers. And sure, that's a conversation that parents and educators should frequently be having. After all, as technology progresses, we need to keep adapting to the new stuff and eliminating the irrelevant and obsolete. What about relevance? Fact is, most adults don't use complex numbers in their daily lives. So I'd teach it only to those who will (eventually) need it-aspiring mathematicians, scientists, engineers etc. For most other professions-including management, accountants, lawyers, performing arts, humanities etc.-it's unnecessary. I'd rather teach students something useful to their daily lives. Or directly relevant to their future career. Or even give them back their free time so that they can be creative. In short, I wouldn't teach complex numbers to schoolchildren that early, nor to everyone. I agree about one thing though-you shouldn't tell children something is impossible when, in fact, more powerful tools _can_ solve the problem. When I come to such roadblocks while teaching, I'm very careful to explain that the tools _we currently have_ cannot solve this particular problem. At some point in the future, we'll learn how to use more powerful tools, but for now, leave them alone. For all you know, there could even be a family of numbers-which we haven't been taught yet-that can divide by zero!
Little suggestion: what if you dig into the rabbit hole of multi variables? Maybe with limits, gradients divergence and curl (or rotor as we italians call it)
Isn't this functionally the same as y = X^1 + X^-1 (in this case solving for when y=1), which I believe the graph of which has a slant asymptote at y=X and a vertical asymptote at X=0. This would imply that there are no real solutions when y=0 and when y=1.
In the real numbers, no. For any positive real x, either x > 1, which means we need 1/x to be negative (impossible), or x < 1, which means that 1/x > 1, therefore 1/x + x > 1 as well. For negative x, we'll be adding two negative numbers, which will never be 1. Now, in the complex numbers, we have i + 1/i = 0, which is new (not what was asked, but also previously impossible). If you look at an equilateral triangle (with sidelength 1), that actually exposes one possible answer. e^(i ⋅ 60°) + e^(i ⋅ -60°) = 1/2 + i ⋅ root(2, 3)/2 + 1/2 - i ⋅ root(2, 3)/2 = 1 To cover all solutions though, z + 1/z = 1 => z² - z + 1 = 0 => z ∈ 1/2 ± root(2, 1/4 - 1) .... which will just give the original solution, plus its conjugate/inverse.
Same problem in DESMOS, I noticed. I ought to be a line. Try 2.1 instead of 2, and then 2.01 etc However, when you "clear" the fraction in DESMOS and rewrite the equation, you obtain the desired line. Not sure why
It can be followed by the ancient Formula to calculate Quadratics. It was used by Babylonians and is older than quadratic formula This produces the reduced quadratic equation which is equivalent to quadratic formula i.e., -P/2 ±(√P/2)²-q Request: Please make a video on explaining the PQ formula for quadratic equation.
the first part gives you two solutions right away: notice you have x=y*( (1+-sqrt(3)i)/2), just notice that x/y = (1+-sqrt(3)i) / (2) and so y = 2 and x = 1+sqrt(3)i or 1-sqrt(3)i
multiply by xy: x^2+y^2=xy notice: (x-y)^2 = x^2+y^2-2xy notice: x^2+y^2=xy+xy+(x-y)^2 since both are true, we can combine them to get: (x-y)^2 = -xy. if we assume these are real numbers, we know -xy must be positive, except in the case x=y=0, which is not allowed by the original equation. Returning to the first step, knowing xy must be negative: x^2+y^2 = xy < 0, which is impossible in the real numbers. Therefore the solution must be complex, and i'm too lazy to continue from that.
I may be catching on to the trick behind Dr. Peyam's "Impossible" videos ... they're impossible with _real_ numbers, but solvable with _complex_ numbers. Right? Right??? Anyway, my real question is: Are there equations that cannot be solved with _complex_ numbers, but solvable with other, even more exotic types, such as quaternions, octonions etc.? I've never used-or even formally studied-the latter types; just read them up out of interest.
So at 0, the values are at right angles and at 2, the coincide. This implies that between 0 and 2 there is a continuum or angles between 0 and 90 degrees. Cool.
@@arturcostasteiner9735 Not really. It's still the same quadratic equation: x² - axy + y² = 0, and has the same quadratic solution for _x_ in terms of _a_ and _y._ As before, the solutions will be complex. In some cases, they'll be real. (Complex numbers include real numbers.) Note that _a_ and _y_ can also be complex-the quadratic formula will still work. Of course, if _a_ is some other type of number-quaternion, octonion etc.-then yes, it does get more complicated, because the high-school quadratic formula no longer holds.
