beyond real numbers: solving x² + 1 = 0 in different number systems
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- เผยแพร่เมื่อ 7 ส.ค. 2023
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Error at 2:30: i×j×k= -1, but you wrote +1.
I love this kind of exploration of maths. It's a good way of revisiting fields and polynomials, but from a slightly different angle.
In the quaternion case, the requirement that the imaginary parts vanish does not trivially imply a=0 without backtracking. It could still be that b=c=d=0. Only if you try that, you'll find you're left with an unsolvable a² = -1.
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
8:30 I believe you need the closed interval [0, pi] for the angle phi. With an open interval you don't have coordinates for the pole corresponding to that end of the interval.
The imaginary part of the quaternions being on the surface of the sphere for the solution is analogous to the complex number case where the imaginary part is on the 1D sphere i.e. ±i
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)?
In the case of Zp, I suppose Michael doesn't prove that there are exactly 2 solutions because he thinks some algebraic background is required for that. However, there is an elementary approach.
First, show that a solution exists as in the video. Call that solution b. Notice that there is at least one more, namely -b. Therefore, we can factor as so: (x - b)(x + b) = x^2 + 1. Suppose x is another solution to x^2 + 1 = 0. Therefore p | (x - b)(x + b), but then p | x - b or p | x + b, showing that we must have x = b (mod p) or x = -b (mod p). Done.
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
i really love these exploratory videos, there's something wonderful about just playing around with math and especially with abstract algebra!
You could also solve in the p-adics, for p = 1 (mod 4).
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.
In the Zp you missed that x²≡-1(mod p) have a solution when p=2 and thats not congruent to 1(mod4). The error is when you say if (-1)^((p-1)/2)≡1(mod p) then (p-1)/2=2n. Thats true if p>2 beacause 1 is not congruent to -1 (mod p) and ( -1)² =1, so only in even powers you will have( -1)ⁿ≡1(mod p), but if p=2 its true that -1≡1(mod p) then for every power you will have the congruence (-1)ⁿ≡1(mod 2).
x²=-1 also has a solution in Z2, because -1≡1.
In Fq, where q is a power of an odd prime, you get the same number of solutions as in Zp (which is the same as Fp): if q≡1 (mod 4), there are two solutions; if q≡3, none. So F3 has no solutions, F9 has two, and F27 has none.
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.
26:32 Homework
26:58 Good Place To Stop
I usually take Zp to be the p-adic integers. Z/p or Z/pZ is how I'd write the finite field of Z mod p. The p-adic numbers are actually another good place to look for solutions to x^2=-1. It's a fairly simple part of p-adic analysis and extends really nicely from the Z/p case. Good way to introduce people to the subject. =]
nice exploration! also glad to see the ol’ button click illusion again
Love it. beautiful and with a perfect twist.
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 love this!
Is there a system of x^2 =1?
You did one of my favorites: x^2 =0, the Duals.
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.
- Loved it
At 18:18 it is not true that (p-1)/2 is a whole number because p = 4n + 1, as that's what we're trying to prove, so the logic would be circular. It's a whole number because we're (implicitly) supposing that p is odd.
Also, here is a more algebraic and direct, but less elementary proof of the other direction: Zp is a field, so it has a cyclic multiplicative group. Let g be the generator and consider a := g^((p - 1)/2). Notice that a^2 = 1, but that means a = ±1 because we're working in a field. a != 1, since g^(p-1) = 1. Therefore, a = -1. Now consider b := g^((p-1)/4). b is well-defined because p = 4n + 1 and b^2 = a = -1. The other solution is -b. Done.
Conclusion: in Z[n], the equation x²=-1 has 0 or more solutions.
Very useful, interesting and non-trivial content. Thanks!!!!
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.
Wow maybe I'm going too far but the solutions in the quaternion is a circle of radius 1 in the imaginaries axies (3D) mirror with the complex numbers because i and -i IS the circle of radius one in the imaginary axis (1D) !
So elegant
Great observation. Reminds me of the radius of convergence for analytic functions in R and C.
In the quaternions, shouldn't b = c = d = 0 be ruled out before concluding a = 0? (I mean it's just one line bc a² = -1 has no solutions but still)
Interestingly @12:36 he did not declare a+d must be 0.😂
I was really hoping for a p-adic solution. It also satisfies the p=1 mod 4 condition.
double liked this one. I would like to see this done for 1 and 0 in future vids… if it’s not boring.
