beyond real numbers: solving x² + 1 = 0 in different number systems 

Yeah pretty much, because the complex numbers are 2D and the solutions are on a 1D sphere, and the quaternions are 4D and the solutions are on a 3D sphere, so in both cases the solutions are on a sphere one less dimension than the number system.

yes, but the analogy requires the 1D sphere, which is the most counterintuitive simple shape in all of mathematics

​​@@DeJay7they are in the sphere of the imaginary part. In C the imaginary part is the i axis and in H its the i j k 3d space

@@Pablo360able I mean, it is all points in a 1D space that are equidistant to some center. In 1D, that happens to have only 2 points, but it is a circle.

Okay - I am intrigues by this! Does it mean I can theoretically say something like ... Problem in n-space will have a solution in n+1-space? For example a problem with or without solution in 2-space will always have a solution in 3-space a problem with or without solution in 3-space will always have a solution in 4-space (for the purposes of haste on my part take n-space to be same as n-dimensions) And secondary impact thought: will nested non-solutions exist having only solutions living in infinite-space (congruent to infinitely many dimensions)?

I agree in the quaternion space. However, what does the trick to me is the fact that a is real.

Hi, please help me solve this equation A+B+C=100 and 30A+20B+5C=1000 A=? ,B=? ,C=? (A,B,C) are non-null elements and (A,B,C) belong to N- {0}

Hi, please help me solve this equation A+B+C=100 and 30A+20B+5C=1000 A=? ,B=? ,C=? (A,B,C) are non-null elements and (A,B,C) belong to N- {0}

I had to think it through, too If b=c=d=0 then our number is real, which we already have solved.

To be fair, checking b=c=d=0 is easy enough you might as well not mention it in the video

Thanks, this is what I was looking for

well, dont need to see it as a polynomial on Zp. Just notice that if a²==b² (p) then p|a+b or p|a-b. Therefore there cant be more than 2 solutions for x² == r for any residue r, including -1

Oh, here is why there is a solution for p=13. Thanks.

Are you saying that there are two numbers that multiply to …4444444 in the 5-adics? Because I would like to find one of these but didn’t know where to look. I think I determined there wasn’t any solution in 10-adics. Don’t tell me the number that works, just tell me whether it is possible. please

​@@alikaperdue there are no such thing as 10-adics.

@@arsenypogosov7206 The 10-adics are a perfectly well-defined *ring*, the same way as any p-adic is. But they're not used because of the existence of zero-divisors.

@@Jcarr250 and I like the zero divisors. Because no one else does. I was looking at the locations of them in the Sedenions. I imagine they are black holes and see how close I can get without falling in. I prove nothing, so zero divisors don't make a difference to me.

Thats wrong. F9 does not have any solutions. If x² == -1 (p^k) then x² == -1 (p), and this implies p == 1 mod 4. So q == 1 (4) is not sufficient, you need the prime to be 1 mod 4, not just the power. Its actually very easy to prove by induction (on the exponent) that every power of p (where p == 1 mod 4) has exactly 2 solutions. So in general, for any integer k, and any odd prime p x² == -1 (p^k) has exactly 1 + (-1/p) solutions, where (a/p) stands for "a legendre p"

F_9 is not Z_9, i.e. it is not the integers mod 9, as that is not a field. F_9 is the field with 9 elements, and it can be represented as Z_3[x]/(x^2+1), that is, the polynomial ring with coefficients in Z_3 modulo the ideal generated by x^2+1 (or either of the other two irreducible quadratic polynomials over Z_3). By its very construction you can see that in F_9 we have elements that square to -1 (which is still equal to 2), namely x and 2x.

Interestingly @12:36 he did not declare a+d must be 0.😂

I believe there are the hyperbolic numbers that exist where u^2 = 1 and u != 1 so you could start looking for solutions using those

Let y = x-1 then x+1=y+2 So y (y+2) = (x-1)(x+1)=x^2-1=0 so y^2 + 2y = 0 I forgot the point I was making... Well, I can at least say that there is such a system. You can let x be the matrix [[0,1], [1,0]] this squares to 1. This system, like the dual numbers, and as shown above, has zero divisors. Idk if it has any nilpotent elements.

I think those are the split complex numbers.

@@quanquin3822 But bicomplex numbers are even more interesting because they contain both split complex numbers and complex numbers. They are exactly like quaternions, except that they are commutative and they have zero divisors.

@@christopheriman4921 TY! I appreciate the leads.

