what fractions dream of

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[@usptact]

Dreams turn out to be complex. Who would’ve thought.

[@TerraBlo]

Well, if it's a fractions "dream" then imaginary numbers should be allowed

[@Retroist2024]

Imagine there is no heaven

[@DarinBrownSJDCMath]

There's a nice geometric interpretation of the complex solutions. The two numbers differ by a rotation of 120 degrees, and the equation says if you invert each of them separately (i.e. reflect about the real axis and "reflect" about the unit circle by inverting the modulus), and then add, you get the same result as if you had added them first and then inverted the sum (by the same two "reflections").

[@DeJay7]

That's really beautiful, seeing how that complex number that we derived y/x to be is not just one of the infinitely many complex numbers, but one with a real meaning.

[@MenloMarseilles]

After seeing it was a complex number, I suspected there was some interesting geometrical property implied by the equation. Thanks for providing it!

[@JarkkoHietaniemi]

Wouldn't 1/x+1/y = 2/(x+y) be more "natural" dream?

[@monishrules6580]

Sodium hydride

[@klembokable]

Anything but a one in the numerator is cringe

[@JarkkoHietaniemi]

@@klembokable 355/113 begs to differ.

[@AverageCommentor]

Yeah, and the solution is much nicer too. x = ±iy y = ±ix

[@reminderIknows]

@@klembokable22/7 also begs to differ

[@bsmith6276]

I did this a bit more direct. First I will note that if x=y then 1/x+1/x = 1/(2x), which makes (3/2)/x = 0, which has no solutions. Then x!=y. Multiply through by the common denominator to get y^2+xy+x^2+xy=xy, which simplifies to y^2+xy+x^2=0. Now multiply each side by x-y, which is nonzero, to get x^3-y^3=0, or x^3=y^3. Since x and y are unequal then there are no real solutions, but we can have a complex solution if the ratio of x/y is one of the complex cube roots of unity.

[@burk314]

I like this method, and it does extend to other algebraic systems. You're just looking for solutions to a^3=1 in whatever system you're working in. You should probably specify that you're looking for cube roots of unity except for 1, since that was introduced when you multiplied by x-y and you've already shown it does not lead to a solution.

[@aditya-u7s9v]

an easier way to finish is to multiply the final equation by 2 and notice that that simplifies to x^2 + y^2 + (x+y)^2 =0 which has no real solutions apart from (0,0)

[@Simpson17866]

That is beautiful :)

[@marcushendriksen8415]

There's an even more direct way once you get to x^2+xy+y^2=0: the quadratic formula. Using it to solve for x, you get x=(-y±y*sqrt{-3})/2, or x=y(-1±3i)/2. But this can only be real if y is either 0 (not allowed, since 1/y is defined) or if it has a nonzero imaginary component; i.e. x and y cannot both be purely real numbers

[@konraddapper7764]

Your solution also directly gives you an answer for the 2x2 matrix case: just take x=c*R (phi) and y=c*R(phi+-2/3*pi), with R being the 2 D rotation matrix. However, I am not sure if that is all the solution

[@jasekochhar8336]

I would like to say that you have come a long way over the course of your channel in explaining well the topics in your videos. I recently watched almost all of your group theory videos from 4 years ago as I’m doing research in that area of math before I’ve even taken the course l, and I wanted to get a good foundation on the material. Those videos were still very, very good, but now watching this I can see that you’ve gotten much clearer and thorough and more accessible to viewers with less general understanding of what you’re teaching. Your videos are excellent and I look forward to watching a few more of your course playlists for upcoming courses!

[@PRIYANSH_SUTHAR]

Indeed he is phenomenal now 😊

[@crosseyedcat1183]

For matrices you'd have A^2 + A + I = [0]. So really you want A^2 + A = -I. You can then apply the Jordan Canonical Form to get (PJP^-1)^2 + PJP^-1 = -I which you can expand out to: (PJP^-1)PJP^-1 + PJP^-1 = -I. The P and P^-1 cancel on the matrix square so you get PJ^2P^-1 + PJP^-1 = -I. You can then factor out P and P^-1 on both sides on the left to get: P(J^2 + J)P^-1 = -I. Finally you can multiply the P and P^-1 across to get: J^2 + J = P^-1 (-I) P = J^2 + J = -I. This is impossible in general since if J has any nilpotent parts, J^2 will have terms off diagonal (and so will J) they won't be on the diagonal; however, if A is diagonalizable, we'll just get n versions of the original a^2 + a + 1 = 0. In that case we'll see that A has to have complex eigenvalues. Over real matrices, complex eigenvalues come in conjugate pairs, so A^2 + A + I = [0] has no solution over the reals in odd dimensions, but has solutions in even dimensions. In general, this shouldn't be that surprising since the solution originally requires complex numbers to exist and we know that C is homeomorphic to the space of 2x2 orthogonal matrices. In even dimensions, you just get duplicates of the complex solution for each 2x2 block. Another way to think about it is to use the Caley-Hamilton theorem. In that case, A^2 + A + I = [0] will be satisfied by any matrix with characteristic equation x^2 + x + 1 = 0 (which is 2x2). I conjecture that in even dimensions above 2, we're seeing powers of this characteristic equation but I haven't really checked.