The 5adic integers have two unique solutions to this equation that take the form of an infinite sequence that diverges in the reals - specifically plus or minus 2^(5^(n)). Note that I am referring to the limit of the sequence as n approaches infinity, rather than any summation, but this number has an infinite number of digits - it just so happens that the “smaller” of these (the ones, tens, hundreds, etc) all end up matching, and as n gets very large the agreement holds for more and more digits to the left. If you square this set of digits, you get an infinite string of digits that are all precisely 4 - this is the representation of -1 in 5adic space.
good to see the return of the magic squares = best youtube transition
8:37 In this case, θ can be π as well. It's the opposite point to θ = 0 on the sphere (which doesn't care about the value of φ)
Moreover, since we thought of d as the z value of a point on a sphere and assuming the north pole is is (0,0,1), then φ is actually the angle measured from the north pole, not the south
great video , thannk sir §
This is a bête noire of mine: I think that in the interest of not being ambiguous, "infinite solutions" should only mean (A) "any number of solutions, including only finitely many, each individual one of which is infinite (i.e. equals infinity)", and the cases mentioned here should be described as (B) "infinitely *many* solutions", (each of which might happen to be finite, as in the cases here). In other words, when I hear "infinite solutions", I want to ask "how many infinite solutions?"
I know that dictionaries (regrettably) allow "infinite" to mean either qualitatively infinite (A) or quantitatively infinite (B), but I think those two should be clearly differentiated when doing rigorous explanations. (I have this same gripe about Matt O'Dowd's PBS physics videos.)
Other than that, all of these videos are EXCELLENT (as are O'Dowd's). Constantly surprising me, who has not done very much number theory or even Abstract Algebra.
ℤₚ where p=2 has only one solution, so stating p is an odd prime is kind of important.
True, x^2+1=0(mod 2) has one solution of multipcity 2, just the number 1! Nice, isn't it?
@MichaelPennMath To answer your last question: p-adics, n-adics for composite n, the solutions of an elliptical curve plus the point at infinite using multiplication as the binary operation...
Dr. Penn, would there be a case where you'd set the matrix determinant to -1 ? Does that even make sense?
Uncountable many solutions !
Yes: Clifford algebras could be another interesting ambient in which to solve that equation.
15:48 "Are there more solutions?" As I'm sure you know, there is a bijection from any two-dimensional open subset of R2 onto all of R2, so the number of solutions is the same.
17:30 When p = 2, there is exactly one solution x = 1.
He added the constraint that p must be odd at some time towards the end.
Take a shot every time Penn says "Ok, nice."
At 19:48, easier to remember is that each field element is a solution of P(x)=x^(p-1)-1, because in the multiplicative group G the order of any x must divide the group order p-1.
Using: if d divides n, then: x^n-1 = (x^d-1) (x^(n/d-1)+x^(n/d-2)+ ... +1), we can factor out (x⁴-1), which in turn factors as (x²-1)(x²+1).
Since G has p-1 distinct elements and they all satisfy P(x)=0, we know that P must have p-1 distinct roots and each root of P must exist in the field! So the factor x^2+1 guarantees there are 2 distinct field elements with x^2=-1.
This has the smell of a 1st class back from summer break for an undergrad abstract algebra class.
Hi dear michael
Please make videos on another math's area's like abstract algebra, differential geometry, algebric geometry and etc...
With regards
Isn't it simpler to use the Chinese Reminder Theorem for the last case?
Do they have to be linear?
Remind me what the Chinese reminder theorem is. ;)
1²≡(-1) mod 2 is also a solution. You discarded it at 18:28 without justification.
Lovely video.
What about p-adic number?
I really enjoy your videos, and My son also, he is calling you Minus Minus guy :)
I quickly checked N, and there are no solutions....
Hi,
26:58 : again a "question for you"? May be functions, where "-1" would be -Id .
find all isometries f of the euclidean plane such that f applied twice is a reflection across a given axis (f•f = -1, as a reflection is understood as analogous to -1)
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 solve (A AND NOT A) = TRUE in Boolean algebra (Z2) it's x(x+1)=1 => x^2 + x + 1 = 0 which is irreducible, and the solution is the field F4, with two new truth values that are complements and their product (AND) is 1
bi and cj but no b..j
I feel like the quaternion derevation would have been faster using the exterior algebra/cross product form.
What about the P-adic numbers?
So we know that x^2 = -1 has no solution over R, one solution over F_2, two solutions over C, you showed Z/65Z has 4 solutions, and H has infinity many.
My professor once said that it is still an open problem to come up with a ring R, such that x^2 + 1 = 0 has exactly three solutions.
(Unless I misremember, in that case I would love to see what example there is)
I don't understand: what is Z_1 supposed to mean?
As usual a really good video, but something is wrong with your sound in the video. I had to turn the volume right up then when an ad kicked in it nearly blew the wax out of my ears!
Yes, I have indeed an interesting idea for other places where x^2 = -1 can be solved: geometric algebra(s). In some of these algebras, some of the basis vectors may square to -1 by construction, i.e. by prescribing the signature of the algebra (the number basis vectors that square to +1, -1 and 0 respectively). For example, in the spacetime algebra G(1,3,0), three of the basis vectors square to -1 (and one to +1) by construction of the algebra. But even in the geometric algebra G(3,0,0) of R^3 where none of the basis vectors square to -1, there are some other vectors that square to -1, if I remember correctly. But can only vectors square to the scalar -1? What about more general multivectors? ...or was it actually some of the bivectors that square to -1 in G(3,0,0) ...have to look it up again....