Can't say for certain, but for c=1 your last formula is atan(sqrt(-12)), I'm sure the result can be written in a better form. I'm getting a=arccos(c/2) "and that's a good place to stop".
@@Milan_Openfeint Argh, within the sqrt expression i somehow used "c^2 - 4" instead of "4 - c^2": Need sqrt(c^2-4) = i sqrt(4-c^2) to get a real value within the sqrt for -2 < c < 2... corrected the above. (Funnily i used the correct expression in the "r^2 =..." line, but nowhere else... . So for c=1 we should get +/-atan(sqrt(12)), though that seems to be false, maybe i shouldn't have done many parts in my head i assume.
I say y = αx solves x/y+y/x = β (where β∈ℝ*), then α∈ℝ iff |β| ⋝ 2. Conversely, α∈ℂ, iff |β| < 2. I just don't like ℂ where all those pesky ℿmi's popping up everywhere! シ
Dr. Peyam's clickbait game is on another level. Nice video, I loved seeing a breakdown of what would've been a very boring math paper into an illustrative and exciting video on a concept that was seemingly indeterminate, can't wait to see more. :)
I know right!
Just goes to show, he definitely knows his audience haha
Thanks so much!!! 😁 Miss you already!!
@@isaackay5887 the equation in the thumbnail becomes 34/15 = 1, which is a contradiction, so then such a pairing will fail.
That’s the first thing I thought of when I saw this video: “Oh, the nerd side of TH-cam is finally getting in on the action! I love it!”
Brilliant video. In the harmonic system in music, rotation by pi/3 is a whole step or going from C to D and pi/2 is a minor third or going from C to Eb. Therefore, both these rotations being roots of unity mean they divide the octave in equal parts thus for a cycle. In this case, because both 3 and 2 divides 6, a tritone is an immediate byproduct of the 6 and 4 cycle all that to say that you can change keys constantly and still end up in the same key you started in without going thorough all 12 possible notes.
Omg this application is mind-blowing!!
@@drpeyam I developed a generator function that takes numbers as inputs and outputs harmonic progressions, chord voicings, Melodie’s or maps of modulation depending on which rhythm you prescribe as the “quantum bit”. Either a measure is the standard unit and your assembling chord progressions i.e. Cmaj7 D-7 G7 or it’s a beat and your assembling chord voicings if all notes are played at once and Melodie’s if all notes are played in temporal succession. Then if a section/phrase is the standard unit then each number represents a key center that you modulate between. You only need 6 angles to consider because of the symmetry of powers of 12. In fact, you can get away with only 3 angles if you allow for negative factors outside and inside the argument. I’ve since written 5 songs with this system.
@@dominicellis1867 then a Major third will become 2π/3 from C to E, or in a higher setting, D to F#, so we will find out whether such an interval will be brighter, like as in A to C#, or darker, like say B to D.
@@byronrobbins8834 exactly, and you can prescribe key center with a delta parameter representing the “new” I chord. It’s essentially a phase shift on an entire musical vector denoting the chords, Melodie’s, or musical movements. Therefore, D maj is equivalent to * d where d in this case = exp(ipi/3).
You've just discovered modular arithmetic, congratulations.
At the end, an alternative way to verify is to recognise that
e^ipi/3 + e^-ipi/3 = 2 cos(pi/3)
Which is obviously = 1
Love your videos Dr. Peyam! And like another comment said, your clickbait game is on another level!
Question though: Do you have a graduate-level textbook or grad-level lecture series that you authored/co-authored or recommend? I just graduated with a B.S. Applied Mathematics degree, and I want to continue sharpening my math skills in areas of Complex analysis, computational linear algebra, etc... I love the way you teach, which - at times in my undergrad studies - has really helped me to solidify my understanding of certain math concepts/principles.
So if you happen to see this and have *_ANY_* recommendations or references, I'd love to take as many as you have - doesn't need to be applied mathematics necessarily! Thank you so much, sir!!
Nice little problem.
Here's my approach.
The problem can be simplified by letting t=x/y, so y/x=1/t (note that, from the given equalion, neither x nor y can be zero, so t≠0, and 1/t is defined), and the equation becomes t+1/t=1.