[@PetraKann]

Also x+y cannot equal zero. In other words x cannot be equal -y where y happens to be equal to x (or the other way around).

[@KeimoKissa]

I'm not sure what you're getting at with "where y happens to be equal to x". It seems to contradict the first part of your comment

[@zecaaabrao3634]

In this situation this implies x or y are 0, so it isn't a new restriction

[@mikey-hm7dt]

@@zecaaabrao3634it does not imply that, eg x=1 y=-1 is the kind thing he’s talking about. But this is not a relevant restriction because we are able to get the much stronger restriction obtained in the video without it

[@gabef9538]

Imagine x = -y = 0. Games at infinity if we could ever reach there.

[@PetraKann]

@@KeimoKissa an example would be x=1 and y= -1 so x+y=0

[@mathboy8188]

Alternate demonstration it's impossible over the reals: Assume there exists real numbers x and y such that 1/x + 1/y = 1/(x+y). 1. Neither x nor y can be 0 (else division by 0), so xy not 0. 2. Clear fractions, i.e. multiply both sides by (x)(y)(x+y), to get: y(x+y) + x(x+y) = xy, which becomes xy + y^2 + x^2 + xy = xy, so x^2 + y^2 = - xy. 3. Add 2xy to both sides, and separately subtract 2xy from both sides, to conclude: (x + y)^2 = xy (x - y)^2 = - 3xy 4. By #3, both xy and -3(xy) are non-negative. By #1 xy isn't 0. Therefore both xy and -3(xy) are positive. 5. Both xy and -3(xy) are positive is impossible, since xy > 0 implies -3xy < 0. This contradiction proves the equation has no real solution for x and y.

[@void7366]

Thank you Michael

[@designingmathematics]

Correct me if i'm wrong but i did this: 1/x + 1/y = 1/(x+y) x,y≠0 i) If x,y are both positive, then 1/(x+y)0 1/x + 1/y >1/x>1/(x+y) and therefore a contradiction ii) If x is positive and y is negative(or vice-versa) we can write y=-k where k is positive then doing some sinplification we have (x-k)²=-xk but (x-k)²≥0 and since x and k can't be 0 we reach a contradiction iii) If x,y are both negative we can write x=-r and y=-k, where k and r are positive, therefore 1/-r +1/-k = 1/(-k-r), multiplying by ×(-1) in both sides we go back to i) and therefore contradiction No solution for x,y real numbers Q.E.D

[@mMaximus56789]

Would have been interesting considering non commutative numbers, like the quaternions

[@peterhemmings2929]

Curious that the calculated scaling factor a is a unit complex number, which simply rotates x by 120 or 240 degrees to get y.

[@RexxSchneider]

If you check the thread posted by @bsmith6286, you'll see that x and y have to satisfy x^3 = y^3 and x ≠ y. That can only happen when a = y/x is a cube root of unity, but not 1.

[@ChristopherBitti]

1/x + 1/y = 1/(x+y) => y/(xy) + x/(xy) = 1/(x+y) => (y + x)/(xy) = 1/(y + x) => (x + y)^2 = xy => x^2 + y^2 + 2xy = xy => x^2 + y^2 + xy = 0 Over the nonzero real numbers, it is quite intuitive that this can not hold. This is because max(x^2, y^2) >= |xy| >= xy and so even if xy happened to be negative, it wouldn't detract enough from x^2 + y^2 to make the sum not positive. In a general field where 2 does not equal 0, we can apply the quadratic formula to obtain: x = [-y +- sqrt(y^2 - 4y^2)]/2 = [-y +- y * sqrt(-3)]/2 Thus, solutions exist and take this form only when the field contains a square root of -3. It is clear then why there are no real or rational roots, but there are complex roots. In the case of Z_p where p is an odd prime (2 = 0 if p = 2 so we can't use the quadratic formula here), the solutions exist in one of two cases by the properties of the Legendre symbol and quadratic reciprocity: 1) -1 and 3 are both individually squares in the field This occurs when p = 1 (mod 4) AND p = 0 or p = 1 (mod 3) 2) -1 and 3 are both individually not squares in the field This occurs when p = 3 (mod 4) AND p = 0 or p = 1 (mod 3) Adding up these cases, we need p = 3 or p = 1 (mod 12) or p = 7 (mod 12) So for example, if p = 7, as demonstrated in the video, roots exist. In the case of 39, as shown in the video, it's not quite so simple, since 39 is not prime. But it is a product of primes, so we can work with it quite easily. Since -3 is a square in Z_3 and Z_13 by the above analysis, we can conclude it is also a square in Z_39 by the Chinese Remainder Theorem.