Great observation. Reminds me of the radius of convergence for analytic functions in R and C.

Do they have to be linear?

Remind me what the Chinese reminder theorem is. ;)

It would be also interesting solving in bicomplex or even multicomplex number systems (where zero divisors are permitted!) I dislike quaternions because they have no commutativity. In their cousins bicomplex numbers, there is commutativity, but there are also zero divisors.

Hey, at least we ruled out a negative number of solutions.

@@radupopescu9977 i don't know bicomplex so idk xd

@@pablomartinsantamaria8689 They are like quaternions, BUT, they are commutative and they have zero divisors.

True, x^2+1=0(mod 2) has one solution of multipcity 2, just the number 1! Nice, isn't it?

He added the constraint that p must be odd at some time towards the end.

He does not assume the conclusion, but I think he did forget to mention that he is look at odd primes p. For p = 2 you will have 1^2 = 1 = -1.

He literally said "Because p=4n+1, (p-1)/2 is a natural number". I put the time stamp for you to listen again. 🙂

@@eaglesquishy ahhh I missed that. Ok, he should have kept silent, haha.

@@eaglesquishy yeah, it was not even subtle 🤣

There are other hypercomplex number systems that further extend the pattern of doubling dimensions, such as the bicomplex, tricomplex, and tessarines. These systems can have various properties and solutions for equations, but they can also become more abstract and less intuitive. It's important to note that as we move beyond complex numbers, the properties of the number systems become increasingly complex and may not follow the same rules as real or complex numbers. This can lead to unexpected and non-intuitive results in solving equations like x² + 1 = 0.

I think that only blades (the outer product of some number of grade-1 elements) can square to -1. In a signature of (p,q), you can probably treat the positive and negative elements separately and find values for a² - b² = -1, and then spread the a values across the positive basis elements in a hypersphere, and spread the b values across the negative basis elements in a hypersphere. The same can be done for pseudovectors (grade (p+q)-1 blades), though for k-grade elements in between things are a little more complicated because their square can be a non-scalar, which is why you need to restrict the domain to k-blades. This can probably be done with a few extra equations to restrict non-scalar outputs to 0. I didn't touch on elements that square to 0 because as long as the square is still a scalar, then you can add as much null elements as you like and it wouldn't change anything.

Hi, please help me solve this equation A+B+C=100 and 30A+20B+5C=1000 A=? ,B=? ,C=? (A,B,C) are non-null elements and (A,B,C) belong to N- {0}

I mean I forgot 1/2n1/2n1/2n1/2n things

These are 4 number systems you have names for. If we call R 1-dimensional numbers, C 2-dimensional numbers, etc., well we can have number systems with 2^k-dimensions for any non-negative integer k. So R C Q and O are the first four members of an infinite set.

@@Qermaq ... No you don't. After the octonions, things are algebraically useless. For example, the sedenions are not even a skew field, because they have zero divisors.

@@michaelaristidou2605 Yet they are number systems. You are entirely correct that we lose stuff as we gain stuff as we increase the complexity of the system, but I'd be reluctant to state "there are just 4" when, for all we know, those 16-dimensional numbers might be needed in the future for something we can't comprehend today. We had no use whatsoever for anything beyond R until we were smart enough to require C. We lose the fundamental theorem of arithmetic in C, but we're ok with that because what's gained is more important. It's entirely possible a mathematical notion we have not yet discovered/invented will require such a system, and I don't think you or I or any mathematicians alive are smart enough to be certain otherwise.

(PS: Probably isomorphic to the M2x2 case shown in the Video. Considering the R vetor space generated by (sin t, cos t, e^it), we obtain a new case.)

Yes he forgot the minus sign. Or he wrote it backwards, as 1 = kji.

Isometries of the Euclidean plane preserve distances between points. In the context of your question, we are looking for isometries \( f \) such that the composition \( f \circ f \) is a reflection across a given axis. An isometry in the plane can be a translation, rotation, reflection, or glide reflection. 1. **Reflection**: If \( f \) is a reflection, then \( f \circ f \) will be the identity transformation (i.e., it maps every point to itself), not a reflection. 2. **Translation**: If \( f \) is a translation, then \( f \circ f \) will also be a translation, not a reflection. 3. **Rotation**: If \( f \) is a rotation by an angle \( \theta \), then \( f \circ f \) will be a rotation by an angle of \( 2\theta \). For \( f \circ f \) to be a reflection, we need \( 2\theta = 180^{\circ} \), so \( \theta = 90^{\circ} \). Thus, a rotation by 90° about any point will satisfy the condition. 4. **Glide Reflection**: A glide reflection is a composition of a reflection and a translation along the reflecting line. It cannot satisfy the condition \( f \circ f \) is a reflection since the composition of two glide reflections would include two translations along the reflecting line, not resulting in a reflection. In conclusion, the only isometries that satisfy the given condition are rotations by 90° about any point in the plane. -GPT4