Multiplying by t (which we can do, as long as we bear in mind that t≠0), we get
t²+1=t
or t²-t+1=0
Let's tweek this slightly more by letting u=-t,
so u²+u+1=0, whose solutions we recognise to be the non-real cube roots of 1, ω & ω², where ω=e^(2πi/3)=-1/2+√3/2i.
So t=-ω or -ω², and y=-ωx or -ω²x.
Why does the solution involve the cube roots of 1, w and w^2?
@@Francesco_Luligo the cube roots of 1 are the solutions of the equation x³=1, or x³-1=0.
Now x³-1=(x-1)(x²+x+1)
So x³-1=0 (x-1)(x²+x+1)=0 x=1 or x²+x+1=0.
So the roots of x²+x+1=0 are the cube roots of 1 except 1 itself, the non-real cube roots of 1.
A similar factorisation argument applies to any positive integer (≥2) order roots of 1, except that if the order is even, we'll also get the real root -1.
For example, x⁴-1=(x-1)(x³+x²+x+1), so the 4th roots of 1 other than 1 itself are the roots of x³+x²+x+1=0.
@@MichaelRothwell1 Thanks!
whenever cancellation of imaginary parts comes up like this, i just think about how someone once pointed out to me... there is no distinction between "positive" and "negative" i. we arbitrarily chose one to be positive, but unlike real numbers, there are no inherent properties you can use to distinguish positive and negative imaginary numbers. you could exchange −i with +i throughout all of mathematics and everything would still work
Pretty much! And it just boils down to reflecting your universe about the x axis
Great Job Peyam. Very Interesting.
So when the constant on the RHS is 0, x is a 90 degree rotation away from y. When that constant is 1, x is a 60 degree rotation away from y. When that constant is 2, x is a 0 degree rotation away from y. Next would be if x/y + y/x = k, find the degrees of rotation between complex x and y in terms of k. Well, let's say when abs(k) < 2
That is so cool!
setting the RHS to k and using the same method as the first solution yields:
x = y/2 ( k ± √(k^2 - 4) )
for k = 0, 1, 2 same results as the video
k = 3:
x = y/2 ( 3 ± √(9-4))
x = y (3 ± √5) / 2
This can be written as φ+1 and 1-Φ (or 2-φ) where φ is the golden ratio and Φ is the golden ratio conjugate, namely 1/φ or φ-1
For k ≥ 2 all the solutions will be real.
k = i
x = y/2 (i ± √(-5))
x = y i (1 ± √5)/2
x = φi y or -Φiy
Seeing why this works is also fun:
x = φi, y = 1
x/y + y/x = φi + (1/i) (1/φ) = φi + (-i)(φ-1)
= i (φ + 1 - φ)
= i
@@drpeyam How about: theta = arctan((sqrt(4-k^2)/k)) * 180/pi
Once again, Dr. Peyam proves that *_Nothing is Impossible!_* 👍
Reminds me of a phase I went through in high school. Having learnt complex numbers, 3D vectors etc., I set out to prove that all hitherto-unsolvable equations can be solved if you invented the right family of numbers. 👨👩👧👦
After all, all the numbers that we're taught-negative, rational, irrational, real, imaginary, complex etc.-had been progressively invented for that exact reason: to solve equations that couldn't be solved otherwise.
I even independently reinvented split-quaternions when a teacher said _"You cannot add scalars to vectors."_ 💡 It's too bad the teacher didn't recognize my genius. 😞
Even now, when I hear _"It has no solution,"_ my response is: _"You mean, using your current toolkit. Get better tools."_ 🧰
In possibly-related news, I'm a very successful engineering manager. 😎
Right now, the only outstanding "impossible" equation is division-by-zero. One of these days, I'll figure out the right family of numbers for that as well! 🔍
I wish there was a solution to x! = 0. I was trying to invent a HyperComplex number system to solve that equation, but I have no idea how algebra would work with that.
And you should watch some videos by BritheMathGuy, he talked about division by 0 in the context of stereographic projection. Sadly, even with that system we can't do 0 * Infinity. But at least we could define 0/0 = 1 (under the right context, not in general)
@@Rudxain Allow me to reply to your two points in two separate comments.
About BriTheMathGuy's video: Yeah, I've seen it. Using a stereographical projection, he defines +∞ = -∞ = ∞, and all a/0 = ∞. But this doesn't solve the problem. As he himself points out, it leads to 0 = 1 = 2 = ...