[@fsponj]

1/0 +1/0 = Error 1/(0+0) = Error 1/0 + 1/0 =1/(0+0) Q.E.D.

[@legendgames128]

Object.is(1/NaN + 1/NaN, NaN) = true Object.is(1/(NaN + NaN), NaN) = true Object.is(1/NaN + 1/NaN, 1/(NaN + NaN)) = true Q.E.D.

[@piedepew]

​@@legendgames128 nan in python?

[@mathboy8188]

( Notation: * = multiplicative inverse in a ring. ) *Proposition:* Let R be a non-trivial commutative ring, and consider the equation x* + y* = (x + y)* for some x, y in R. 1) -3 is a quadratic residue in R is a necessary condition for a solution to exist in R. 2) If R = F, a field of characteristic not in { 2, 3 }, then -3 is a quadratic residue in F is a necessary & sufficient condition for a solution to exist in F. 3) If R = F, a field of characteristic 3, then always have the solution x = y = 1. *proof:* *1)* If a solution exists, then x* + y* = (x+y)* for some x,y in R. The algebra is to "clear fractions" by multiplying both sides by (x)(y)(x+y), to get x^2 + y^2 = - xy, and then both add & subtract 2xy from both sides to get: (x + y)^2 = xy and (x - y)^2 = - 3xy. Thus (x - y)^2 = - 3 (x + y)^2, and since (x + y)* exists (from the initial equation), get [ (x - y) (x + y)* ]^2 = - 3. Therefore -3 is a quadratic residue in R. *2)* By #1, suffices to show that a solution exists when -3 is a quadratic residue in F. Let q in F s.t. q^2 = -3. Will show that x = q + 3 and y = q - 3, are solutions to the equation. Obviously x + y = 2q. First will show that x(y)(x + y) is not 0, which will give that x* and y* and (x + y)* exist. x(y)(x + y) = (q + 3)(q - 3)(2q) = 2 q (q^2 - 9) = 2 q ( (-3) - 9) = -24 q = (24)(-q). Since char(F) not 2 and char(F) not 3, have that 24 = (2)(2)(2)(3) is not 0. Since (-q)(-q) = q^2 = -3, and -3 is not 0 (because char(F) not 3), have that -q is not 0. Therefore x(y)(x + y) = (24)(-q) is not 0, and so x* and y* and (x + y)* exist. Now, from q^2 = -3 get that 3q^2 = -9, so 4q^2 = q^2 - 9, and so 2q(q - 3) + 2q(q + 3) = (q - 3)(q + 3). Multiply both sides by (q + 3)* (q - 3)* (2q)* (which is the product x* y* (x + y)* ) to get: (q + 3)* + (q - 3)* = (2q)* , hence (q + 3)* + (q - 3)* = ( (q + 3) + (q - 3) )* , hence x* + y* = (x + y)* hence the equation has a solution in F. 3) If char(F) = 3, have 2(2) = 4 = 1 + 3 = 1, so 2* = 2. Thus (1 + 1)* = 2* = 2 = 1 + 1 = 1* + 1*. *Applying to this to F = Z/(7Z) as per Dr Penn's example:* 2^2 = 4 = 7 - 3 = -3, so -3 is a quadratic residue, and char(F) = 7, so not in { 2, 3 }, so the equation must have a solution. Get all the multiplicative inverses here: 1(1) = 1, 2(4) = 8 = 1, 3(5) = 15 = 1, 6(6) = 36 = 1. From the proof, q = 2, so x = q + 3 = 2 + 3 = 5 and y = q - 3 = 2 - 3 = 1 = 6. Thus x = 5, y = 6 is another example in F. Check: 5* + 6* = 3 + 6 = 9 = 2 = 4* = 11* = (5 + 6)* Also, 5^2 = (-2)^2 = 2^2 = -3, so can also use q = 5, giving x = q + 3 = 5 + 3 = 8 = 1 and y = q - 3 = 5 - 3 = 2. Thus x = 1, y = 2 is another example in F. Check: 1* + 2* = 1 + 4 = 5 = 3* = (1 + 2)* .

