Discrete formula for the square root of an Eisenstein    real

 

 

 

To get some better familiarity with the geometry involved here, I made up the graphs below.  The first graph depicts the complex numbers with basis , which is overlaid by the lattice of the Gaussian integers, that is, the fundamental lattice:

 

This next graph depicts the depicts the complex numbers with basis , which is overlaid by the lattice of the Eisenstein integers, that is, a triangular, or hexagonal lattice:

 

 

The fundamental lattice of the Gaussian integers and the triangular, or hexagonal, lattice of the Eisenstein integers are related lattices known as rhombic lattices.  This diagram depicts some of the common types of lattices of the imaginary quadratic rings, of which both the Gaussian integers and Eisenstein integers are members:

 

 

 

Note:   is the same as .

 

For those of you who may not be unfamiliar with the procedure for deriving the discrete formula for the square root of a Gaussian real, I’ve included this link to this MathType gif showing it’s algebraic derivation… Discrete formula for the square root of a Gaussian real. (This is just a GIF file.  So, like with IE, you may have to enlarge it to be able to read it.)

 

Before continuing, I want to include the change of basis formulas and some facts about Eisenstein reals:

 

 

 

Now, on to the derivation of the discrete formula for the square root of an Eisenstein real…

 

 

 

 

 

 

 

 

 

You need to take the positive root on the inner square root , as indicated by the dropping of the  and it’s replacement with  before the inner radical.

 

Unfortunately, unlike the square root formula for Gaussian numbers, I only seem to be able to get the imaginary portion of  from this quadratic, presumably because one of the relations between  of the root  is not a simple symmetric one, as with the relation  in the derivation of the discrete square root formula for a Gaussian real.  That is, solving eq.  in the discrete square root formula for a Gaussian real for  yields  and solving for  in  yields , a simple interchange of the values of  and .  Whereas, the relation between  in  for the discrete square root of an Eisenstein real is not a simple symmetric one.

           

So, now we could attempt a back substitution to get , the real portion of  utilizing equation  once  has been determined with equation :

 

 

 

However, at this point I noticed a serious problem…  What if , is a nonnegative real number?  That is, , where ?  

 

 

 

A result one might expect and desire.  However, now if one plugs the value  into eq. , one gets an indeterminate result.  So, these formulas, by themselves, won’t give us any result if  is a nonnegative real number.

 

Eq.  can also be reformulated to solve for :

 

                                     

 

However, this is not an appealing option, largely because, if  and , we’ll be forced to define a new definition for the square root of a negative real number… which really doesn’t make sense anyway since we’re attempting to extract a square root of a real number .

 

I’ll show that it’s not necessary to define a new meaning for the square root of a negative real number to utilize these formulas (in their final form) since we can keep all quantities under the radicals positive and still extract the square root of any Eisenstein real.

 

I originally decided to solve for  in the original equations,  and  simply because it appeared solving for  would be easier than solving for .  In fact, I think it is much easier to solve eliminate  from these equations, but I decided I needed to go back and see if I could find a way to eliminate  instead… and see where that might lead:

 

 

 

 

 

 

 

 

 

 

You need to take the positive root on the inner square root , as indicated by the dropping of the  and it’s replacement with  before the inner radical.

 

Again… unfortunately, unlike the square root formula for Gaussian numbers, I only seem to be able to get the real portion of  from this quadratic, again, presumably since  is not a simple symmetric relation of .

 

So, we could use back substitution to get , the imaginary portion of  utilizing equation  or  once  has been determined with equation :

 

 

 

Neither option is particularly appealing, however.  In , if ,  … which, at the very least, would again require a redefinition of the taking of a negative real quantity under the square root radical.  And, in , again, now  will be indeterminate if .