He goes on to talk about Wheel Algebra, wherein indeterminate forms like 0/0, 0^0, 0.∞, ∞/∞, ∞ - ∞ are all _defined_ to be a new element, the _bottom_ element ⊥. And all operations involving the bottom element result in the bottom element.
I'm okay with that. As a computer programmer, I occasionally deal with NaN, which is basically the same thing.
The real problem why Wheel Algebra can't be used in practice is because basic rules like a/a = 1 and a(b+c) = ab + ac don't hold. 😨
I've also seen his other videos he defines and/or proves identities like:
👉 0/0 = ∞ (not 1 like you suggested)
👉 0^0 = 1
Unfortunately, those are neither convincing nor useful; I'm pretty sure they were just for fun.
I took a different approach in high school: I invented an entirely _new_ family of numbers that I called *_Circled_* numbers.
I described it in a different comment, so let me copy-paste myself:
All of us know that you shouldn't divide by zero, and/or if you do, you get infinity. And none of us is happy with that. Admit it-we all occasionally wonder if that problem can be solved. 😜
And so-inspired by number families like *hyperreal, infinitesimal, ordinal, dual* etc.-I invented a new family of numbers that are solutions to division by zero. I write them as _circled_ numbers.
For example: 1/0 = ①, 2/0 = ②, and in general, a/0 = ⓐ.
I even convinced my engineering colleagues at work that such numbers exist. It was easy-I had already blown their minds by teaching them all those other number families I mentioned earlier, plus *surreal, split-complex, quaternion, octonion* etc. So they were ready to believe in anything! 🤣
@@Rudxain Solutions to x!=0, or more generally, Γ(z)=0, is *_exactly_* the kind of problem I'd like to see Dr. Peyam solve using a hitherto-unknown family of numbers!!! 👍
Dr. Peyam's usual trick of using complex numbers to solve seemingly impossible equations _won't work here!_ ☹ After all, it's well-known that there's no solution even among complex numbers.
What else ya got, doc? 🤔
I thought a lot about this. I didn't make much progress. So, yeah, I think it's a very interesting problem!
@@nHans All of that is true. The circled numbers seem like a good idea to preserve numerical data. I also tought of something similar for Infinity, like ♾ = ♾ but ♾+1 != ♾, this would preserve a lot of algebraic properties at the cost of making each Infinity unique. This also meant that sqrt(♾)^2 = ♾ and sqrt(♾^2) = +-♾ (because sqrt has 2 solutions)
I also thought about how negative numbers are actually positive numbers multiplied by -1, and we could call -1 the "inverted unit" and give it a special symbol like "@", so -5 in this algebraic system would be written as 5@ and read as "5 inverse units". @ is the solution to the equations x + 1 = 0, x^2 = 1 && x != 1 (this one is similar to epsilon, the 2nd solution to sqrt(0), which is x^2 = 0 && x != 0)
This algebraic way of representing negatives seems useless, but it actually makes it easier to understand some of the properties of negative numbers in a different perspective. It also made me realize that @@ = 1 is *actually an axiom,* not a theorem, so we could have a different kind of arithmetic where "negative times negative is negative" instead of positive, but nobody uses that axiom because it's more useful for @ to act as a sign flipper instead of a black hole
@@nHans (replying to the factorial thing) exactly! But it really depends on which factorial extension we're using as "base". We could try using the Pi function, spline interpolation, analytic continuation, etc.
But since 0 is an integer, we could use a definition that only works in the domain of ints. There's a factorial extension that returns (-1)^n * n! for negative int values of n. And there's another definition that's based in terms of divison, so n! = -1/n! (again, for negative n). Under that definition we can calculate the limit, and the solution to x! = 0 becomes -♾
can you take x/y + y/x = 1 as x² + y² = xy, then take x² - xy + y² = 0, then under the condition that x ≠ -y multiply both sides by x + y to get (x + y)(x² - xy + y²) = 0 or x³ + y³ = 0?
thus you'd get, as x ≠ -y, x = ³√y³ with either one of the two complex branches of the cube root function?
Watch the video and find out
@@drpeyam I did, but you went in a different direction and I'm not profecient enough with the complex roots of a number to tell whether my solution set is equivalent to yours.
@@drpeyam hmm I went back and referred my old high-school mah textbook, and I think my answer set is equivalent, if you take x = (ω/ω²)y where ω and ω² are the complex cube roots of unity
Just one small comment, the operation on irreducible fractions where you add the numerators and the denominators to find a new fraction actually has some interesting mathematical properties, and it's called the mediant.