[@charlottedarroch]

Nice work! I think you can show something a little more general. Suppose you have a commutative ring such that -3 is a quadratic residue and such that both 2 and 3 are units. Then there exists units x,y, such that x+y is also a unit and such that 1/x+1/y = 1/(x+y). Again as you had before, take q in the ring with q^2 = -3, let x = q+3, y = q-3 and as before x+y = 2q. Then 1/x = -q/12+1/4, 1/y = -q/12-1/4 and 1/(x+y) = -q/6. So indeed, as long as 2 and 3 are units, 1/12, 1/4 and 1/6 all exist, so 1/x, 1/y and 1/(x+y) all exist. This allows us to include lots of rings which aren't fields. For example, letting Z_p be the ring of p-adic integers, then 1/x+1/y = 1/(x+y) has a solution in Z_p for all primes p congruent to 1 mod 3, as in such rings -3 is a quadratic residue (can be seen from quadratic reciprocity and Hensel lifting) and both 2 and 3 are units (again by Hensel lifting).

[@mathboy8188]

@@charlottedarroch Nice indeed! ( q + 3 )( -12* q + 4* ) = -12* q^2 + [ 4* - 3(12*) ] q + 3( 4* ) = - ( 3* )( 4* ) (-3) + [ 4* - 3( 3* )( 4* ) ] q + 3( 4* ) = 4* + ( 4* - 4* ) q + 3( 4* ) = ( 0 ) q + 4( 4* ) = 1. ( q - 3 )( -12* q - 4* ) = -12* q^2 + [ - 4* + 3(12*) ] q + 3( 4* ) = - ( 3* )( 4* ) (-3) + [ - 4* + 3( 3* )( 4* ) ] q + 3( 4* ) = 4* + ( - 4* + 4* ) q + 3( 4* ) = ( 0 ) q + 4( 4* ) = 1. x* + y* = ( -12* q + 4* ) + ( -12* q - 4* ) = - 2 ( 12* ) q = - 2 ( 2* )( 6* ) q = - 6* q. (x + y)* = (2q)* = 2* q* = 2* q* [ q^2 (-3)* ] = - 2* 3* ( q q* ) q = - 6* q. ~~~~~~~~~~~~~~~~~~~~~~~~ I assume you found the inverses this way: 1 = ( q + 3 )( a q + b ) = a q^2 + (3a + b) q + (3b) = -3a + (3a + b) q + (3b) = (3a + b) q + 3(b - a). Then b = -3a, so 1 = 3(-4a), so a = -12* and b = 4*. It's an obvious trick in hindsight (ain't it always)... one I'll remember. When I was doing this, I didn't see an immediate way to check for inverses in a general ring, so just jumped into the easier field case. As for p-adic numbers and Hansel Lifting, I've never had cause to learn about either of them. From what I just read on Wikipedia about them, your comment makes sense on its face, although, again, I don't actually know the topic. Your comment taught me a few new things - thanks!

[@TastySalamanders]

I already suspected there would be a modulo answer since Numberphile recently did a video on the "Freshman's dream". However what I never realised until now that when doing arithmetic under a modulo operation was that non-integers can be represented multiple ways under it. Using the example in the video about 1/3 = 5 under Mod 7: 5 * 6 = 1/3 * 6 30 = 2 30 mod 7 = 2 I experimented a bit and made a realisation after finding that 6.5 under mod 12 = 0.5 6.5 is 13*0.5 and of course 13 mod 12 = 1. So 13*0.5 = 1*0.5 so of course 6.5 is also a half under mod 12. And same applies in the example used in the video. 5 is 15*(1/3). 15 mod 7 = 1. So again 15*(1/3) = 1*(1/3). So of course 5 is 1/3 under mod 7.

[@patrickhickey7673]

Fun and simple explanation, and a nice “homework” exercise 👍 👍

[@ArnoldOverwater]

Although it doesn't work with the integers, it should work with the Eisenstein integers, since they have a (-1+i*sqrt(3))/2 built in.

[@monishrules6580]

Umm yeah true

[@tiripoulain]

If you take any ring R, and you form S = R[a]/(a^2 + a + 1), then it'll work in S. The Eisenstein integers are isomorphic to this ring S where R = Z.

[@yoav613]

For the matrix: A=((1/sqrt3,1),(0,1/sqrt3)) and B=((1/sqrt3,0),(1,1/sqrt3)).

