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I'm not sure if it this is right: Imagen that I have this quaternion $a + bi + cj + dk$ where $a, b, c, d \in \mathbb{R}$.

Then I take $cj + dk$ and make this $(cj + dk)^2=-c^2 +cdi-dci-d^2=-c^2-d^2$ so $(cj + dk)=\pm\sqrt{-c^2-d^2}=\pm i\sqrt{c^2+d^2}$

Then $a + bi + cj + dk = a + bi \pm i\sqrt{c^2 + d^2} = a + i(b \pm \sqrt{c^2+d^2})$

So a Quaternion is just an ordinary complex number?? $a + i(b + \sqrt{c^2+d^2})$ or $a + i(b - \sqrt{c^2+d^2})$ , I mean, it is only an ordinary number $x+yi$

Am I doing something wrong here?

Thank you

Pedro
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    The Quaternions are not a field, hence there can be more than two roots to a quadratic. For example, $x^2=-1$ has more than two roots. Were it otherwise, you could run your argument more simply by remarking that, since $j^2=-1$, we must have $j=\pm i$ and similarly $k^2=\pm i$. – lulu Oct 19 '16 at 18:33
  • oh, I see..yes it is true...Thank you lulu – Pedro Oct 19 '16 at 18:36

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What you can do is render $cj+dk$ as $u\sqrt{c^2+d^2}$ where $u$ is an imaginary unit, that is $u$ is a quaternion solution to $u^2=-1$. But, as @lulu points out, quaternions have more than one additive inverse pair of such units, and in this case $i$ and $-i$ are just not the right pair.

Oscar Lanzi
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