If y is a quaternion, how many solutions does x² = y have? That is, does every quaternion have a square root, and if so, how many square roots does it have?

A quaternion can have more than two roots. There is a example right in the definition of quaternions:

i² = j² = k² = −1

and so −1 has at least three square roots. It follows that every negative real number has at least three square roots.

Negative reals

In fact, every square root of −1 corresponds to a square root of any other negative number. To see this, let p be a positive real number and let r = √p > 0. If x is a square root of −1, then rx is a square root of −p. Conversely, if x is a square root of −p, then x/r is a square root of −1.

So all negative real numbers have the same number of square roots in the quaternions, and that number is at least 3. In fact, it turns out that all negative real numbers have infinitely many quaternion roots. A calculation shows that

(a + bi + cj + dk)² = −1

if and only if b² + c² + d² = 1. So there are as many square roots of −1 as there are points on a sphere.

Not negative reals

If a quaternion is not a negative real number, then the story is much more familiar: the equation x² = y has one solution if y = 0 and two solutions otherwise.

To summarize, x² = y has

• one solution if y = 0,
• infinitely many solutions if y is negative and real, and
• two solutions otherwise.

It’s easy to show that 0 has only one root: |x²| = |x|², and so if x² = 0, then |x| = 0 and so x = 0. Here the norm of a quaternion is defined by

|a + bi + cj + dk| = √(a² + b² + c² + d²).

If p > 0, the solutions to x² = −p are bi + cj + dk where b² + c² + d² = p.

What’s left is to show that if y is not zero and not a negative real, then x² = y has two solutions.

Polar form and calculating roots

Quaternions have a polar form analogous to the polar form for complex numbers. That is, we can write a quaternion y in the form

y = r(cos θ + u sin θ)

where r is real and positive, θ is real, and u is a unit quaternion with zero real part. We can find r, θ, and u as follows. Write y as

a + bi + cj + dk = a + q

so that a is the real part and q is the pure quaternion part. Then we can find r, θ, and u by

r = |y|
cos θ = a/|y|
u = q / |q|

Then the two roots are

x = ±√r (cos θ/2 + u sin θ/2).

Python code

First we’ll implement the algorithm above, then show that it produces the same results as the Python package quaternionic.

First, our code to compute x satisfying x² = y.

    import numpy as np

    np.random.seed(20210106)

    def norm(q):
        return sum(t*t for t in q)**0.5

    y = np.random.random(size=4)
    r = norm(y)
    theta = np.arccos(y[0]/r)

    u = y + 0 # make a copy
    u[0] = 0
    u /= norm(u)

    x = np.sin(theta/2)*u
    x[0] = np.cos(theta/2)
    x *= r**0.5

And now we check our results using quaternionic.

    import quaternionic

    qx = quaternionic.array(x)
    qy = quaternionic.array(y)

    print(f"y   = {y}")
    print(f"x*x = {qx*qx}")       # should equal y
    print(f"x   = {x}")           # our root
    print(f"x   = {np.sqrt(qy)}") # root via package

This produces

    y   = [0.61615367 0.07612092 0.09606777 0.11150865]
    x*x = [0.61615367 0.07612092 0.09606777 0.11150865]
    x   = [0.79189641 0.04806243 0.06065678 0.07040609]
    x   = [0.79189641 0.04806243 0.06065678 0.07040609]

More quaternion posts

• Finnegans Wake and quaternions
• How far is xy from yx on average for quaternions?
• Probability of commuting

For an exploration of the roots of general polynomial equations over polynomials, see Equations in Quaternions by Ivan Niven. The American Mathematical Monthly, Dec., 1941, Vol. 48, pp. 654-661.

To contemporary ears “quaternion” refers to a number system discovered in the 19h century, but there were a couple precedents. Both refer to something related to a group of four things, but there is no relation to mathematical quaternions other than that they have four dimensions.

I’ve written before about Milton’s use of the word in Paradise Lost. Milton is alluding to the four elements of antiquity: air, earth, fire, and water.

I recently found out the word quaternion appears in the Latin Vulgate as well. Acts 12:4 records that Herod had Peter guarded by four squads of four soldiers each, “quattuor quaternionibus militum.”

Update: Thanks to Phil for pointing out in the comments that quaternion appears in the King James as well, “four quaternions of soldiers.”

More quaternion posts

• How far is xy from yx on average for quaternions?
• Dot product, cross product, and quaternion product
• Beyond quaternions: octonions and sedenions

The blog post that kicked off the recent series of posts looked at how far apart xy and yx are for quaternions. There I used the Euclidean distance, i.e. || xy − yx ||. This time I’ll look at the angle between xy and yx, and I’ll make some analogous graphs for octonions.

For vectors x and y in three dimensions, the dot product satisfies

where θ is the angle between the two vectors. In higher dimensions, we can turn this theorem around and use it as the definition of the angle between two vectors based on their dot product.

(The plots below are jagged because they’re based on random sampling.)

Here’s the Euclidean distance between xy and yx for quaternions from the earlier post:

And here’s the corresponding angular distance:

The range has changed from [0, 2] to [o, π], and the distribution now shifts left instead of right.

Here’s a similar graph, looking at the angular distance between xy and yx for octonions, something I haven’t plotted before in Euclidean distance.

This graph is more symmetric, which we might expect: since octonions have less algebraic structure than quaternions, we might expect the relationship between xy and yz to behave more erratically, and for the plot to look more like a normal distribution.

Finally, let’s revisit the distance between (xy)z and x(yz) for octonions. Here is the distribution of the Euclidean distance from a previous post:

And here is the corresponding histogram based on angular distance.

These plots are based on uniform random samples of quaternions and octonions of length 1, i.e. points from the unit spheres in 4 and 8 dimensions respectively. Quaternions and octonions have the property that the product of unit length vectors is another unit length vector, and the angle between to unit vectors is the inverse cosine of their dot product.

I thought that sedenions also had this norm property, that the product of unit length vectors has unit length. Apparently not, as I discovered by trying to take the inverse cosine of a number larger than 1. So what is the distribution of lengths that come from multiplying two sedenions of length 1? Apparently the mean is near 1, and here’s a histogram.