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}
Another number system is triplex numbers where i^2 = j, j^2 = i, ij = ji = 1. It can be proven that there is no solutions in the triplex numbers. Not to Fermat the proof, but it leads to b^2 = sqrt(-5/4) which is impossible over reals.
"So, let's recall what that is," when referring to quaternions. Haha. You, sir, think I know way more than I do. But thanks. :D
6:22 no it does not follow that a has to be 0, another solution is b=c=d=0. (Of course that would give a^2=-1 with real a, but he did not say that.)
fun!
13:38 "But let's just not do that here."
The fact that d^2/dx^2 cos(x) = - cos(x) comes to mind.
So the difference operator can be a solution to the equation in the space of Linear Operators on the vector space generated by (sin, cos).
Admittedly, not a number system though… 😅
(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.)
FFS, already 3 non skip-able ads interruptions and I'm not yet at the mid point of the video.
I think that 2x2 matrix in C goes to Pauli's matrix ?
12:10 a²+d²=
Using wilson's theorem to prove that -1 is a square mod p when p is a p is a odd prime is very chaotic evil lmao.
I understand that you want to avoid mentionning field and group theories but wilson theorem doesn't help understanding what's going on at all if you don't see it from the group theory pov. At this point you could use fermat's little theorem + prove that a degree 2 polynomial has at most 2 roots in Zp to get the same result and it wouldn't take much more time
3:24 ijk=-1
I thought you were gonna start from the Peano axioms to prove there exists no solutions in the real numbers 😂 Wasn’t expecting it to be so brief
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.
In my favorite field, the nimbers, there's only one solution to x^2 + 1 = 0 (with multiplicity two). In fact, because the field is algebraically closed with characteristic two, every nimber has exactly one square root (repeated twice).
The proof for the finite field case is invalid for a field of two elements, which of course has one solution.
Do Hamilton quaternions make anybody else think of rock paper scissors?
Don't see it stated in the comments, but the rightward proof is faulty because you assume the result (both sides of the equation to power (p-1)/2, because we know, p=4n+1; which is the result of proof)
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
I’ve been studying geometric algebra. It wasn’t in the syllabus really when I did maths in the 80s, I think one course mentioned Clifford Algebra as an aside. But it seems to generalise a lot of concepts and should be a fruitful source of examples.
15:37 b must not be equal to 0 though
isn't it ijk=-1?
Yes he forgot the minus sign. Or he wrote it backwards, as 1 = kji.
@15:45 it shouldn’t be all real number, b can’t be 0
There is a nice parallel with the video of some days ago about factorization of primes in the Gaussian integers Z[i].
The equation x^2+1= 0 mod p has:
- 1 solution with multiplicity 2 for p = 2
- 2 solutions with multiplicity 1 for p=4n+1
- 0 solutions for p=4n+3.
The prime number p in Z factorizes in Z[i] as:
- the square of a Gaussian prime (times a unit) for p = 2
- the product of two Gaussian primes for p=4n+1
- itself, i.e. p remains prime for p=4n+3.
And finally, in binary, -1= (2^n)-1
18:27 nah bro it's just a prime why would it necessarily be 4n+1, nєN
8:35 all half integers should work
Random part in vid
I mean I forgot 1/2n1/2n1/2n1/2n things
Try M_2(C) and M_2(H)
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.
18:23 Sir, I think you committed one of the mortal sins in math, which is assuming the conclusion. 😝 Though it could be easily remedied by separating into cases where p=2 and p is odd. Nonetheless, a very interesting video.
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 🤣
I guess the point you are trying to make if the problem doesnt state the number system you find a lot of answers without being "wrong".
These are not all number systems. There are only 4 number systems, R, C, H and O. The latter it's not discussed in the video.
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.
Take a look at problem 5 of IMO 2022
And Problem 1 of IMO 2023
Both are number theory problems :)
I love quaternions. They aren't very useful in mathematics, I believe, though they have found some favour in computer graphics, which allows me as a programmer to import a library of them and play with it at times.
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.
i^2 = i * i = (square root of -1) * (square root of -1)= square root of (-1 * -1) = 1.
the hamiltonian seems to define a self-contained infinite 3d virtual universe for every 0th dimensional point (singularity) in our universe...the set of mutliverses would be undefinable...the only problem is time...can time be a complex number?
if not, then it's a static universe with no entropy (no time arrow), let's look at it another way:
replace the word 'multiverse' with 'power and/or data storage' and we've really got something...
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.
x^2 = -1 over the ring of polynomials K[x] would be interesting.
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.
For complex numbers, all quadratic equations except x²=0 have 2 solutions, which only has one solution
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.
Lol bro there are only two numbers even and odd just like God intended them to be stop being woke
20:50 If you want to do the simple other case of p=2, just note that 1 ≡₂ -1 so 1² ≡₂ -1 is the only solution in Z₂
Don't we also have infinitely many solutions in the complex numbers? Since we can always multiply in an i⁴ (or if we write it in polar coordinates, 2Pi)?
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.
ijk=-1
3:15