Quaternion
Square roots of −1

In the complex numbers, C, {\displaystyle \mathbb {C},} there are exactly two numbers, i and − i, that give −1 when squared. In H {\displaystyle \mathbb {H} } there are infinitely many square roots of minus one: the quaternion solution for the square root of −1 is the unit sphere in R 3. {\displaystyle \mathbb {R} ^{3}.} To see this, let q = a + b i + c j + d k be a quaternion, and assume that its square is −1. In terms of a, b, c, and d, this means

a 2 − b 2 − c 2 − d 2 = − 1, x | 2 a b = 0, 2 a c = 0, 2 a d = 0. {\displaystyle {\begin{aligned}a^{2}-b^{2}-c^{2}-d^{2}&=-1,{\vphantom {x^{|}}}\\2ab&=0,\\2ac&=0,\\2ad&=0.\end{aligned}}}

To satisfy the last three equations, either a = 0 or b, c, and d are all 0. The latter is impossible because a is a real number and the first equation would imply that a = −1. Therefore, a = 0 and b + c + d = 1. In other words: A quaternion squares to −1 if and only if it is a vector quaternion with norm 1. By definition, the set of all such vectors forms the unit sphere.

Only negative real quaternions have infinitely many square roots. All others have just two (or one in the case of 0).
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