[@tition1]

If you convert 1 and (sqrt(-3)-1)/2 to 4x4 rational matrices you get:: 1/2[[-1,0,0-3],[0,-1,3,0],[0,-1,-1,0],[1,0,0,-1]] and the 4x4 identity matrix. Should be possible to do with a 2x2 rational matrix [Edit: A=1/2[[-1,-3],[1,-1]] and B= id]. For an example that's "brand new" compared to your exposition, you'd have to find 2x2 rational matrices that *do not commute* - if they commute, then your entire discussion holds and pick a 2x2 matrix with characteristic polynomial a^2+a+1=0

[@tition1]

For the 2x2 matrices, this should be A=1/2[[-1,-3],[1,-1]] and B= id. Here, A = 1/2[[-1,0],[0,-1]] + 1/2[[0,-3],[1,0]], where the first summand represents -1/2 and the second summand represents sqrt(-3)/2.

[@tition1]

Here we go, a 2x2 rational matrix solution with A and B not commuting, so this is not covered by the exposition here: A= [[1,1],[0,-1]], B= [[-1,0],[1,1]]. Notice how A B eq B A. Both A and B have characteristic poly q^2-1=0.

[@lemmonlemon]

Shouldn't there be three domain restrictions, x, y, and x+y not being zero?

[@MarceloResegue]

As someone commented here before, 'a' is just a cubic root of unity (because if a³=1 and a=/=1, that makes a²+a+1=a(a²+a+1) and therefore a²+a+1=0). If we are in a ring that has a cubic root of 1, we can find elements that satisfy the "dream equation". So, for the 2x2 matrices, take 'A' as the 120° rotation matrix, ie, A={{-1/2, -√3/2},{√3/2, -1/2}} and keep your matrices invertible. Testing... X={{2,0},{0,1}} ; Y=AX={{-1, -√3/2},{√3, -1/2}} 1/x+1/y -> inverse(X)={{1/2, 0}, {0, 1}} ; inverse(Y)={{-1/4, √3/4},{-√3/2, -1/2}} -> inverse(X)+inverse(Y)={{1/4, √3/4},{-√3/2, 1/2}} 1/(x+y) -> inverse(X+Y)=inverse({{1, -√3/2},{√3, 1/2}}={{1/4, √3/4},{-√3/2, 1/2}} It checks! (The matrix notation is the one used by WolframAlpha because I'm too lazy to invert even 2x2 matrices by hand.)

[@bartekabuz855]

What we need is a field with sqrt(-3). In Z_p we have sqrt(-3) iff p = 1 mod 3

[@cosimobaldi03]

Cool

[@RonWolfHowl]

How did you obtain that result?

[@RonWolfHowl]

Also, that implies that Michael is wrong to state that there is a solution in Z_39, as 39 !== 1 (mod 3).

[@bartekabuz855]

@@RonWolfHowl en.wikipedia.org/wiki/Legendre_symbol

[@glowstonelovepad9294]

1/x + 1/y = 1/(x+y) y/xy + x/xy = 1/(x+y) (x+y)/xy = 1/(x+y) (x+y)^2 = xy x^2 + 2xy + y^2 = xy x^2 + xy + y^2 = 0 (This is similar to x^2 + xy = y^2 where y/x is the golden ratio, but in this case it's x^2 + xy = -y^2.) x^2/xy + 1 + y^2/xy = 0 x/y + 1 + y/x = 0 x/y + 1 = -(y/x) Stuf

[@mihaigabrielbabutia4595]

A quick manupulation of the fractions and a small trick also leads us to the conclusion, as shown below: (x+y) / xy = 1/(x+y) (x+y)^2=xy x^2 + xy + y^2 = 0 / *2 x^2 + y^2 + (x+y)^2 = 0 Since x,y are real numbers, and we have a sum of squares, it is necessary that all terms are equal to zero hence x=y=0 which cannot be true, since they are denominators. Thus we have no real solutions.

[@Oberatous-Udurabas]

My favorite one is (x-y)*(y-x)=(x/y)+(y/x)

[@evansaschow]

Isn’t this the n=-1 case of the extension of The Freshman’s Dream?

[@JamesLaFleur]

Also interesting ist the Farey Addition!

[@geertb4341]

I'm probably overlooking something, but for the matrix case, why would the matrixes be a (scalar) multiple of each other? Or did you mean a 'matrix multiple'?

[@cosmosapien597]

The dream: X and Y strolled through the lush meadow. X's heart skipped a beat. "You're the one, my love!" X exclaimed, hoping to convey the depth of their emotions. Y's eyes sparkled with a sudden revelation. "Ah, X, I see what you mean. Our connection is like the equation we've been pondering:- 1/x + 1/y = 1/(x+y) When I'm 1, your values become the cube roots of unity!" X's mind was amazed. "You mean... our love is like a fundamental building block of the universe?" Y nodded. "Exactly! Just as the cube roots of unity - (-1 ± i√3) / 2 - form a perfect equilateral triangle in the complex plane, our bond forms a perfect harmony in the fabric of reality." X's heart swelled with emotion. "Y, you're the mathematician of my heart. I never knew our love could be expressed so beautifully." Y smiled, and together they gazed out at the meadow, their love resonating with the underlying structure of the universe, a perfect union of heart and mind.