Interesting!
Can you find a general solution, solving for any integer?
Nice! I had been expecting a video about Farey addition (and, possibly, Ford circles), which obviously gives x/y ⊕ y/x = (x+y)/(y+x) = 1. It's always fun to be surprised.
Ooooh I like that too!!
@@drpeyam I would tend to Agree with this, as that equation then becomes X + Y = X + Y, and that equation is a tautology.
Perhaps we should teach kids complex mathematics from and early age. Up to the end of secondary school 15-16 years old, we were being taught that the solution for a quadratic doesn't exist if (bb-4ac)
Seeing how many people struggle with negative numbers, I don't think it would work.
I totally agree with this!
Gosh, I remember in elementary school (I think it was 1st grade) I asked my teacher about a simple addition/subtraction problem in which the answers were negative, and I kid you not she told me that we can't have negative numbers and that we can't solve equations like this because "we can't have negative numbers..." I responded by pointing to the thermometer on the wall which clearly showed it was 29-30˚F today (i.e. -1˚C), which she justified as an exception and that I didn't need to worry about it. Lol
I don’t know if you need to teach complex mathematics, but you can hint at it. For example, they are going to know about whole numbers and fractions. So you can do something like, can you divide 5 by 3? No, if the answer has to be a whole number, but yes if it can be a fraction.
Can you find the square root of -3? No, if it has to be a real number, but there are other numbers where the answer is yes. In fact, there are lots of other types of interesting numbers that the students would want to learn about in the future.
If you try to teach complex numbers really early, then where do you stop? Complex, quaternions, matrixes, infinities, etc. But I think it is bad to say there is no solution. Better to say, there is no solution with the numbers that we have studied so far. Sad thing is, some that teach ages up to 15 or so, may not even know about complex numbers themselves.
No, I'm afraid I disagree about teaching complex numbers to kids. In fact, a lot of well-intentioned people-like yourself-often suggest that some particular topic or another should be taught to all schoolchildren at an early age, because they found it very interesting or useful for themselves. Not just math topics, but also subjects like philosophy, cursive writing, farming, quantum mechanics, foreign languages, touch-typing, ethics, music etc.
There are several reasons why I disagree. Let me discuss the two most important ones: Time and Relevance. (The others are Pedagogy and Teaching Resources.)
It's not like schoolchildren are idling on standby with nothing to do. Rather, their curricula and schedules are already overcrowded. They don't have enough time even to absorb what they're being taught, let alone play and just enjoy being children.
Every new topic that you want to add has to take time away from something else-eating, playing, sleeping, or a subject that you think is less important than complex numbers.
And sure, that's a conversation that parents and educators should frequently be having. After all, as technology progresses, we need to keep adapting to the new stuff and eliminating the irrelevant and obsolete.
What about relevance? Fact is, most adults don't use complex numbers in their daily lives. So I'd teach it only to those who will (eventually) need it-aspiring mathematicians, scientists, engineers etc. For most other professions-including management, accountants, lawyers, performing arts, humanities etc.-it's unnecessary. I'd rather teach students something useful to their daily lives. Or directly relevant to their future career. Or even give them back their free time so that they can be creative.
In short, I wouldn't teach complex numbers to schoolchildren that early, nor to everyone.
I agree about one thing though-you shouldn't tell children something is impossible when, in fact, more powerful tools _can_ solve the problem. When I come to such roadblocks while teaching, I'm very careful to explain that the tools _we currently have_ cannot solve this particular problem. At some point in the future, we'll learn how to use more powerful tools, but for now, leave them alone.
For all you know, there could even be a family of numbers-which we haven't been taught yet-that can divide by zero!
@@nHans L’Hôpital’s Rule for divide by zero.
Sir please make a video on explaining the PQ formula for quadratic equation.
چه روش جالبی، ممنون دکتر!
Little suggestion: what if you dig into the rabbit hole of multi variables? Maybe with limits, gradients divergence and curl (or rotor as we italians call it)
There’s a whole playlist
@@drpeyam sry, didn't notice
What a strange coincidence he uses variables instead of the actual numbers that he chose.
Cool video as always
Etudes in x/y+y/x. Amazing 😻
I like that title!!
Nice shirt! Where'd you get it?