 

I believe it’s best to combine the two formulas,  and :

 

 

 

 

In any reformulation of eqs.  one must be mindful of the sign of the quantity inside the outer radical, i.e. in  that is,

 

 

 

since certain reformulations will result in an inversion of the sign of the numerator in , or likewise in the outer radical quantity of  necessitating negating the positive root on the inner square root  and taking the absolute value of the quantity contained within the whole of the numerator, or in the outer radical in , in order to procure the appropriate quantity for the satisfactory extraction of the square root.

 

The only remaining pragmatic matter for utilization of these formulas is the taking of the signs of the real and imaginary portions.

 

First however, I wish to prove, as asserted previously, that all quantities under both the inner and outer radicals will always be nonnegative, avoiding the requirement of some redefinition of the taking of the square root of a negative quantity under the radical.

 

First, it’s necessary to prove that  will always be nonnegative:

 

 

 

 

 

 

Next, it’s necessary to prove that:

 

 

 

Since we adhere to the convention that all , the above inequality will hold if:

 

                         

 

Finally, it must be proven that:

 

                                 

 

Again, since we adhere to the convention that all , the above inequality will hold if:

                                               

                                     

 

Now, without loss of generality:

, so:

 

 

 

Now, on to the taking of the signs of the real and imaginary portions.  I believe the most convenient equation for this purpose is a variant of equations  … .  Again, unfortunately, unlike with the Gaussian reals, the relation of the signs of the root to the components of the square is a somehow more complex one.  However, I think the followed sequence of implications is relatively easy to follow:

 

 

You’ll notice that when , the implication is that either one, or both,  are zero, so we simply take  of whatever portion, real or imaginary, we may have.

 

Next, the derivation of conditions for a pure real and pure imaginary root:

 

First, the conditions for a pure imaginary root:

 

Taking inequality  from proof  above, setting it to zero and solving we find:

 

 

 

However, an extraneous root was added in the squaring operation, so since we abide by the convention ,

 

 

 

Combining these two conditions, we find the necessary and sufficient conditions for a pure imaginary root is:

 

 

 

* Zero is considered a real number here.

 

This set of Eisenstein numbers is illustrated with the diagram below:

 

 

Now, the conditions for a pure real root:

 

Taking inequality  from proof  above, setting it to zero and solving we find:

 

 

 

However, again, an extraneous root was added in the

squaring operation, so:

 

Since we abide by the convention ,

 

,

 

 

 

Combining these two conditions, we find the necessary and sufficient conditions for a pure real root is:

 

 

 

Another question that one might have is:  What, or where, are the square roots of negative real numbers?

 

In the basis , the square roots of  is of the form .  Using the change of basis formula  we find  in  coordinates is:

 

 

 

Notice that all these are real multiples of .

That is:

 

 

 

Hence, the square roots of negative real numbers are precisely where one would expect them to be.  That is, they lie on an axis perpendicular to the real axis passing through the origin.  Which, of course, is the  -axis in the basis .  This is depicted in the graph below:

 

 

You might also notice that the purely imaginary numbers in the basis , that is, the Eisenstein reals lying on the  -axis, along with the Eisenstein reals lying on the  -axis (those on the  -axis are not, however, purely imaginary) are precisely the genuinely complex cube roots of the real numbers.

 

Now, some examples of using of the discrete formulas for the square root of an Eisenstein real:

 

 

 

 

 

 

 

 

     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

References:

 

An Introduction to the Theory of Numbers, G.H. Hardy, E.M Wright.   Oxford University Press 1979.

 

The Book of Numbers, John H. Conway, Richard K. Guy.   1996 Springer-Verlag New York, Inc.

 

A Course in Computational Algebraic Number Theory, Henri Cohen.   Springer-Verlag Berlin Heidelberg 1993.

 

Problems in Algebraic Number Theory, Jody Esmonde, M. Ram Murty.   1999 Springer-Verlag New York, Inc.

 

A Classical Introduction to Modern Number Theory, Kenneth Ireland, Michael Rosen.   1972, 1982, 1990 Springer-Verlag New York, Inc.