[@mathmachine4266]

Multiply both sides by x, y, and x+y y(x+y)+x(x+y)=xy (x+y)²=xy x²+2xy+y²=xy x²+xy+y²=0 Whatever the value of y, if we multiply x by some k, y would also multiply by k. So, we can say x is 1, and then solve for the case when it isn't 1 by multiplying by k. y²+y+1=0 (-1±√(1-4))/2 y=(-1±√(3)i)/2 So, whatever x is, y is just x(-1±√(3)i)/2. So, no matter what, unless x and y are both 0 (which they're not), at least one of the two must be complex. Which is fine by me. Although now that I look at 0:00, I can see that x and y have to be real, so...wap wap

[@davidseed2939]

2:29 at this point just multiply the equation by x 1+1/a =1/(1+a) multiply by a(1+a) a+a² +(1+a) =a a²+a+1=0 2a=-1±√(1-4) a=(1±i√3)/2

[@cicik57]

Prove that solutions are impossible: having x²+xy+y² = 0; you have t²+t+1 =0 for both t = y/x AND x/y so x/y = y/x => x² = y². x = -y is not possible since x+y is below the fraction. then try x = y: 2/x = 1/2x is impossible.

[@renesperb]

You can rewrite the equation as (x +y)^2= x*y and then (x/y)^2+(x/y)+1 = 0 .Solve this quadratic equation to get z = x/y =1/2 (-1 + or - i √3 ). Hence you can choose y and take x = z *y , one has infinitely many complex solutions.

[@samdean1966]

This is okay but its a lot of mucking around: just rearranging gets hou to x²+xy+y²=0, but its easy to see (double it to make x²+(x+y)²+y²) that x²+xy+y²>0.

[@celestialTangle]

Notice that the solutions to this equation are at 120° angles around a circle on the complex plane!

[@papasalt8823]

Obligatory reference to Project Euler's Problem 100 (Diophantine Reciprocals 1) where you solve the equation 1/x + 1/y = 1/n where x, y, n are positive integers Solving the aforementioned problem would probably also answer this one too since you can simply set n = x+y.

[@Happy_Abe]

Would there be solutions in some abstract space where we don’t have that y=ax. For example in an arbitrary ring, where not every element are multiples of one another. Can we find examples there too? We can’t use this quadratic trick there though.

[@mathboy8188]

For a commutative ring R: i) -3 being a quadratic residue is a necessary condition for a solution to exist. ii) -3 being a quadratic residue, and 2 and 3 both being units in R, is a sufficient condition for a solution to exist. That's on my comment beginning _( Notation: * = multiplicative inverse in a ring. )_ _Proposition: Let R be a non-trivial commutative ring, and consider the equation x* + y* = (x + y)* for some x, y in R._ and the extension of it found in the reply comment there left by @charlottedarroch.

[@Happy_Abe]

@@mathboy8188 thanks

[@eveeeon341]

I'm coming from this with a really vague understanding of the mathematics around the Langland's program (that I'm trying to improve), but the whole "in which number systems is this function solvable in" really gives me Langland/L function vibes, am I correct in relating this?

[@WRSomsky]

Since you always had an overall scaling that factors out, and exchanging x & y doesn't really do anything, I'd say that there is really only one primitive solution in the complexes. You could call it {x,y} = { 1, (-1+iSQRT(3))/2 } or maximizing symmetry {x,y} = { (1+iSQRT(3))2, (1-iSQRT(3))/2 }

[@edwardlulofs444]

As soon as you went past the complex numbers, I immediately thought of the Clifford algebra.

[@hassanalihusseini1717]

I wish it was like 1/x + 1/y = 2/(x+y). That would be the real dream! 🙂

[@samwalko]

By symmetry of x and y, if z works as a value of a, then so does z^-1. In the second example, 2 and 4 are inverses, and are also the only two solutions.

[@sirlight-ljij]

What fractions dream of even half of Farey can achieve

[@bwwwwwww3425]

Looks a bit like a field of characteristic -1, doesn't it?

[@conrad5342]

So in complex numbers x and y need to have the same absolute value and differ in phase by pi*2/3 ..Z7 looks wild

[@thiagodonascimento7926]

11:55 I don't understand why 1/6 is equal to - 1 in Z7. Can someone explain what that means? I do understand though that 1/6 equals 6 in Z7.