It‘s for sale on my Teespring store 😁
can you please make a video on the GAMMA FUCNTION and (1/2)! is sqrt(pi) / 2 ??
Already done
Isn't this functionally the same as y = X^1 + X^-1 (in this case solving for when y=1), which I believe the graph of which has a slant asymptote at y=X and a vertical asymptote at X=0.
This would imply that there are no real solutions when y=0 and when y=1.
just amazing as always
Always so cool and interesting!😊
I once had to prove that x/y+y/x >=2 for all real x,y. Thanks for the reminder! :)
This is certainly true if x and y are both positive or both negative. If they have opposite signs, then x/y + y/x
Math is cool! Thanks Dr. Peyam!
So no real solutions
Very nice thank you 😊
1:53 when a maths professor writes obfiscated notation
If we add in the same way as we multiply it will be correct answer but this addition would be wrong
Its amazing 👏. Nothing is impossible again you have proved
In the real numbers, no. For any positive real x, either x > 1, which means we need 1/x to be negative (impossible), or x < 1, which means that 1/x > 1, therefore 1/x + x > 1 as well. For negative x, we'll be adding two negative numbers, which will never be 1.
Now, in the complex numbers, we have i + 1/i = 0, which is new (not what was asked, but also previously impossible). If you look at an equilateral triangle (with sidelength 1), that actually exposes one possible answer. e^(i ⋅ 60°) + e^(i ⋅ -60°) = 1/2 + i ⋅ root(2, 3)/2 + 1/2 - i ⋅ root(2, 3)/2 = 1
To cover all solutions though, z + 1/z = 1 => z² - z + 1 = 0 => z ∈ 1/2 ± root(2, 1/4 - 1) .... which will just give the original solution, plus its conjugate/inverse.
ln(3/5)+ln(5/3)=ln1, mabye ln has the same color as the paper!!😀
LOL
@@drpeyam of course ln(1) is zero, so ln(0.6) is also -ln(5/3).
If you put x/y+y/x=2 it doesn't show the graph in geogebra, why?
Same problem in DESMOS, I noticed. I ought to be a line. Try 2.1 instead of 2, and then 2.01 etc However, when you "clear" the fraction in DESMOS and rewrite the equation, you obtain the desired line. Not sure why
It can be followed by the ancient Formula to calculate Quadratics. It was used by Babylonians and is older than quadratic formula
This produces the reduced quadratic equation which is equivalent to quadratic formula i.e., -P/2 ±(√P/2)²-q
Request: Please make a video on explaining the PQ formula for quadratic equation.
Check out my most popular video
@@drpeyam thanks
.
the first part gives you two solutions right away: notice you have x=y*( (1+-sqrt(3)i)/2), just notice that x/y = (1+-sqrt(3)i) / (2) and so y = 2 and x = 1+sqrt(3)i or 1-sqrt(3)i
Thank you.
When I looked at the first equation, I assumed that the answer was in real numbers, not imaginary numbers.
multiply by xy: x^2+y^2=xy
notice: (x-y)^2 = x^2+y^2-2xy
notice: x^2+y^2=xy+xy+(x-y)^2
since both are true, we can combine them to get: (x-y)^2 = -xy.
if we assume these are real numbers, we know -xy must be positive, except in the case x=y=0, which is not allowed by the original equation. Returning to the first step, knowing xy must be negative: x^2+y^2 = xy < 0, which is impossible in the real numbers. Therefore the solution must be complex, and i'm too lazy to continue from that.
Very cool, thank you.
I may be catching on to the trick behind Dr. Peyam's "Impossible" videos ... they're impossible with _real_ numbers, but solvable with _complex_ numbers. Right? Right???
Anyway, my real question is: Are there equations that cannot be solved with _complex_ numbers, but solvable with other, even more exotic types, such as quaternions, octonions etc.? I've never used-or even formally studied-the latter types; just read them up out of interest.
I’m trying to work on a couple of them, but I have a nice quarternion equation video that I published
that's a super question @Niranjan Hanasoge
So at 0, the values are at right angles and at 2, the coincide. This implies that between 0 and 2 there is a continuum or angles between 0 and 90 degrees. Cool.
I feel violated. U click baited me. But I learnt something today
Another mind-blowing question 😁
Complex numbers are unbelievable!
You have just taken a simple mathematical problem, which was clearly incorrect, and turned it into a gymnastic event.