If you want a rigorous argument then the reason x^2 = -1 has no solutions in *R* is because *R* is a (nontrivial) linearly ordered ring. In any such ordered ring we have the familiar properties that a>0 & b>0 ⇒ ab>0, a 0 (since 1 ≠ 0, and if 1 < 0 then we would get the contradiction 1 = 1^2 > 0). Thus -1 < 0 and so x^2 = -1 is impossible; whether x ≥ 0 or x ≤ 0 we always have x^2 ≥ 0.

@@schweinmachtbree1013how does one prove those “familiar properties”?

​@@synaestheziac the axioms for an ordered ring say that the order is compatible with the ring operations, i.e. a

@@synaestheziac One doesn't prove any properties for the so called real numbers. One axiomatically declares them. The declaration that real numbers form a field is obviously false. Non-demonstrable and non-computable "numbers" can't do arithmetics. Formalism is just post-modern language game telling us that Emperor's New Cloths are really real.

Isn't it much simpler, though? We know that when a positive real number is multiplied by a positive real number, the product is also positive. We also know that when a negative real number is multiplied by a negative real number, the product is positive. Thus the only case where the square of a real number is not positive is when we consider 0. This excludes a negative real number from having any square root among the reals. Yes, you can go full into the weeds on this if you like, but Michael's point is it is the trivial example.

He only assumes that p is odd (which I think he forgot to state). -1 = x^2 and p odd, then you can raise to the power (p-1)/2, so (-1)^((p-1)/2) = (x^2) ^ ((p-1)/2) = x^(p-1) = 1. This means that (p-1)/2 is even (where you again use that p > 2, since otherwise -1 = 1), so p = 1 mod 4.

@@ronald3836 Right, and assuming p is odd is critical, because x²=−1 does have a solution in Z_2, but 2 is not congruent to 1 mod 4.

@@twwc960Indeed

When K is a field, K[X] has no zero divisor, so 1 and -1 are the only solutions.

If characteristic of the field K is different from 2, 1 and -1 are the two different solutions. If characteristics of K is 2, there is only one solution 1=-1.

Those are not so interesting cases, because ℝ ⊆ ℍ (or C). In case of ℝ you have infinite many solutions, so obviously you have infinitely many solutions in other two cases.

@@DastarToRon sure, but i'm not paying attention to number of solutions, but in which solutions. At a first sight, who can say the are no other solutions extending to C or H?

@@loloplatz There are another solutions. On 13:34 he talks about case 1 on solving the matrix under real numbers. He got there a^2=b^2=-1, and this case clearly gives another two matrices under complex numbers. This case also gives you infinite many solutions under H, exactly like on solving under H.

Quaternions? Not useful in mathematics? What are you smoking

@@Pablo360able I think he means outside of some more specialized mathematics. Few learn them unless they're a dedicated math major. However, computer 3D animation uses them under the hood, so anyone doing graphics will at least encounter them. At least I think that's what he meant.

In fact, I prefer the descrition of the Quaternions as a pair consisting of a real number (the "real" part) and a 3-d real vector (the "imaginary" or "vector" part). Addition is still componentwise, and multiplication can be beautifully described using the scalar product and the cross prodct of the two vector parts.

How about x^+2x+1 = 0?

@@ronald3836 that's x²+2x+1, not x²

@@tamoozbr ok, but it is a quadratic equation with 1 solution (having multiplicity 2).

x^2 = 0 still has 2 solutions, and both are 0. I get why it seems tricky, as the sign of 0 is kinda meaningless.

Yes but those are not different numbers since your multiplying by 1

no, because those are just different ways of writing the same solution; If that counted, every solvable equation would have infinite solutions .

In the complex numbers you can write x^2+1 = 0 as (x+i)(x-i) = 0 because of commutativity. Since C has no zero divisors, this means x+i = 0 or x-i = 0. So the only solutions are -i and i. In the quarternions this does not work because you don't have commutativity.