[@darksecret965]

-1 × -1 = 1 6 × 6 = 1 6 = 6^-1 -1 = 6^-1

[@yt2979a]

X^2 + X + 1 is a primitive polynomial over Z2, so it can be used to construct GF(4), in which this polynomial is reducible (i.e. has 2 roots). So in GF(4) we also can find x,y satisfying the dream relation. Furthermore, any field GF(2^m) for m even contains GF(4), so the polynomial is reducible in those fields too, meaning they also have solutions for the dream relation.

[@yoyostutoring]

What does "GF(n)" mean?

[@jaxoncr]

@@yoyostutoringgalois field of order n. it is the finite field of order n, but n must be a power of a prime

[@zygoloid]

Hm, does this give a method for computing a table of multiplicative inverses mod p? If you cam find a primitive cube root of unity, then you can compute one inverse from another.

You can already do that without any of this machinery. Assume 2 is a generator of your multiplicative group. Find k=1/2. Then you can generate subsequent powers of 2 and subsequent powers of k as their inverses, by simply multiplying by either 2 or k, and eventually you'll cover the whole field.

[@charleyhoward4594]

mike is so smart !

[@Bodyknock]

Just fiddling around, if you assume x, y, and x+y are all not zero, then you can multiply both sides of the equation by their combined product and simplify to get: y² + x² + xy = 0 Which has kind of a nice symmetry to it. 🙂 If you then as in the video write y=ax you get a²x² + x² + ax² = 0 x²(a² + a + 1) = 0 Leading to the same solutions for a as in the video. The nice thing here, though, is if you are calculating y² + x² + xy = 0 for a given x and y for an example then you're only doing multiplication, you don't have to calculate inverses which can be slightly tricky depending on the number system.

[@janMaja]

nice!

[@patrickfrazier5740]

Why is this called a dream equation?

[@marc-andredesrosiers523]

what about p-adic numbers? it'd be neat to explore an equation with no real solution, but solutions in the p-adic

[@mathboy8188]

@charlottedarroch showed that it has a solution in the p-adic ring provided p is congruent to 1 mod 3. She left her explanation on my comment that begins: _( Notation: * = multiplicative inverse in a ring. )_ _Proposition: Let R be a non-trivial commutative ring, and consider the equation x* + y* = (x + y)* for some x, y in R._

[@5alpha23]

Haha sure, let's just take Z_39

[@arekkrolak6320]

Your calculations are correct but inference has a flow/shortcut. If assumption is made the solutions are real and complex solution is found, it doesnt automatically prove complex solution is correct, as wrong assumption was fundament to all calculations.

[@Eyalkamitchi1]

Is there something that can be described as a complex Z number system?

[@pierreabbat6157]

Z39?? You can't grow wheat there! Try Z13.

[@S8EdgyVA]

Then there are those fraction poets who bask at the complexity of the reality that y1/x+1/y=(x+y)/xy

[@bendunselman]

At 5:13 I disagree that x can be any point in the complex plane since it must (as in the reals) not be the origin.

[@grchauvet]

Yes, but it does actually work on the Riemann sphere, where x = y = 0 and 1/0 + 1/0 = 1/(0+0)

[@bendunselman]

@@grchauvet iff addition is defined in the extended complex numbers (C×) for infinity plus infinity and equal to infinity, which I dont know. I did see on wikipedia multiplication of infinity to be defined and equal to infinity. N.B. and saw the additive and multiplicative inverses to be undefined in C× according to wikipedia.

[@saaah707]

Wow cool. Does anyone know if there is some terminology for that equation a^2+a+1=0? So I can search. I'd like to read more about this sort of method

[@andreyfom-zv3gp]

uhm... quadratic equation?

[@ojas3464]

Primitive Roots of Unity could be a try

[@andreyfom-zv3gp]

@@ojas3464 maybe but there are ROOTS but not equation + it's a bit not about this equation. It's about some CERTAIN root of the equation xⁿ=1, and we technically can that its also a root of corresponding equation without (x-1) multiplier: x^(n-1)+...+x+1=0. So this equation is just a CERTAIN example of the EQUATION generating ALL ROOTS of unity besides 1.

[@saaah707]

@@andreyfom-zv3gp yes it's quadratic, but what i mean is, the fact that it's used to generate solutions for the original 2 variable equation

[@mitchwyatt9230]

Cyclotomic polynomials

[@tatfr0guy]

Tamely ramified extensions have entered the chat

[@tensorix]

Could you please elaborate further? I am not sure how it relates to the video

[@edwardlulofs444]

Oh, that’s a lot of fun! 😄

[@lormador8493]

have youu tried sllving the millenium prize problems

[@cosmosapien597]

I've solved them in my mind. But the proof is too long to publish in text.

