Is there such a thing as an equation with noncomplex quaternion solutions?

Question

I'm familiar with equations with real solutions and equations with nonreal complex solutions. Examples:

$x^2-3x+1=0$

has the real solutions

$3\pm \sqrt{5} \over 2$

and this other equation:

$3x^2-x+2=0$

has the nonreal complex solutions

$1\pm i\sqrt{23} \over 6$

but to my understanding, complex numbers are just a special kind of quaternion numbers, so I wonder if there's such a thing as an equation (not necessarily a polynomial equation) with noncomplex quaternion solutions. If there's such a thing, I would like to see an example.

Equation $x^2=-1$ has infinitely many quaternion solutions. Solutions include $i,j,k$ but many more as well. — GEdgar, May 21 '16 at 01:13
Equations expressing noncommutativity do the trick. Mike's equation, for example, is the same as $xi+ix=0$. — Mariano Suárez-Álvarez, May 21 '16 at 02:36

Asinomás · Answer 1 · 2016-05-21T01:53:40.717

2

Complex numbers have a very special property, they are an algebraically closed field. This means that any polynomial has complex solutions. If $P=a_nx^n+\dots +a_1x+a_0$ is a polynomial with complex coefficients then there exist $n$ complex numbers $\alpha,r_1,r_2\dots r_n$ (not necessarily distinct) such that:

$a_nx^n+\dots a_0=\alpha(x-r_1)(x-r_2)\dots (x-r_n)$.

So suppose $z$ is a non-complex quaternion root of $P$. Then we would have:

$(z-r_1)(z-r_2)\dots (z-r_n)=0$. This is impossible, the quaternions are a domain, this means that if a product of quaternions is zero then one of the factor must be zero. So we have $z-r_k=0$ for some $k$, implying $z=r_k$, meaning $z$ is complex.

edited May 21 '16 at 01:53

answered May 21 '16 at 01:42

Asinomás

105,651

something seems to be wrong with this argument, but I can't see what. – Asinomás May 21 '16 at 01:48
I found the mistake, the thing is the factorization is not well defined because $H$ is non-commutative – Asinomás May 21 '16 at 02:00
1

You shouldn't have deleted this: it is mucho more useful to have this wrong answer , with correction, than not having it! – Mariano Suárez-Álvarez May 21 '16 at 02:20

score 2 · Answer 2 · answered May 21 '16 at 02:16

2

Let's solve $ixi=x$. Explicitly, if $x=a+bi+cj+dk$, the left side is $-a-bi+cj+dk$. So this equation's solutions are those numbers of the form $cj+dk$ - those quaternions with no "complex part".

answered May 21 '16 at 02:16

Is there such a thing as an equation with noncomplex quaternion solutions?

2 Answers2