Ooo - now do the math student’s favorite fraction simplification: 1/x+1/y=1/(x+y)
Already done ✅
Yes - those students were never actually wrong, they just love equilateral triangles :-)
X quadrado? Where are u from? -_-
Thanks for trying Spanish for your lecture. 👍😀
You had me sweating with the thumbnail.
Looks better in polar.
Cute. How about generalizing for =a where a is any number?
Well, then the problem gets way more complicated.
@@arturcostasteiner9735 Not really. It's still the same quadratic equation: x² - axy + y² = 0, and has the same quadratic solution for _x_ in terms of _a_ and _y._ As before, the solutions will be complex. In some cases, they'll be real. (Complex numbers include real numbers.) Note that _a_ and _y_ can also be complex-the quadratic formula will still work.
Of course, if _a_ is some other type of number-quaternion, octonion etc.-then yes, it does get more complicated, because the high-school quadratic formula no longer holds.
0:25 [EN] [EN] [EN]... "dos x son x cuadrado... " [EN] [EN] [EN]... 😁
No the equasion results in 34 over 15
Entendi nada mas achei bonito o desenvolvimento.
Dr Peyam's impossible mission!🤣
Wtf you have to find the LCD making it 2 4/15 or 5/3 = 1.66… and 3/5=.60 and Therefore… is equal to 2.26….
Why is your "y" so similar to 4
i got this result
x/3=3/x
3+3i√3/2 or 3−3i√3/2
Hmm, first x/y + y/x = 1, then x/y + y/x = 0, followed by x/y + y/x = 2.
Somehow i expected x/y + y/x = c next... .
Edit: I couldn't resist, hopefully i didn't made any error:
==> x = 0.5 (c +/- sqrt(c^2 - 4)) y
(interesting) case -2 < c < 2:
r cos(a) := 0.5c and r sin(c) := +/- sqrt(4 - c^2); r >= 0
r^2 = (r cos(a))^2 + (r sin(a))^2 = (0.5 c)^2 + (+/- 0.5 sqrt(4-c^2))^2 = 0.25 c^2 + 0.25 (4-c^2) = 1 r = 1
==> cos(a) := 0.5c and sin(c) := +/- sqrt(4-c^2)
==> a = atan(sin(a)/cos(a))
= atan(+/- sqrt(4 - c^2)/(0.5c))
= +/- atan(sqrt(16/c^2 - 4))
Corrected "c^2 - 4" to "4 - c^2"
Can't say for certain, but for c=1 your last formula is atan(sqrt(-12)), I'm sure the result can be written in a better form.
I'm getting a=arccos(c/2) "and that's a good place to stop".
@@Milan_Openfeint Argh, within the sqrt expression i somehow used "c^2 - 4" instead of "4 - c^2":
Need sqrt(c^2-4) = i sqrt(4-c^2) to get a real value within the sqrt for -2 < c < 2... corrected the above.
(Funnily i used the correct expression in the "r^2 =..." line, but nowhere else... .
So for c=1 we should get +/-atan(sqrt(12)), though that seems to be false, maybe i shouldn't have done many parts in my head i assume.
My head just exploded...
Me thinking we have to solve this over integers...
equis cuadrado
I was inspired by you
r=x/y gives r+1/r = 1....not too hard.
Yes. And this leads to r = exp(pi/3) or r = exp(-pi/3).
Wrong.
5/3+3/5=34/15.
T-shirt, "Körper" -- Body goes here?
Now i am getting dissy 😂
cool
Does 5/3 plus 3/5 equal 1??? NO!! Lol
No. 2 1/15 is correct.
Without watching the video.
?
5/3 + 3/5 = 2 1/15
5/3 = 1⅔ = 1 10/15
3/5 = 9/15
9/15 + 1 10/15 = 1 19/15 = 2 4/15
So i was mistaken. 2 and 4/15 would be correct.
X=1 but Y=2
?
👍
No real solution.
No it doesn't.
Haha, "Körper"....
hahaha you rule
Wow what a nice clickbait
Nice Spanish…
x/y + y/x >= 2sqrt(x/yy/x)=2 by am-gm. x,y > 0. So no real positive solution for thumbnail x/y + y/x = 1.
I say y = αx solves x/y+y/x = β (where β∈ℝ*),
then α∈ℝ iff |β| ⋝ 2.
Conversely, α∈ℂ, iff |β| < 2.
I just don't like ℂ where all those pesky ℿmi's popping up everywhere! シ