[@Ahmed-Youcef1959]

Where are you ???

[@Tletna]

That was an interesting video.

[@finnboltz]

How old are you?

[@bananaforscale1283]

I feel like that's cheating to redefine inverse in the middle of the process.

[@oida10000]

Question is there an easy way to decide if a^2+a+1=0 is solveable in Zn or do we need to check all values in Zn? As for the homework: [(a, b), (c, d)]^2+[(a, b), (c, d)]+[(1, 0), (0, 1)]=[(a^2+a+bc+1, ab+bd+b), (ac+cd+c, bc+d^2+d+1)] and a^2+a+bc+1=0, ab+bd+b=0, ac+cd+c=0 and bc+d^2+d+1=0 solves to a=free, b=/=0, c=-(a^2+a+1)/b and d=-(a+1).

[@dnsfsn]

In Zp we have to have p being 3 more than a perfect square.

[@keniosilva222]

By MeanHarm-MeanArith inequality: (x+y)>=4(x+y) is it impossível.

[@aMartianSpy]

I also got no solutions 😊

[@AlbertTheGamer-gk7sn]

What FRESHMEN dream of: (x+y)^2 = x^2+y^2. If this dream is true, The entire Mathsverse will collapse, As the Pythagorean Theorem will break, As if a^2+b^2=c^2, (a+b)^2 = c^2, And a+b = c, Which violates the Triangle Inequality.

[@DarinBrownSJDCMath]

The entire Mathverse will collapse... unless you're in characteristic 2.

[@moonwatcher2001]

❤ and that's a good place to stop

[@Loyis]

Now do sin^-1(x)=(sin(x))^-1

[@serae4060]

I tryed solving this before watching the video and got y=x((-1(+/-)i√3)/2), so no real solution

[@JacobPlat]

y^2 + xy + x^2 = 0 x,y cannot be 0 and x cannot be -y

[@zanti4132]

Continuing with your first equation, either x² ≥ |xy| or y² ≥ |xy| (one must be true), so x² + xy + y² > 0. No real solutions.

[@robertpearce8394]

Another Michael Penn video that I could follow, although clinging on by my fingertips at times. (Note to self: Don't try this on an actual rockface.)

[@ibazulic]

JavaScript approves!

[@sidharathsharma6197]

Wowww!

[@michelfrance75]

In Z7, a^2+a+1 = 10 not 0

[@johnfoster6412]

There is no such thing as "cross multiplication" in the context here. You're multiplying each side by a/(a+1). You know this, try to be more precise.

[@Alan-zf2tt]

First! But Michael 1/0 +1/0 = 1/(0+0) = 1/(0 +0 + ... 0) is a tautology

[@PetraKann]

Division by zero is undefined (Mathematicians are adamant about this and very sensitive if you bring it up in everyday life 😁)

[@Alan-zf2tt]

@@PetraKann The only reply I can think of at the moment is: a tautology is a tautology otherwise it would not be a tautology. It is a bit like describing isomorphisms as 1 to 1 when indeed they are two to two too. Besides I was hasty to claim #1 EDIT: or is it two to two too too?

[@PetraKann]

@@Alan-zf2tt I think the emphasis is that you can't divide by zero - it's undefined. Logic is a self evident "statement that is true by necessity or by virtue of its logical form." Thus, "all logical propositions are reducible to either tautologies or contradictions". The division of zero is neither

[@Alan-zf2tt]

@@PetraKann As you say Petra - as you say

[@PetraKann]

@@Alan-zf2tt A pseudo-quasi-tautology Alan?

[@RealCasualMath]

Hint and a video I'm kinda proud of: kzbin.info/www/bejne/oIWnY4Wgrt6Mjc0si=SjG9PYUrma3VOzxy&t=359

[@Alan-zf2tt]

I mean c/mon! This is thought provoking stuff. Why modulo ℤ₍ₚᵣᵢₘₑ₎ and why 7 and 39? What is the key that steers such behaviours?

[@mathboy8188]

The key is, entirely, whether or not -3 is a quadratic residue in the field Z/(pZ) when p is a prime. ("-3 is a quadratic residue" means u^2 = -3 has a solution for some u in that field or ring.) In the ring Z/(nZ) when n is not a prime, you still must have that -3 is a quadratic residue if it's going to have a solution, but in this case -3 is a quadratic residue does not guarantee that it has a solution. (The guarantee is that, if -3 is NOT a quadratic residue, then it's certain that the equation has no solution.) You can see the derivation of that in my comment beginning _( Notation: * = multiplicative inverse in a ring. )_ _Proposition: Let R be a non-trivial commutative ring, and consider the equation x* + y* = (x + y)* for some x, y in R._