I am not sure. Based on the definition of a magma, which says $\forall a,b \in M, a \circ b \in M$, if our magma has only $n$ elements $a,b,c,d,...,n$, then by the definition of magma, any product of two elements must be one of the $n$ elements in the magma.

As for the second point, while it is true in general that $(ab)S=T_{(ab)i}\neq aT_{bi}=a(bS)$ for a magma, the converse means if this equality holds, then we can derive the conclusion that all $n^3$ associative laws holds for all elements and hence the magma is associative (and hence a semigroup)

I don't remember from the top of my head the most accurate algorithms to diagonalise the hamitonian (that is the first step to numerically solve any eigenvalue problem). SVD is often good for minimsing errors as having the singular values arranged in descending order, the ones closest to zero can be discarded without adversely affecting the calculated result (unless the 10000th eigenvalue happened to be those near zero ones).

In terms of the error bounds, I never have kept them in my head, thus I am not sure

\begin{align}
H & = -\frac{1}{2}\frac{\partial^2}{\partial x^2 } + x^4\\
H & = \frac{1}{2}i^2\frac{\partial^2}{\partial x^2 } + x^4\\
H & = \frac{1}{2}i^2\frac{\partial^2}{\partial x^2 } -i^2 x^4\\
H & = i^2\left(\frac{1}{2}\frac{\partial^2}{\partial x^2 } - x^4\right)\\
H & = i^2\left(\frac{1}{\sqrt{2}}\frac{\partial}{\partial x } - x^2\right)\left(\frac{1}{\sqrt{2}}\frac{\partial}{\partial x } + x^2\right)+\frac{1}{\sqrt{2}}[\frac{\partial}{\partial x},x^2]\\
H & = i^2\left(\frac{1}{\sqrt{2}}\frac{\partial}{\partial x } - x^2\right)\left(\frac{1}{\sqrt{2}}\frac{\partial}{\partial x } + x^…

[Abstract algebra] Numerous h barers and maths chat people have checked my formalisms.
While it has now been refined again to iron out the inconsistency and improve presentation, it is being pointed out that the row maps to row interpretation of associativity is not the full story, as for the current proof to work, it relies on that a composition table (generalisation of a cayley table to weaker structures than groups) can be constructed from the elements (that means, all elements have the property in that they can be be written in terms of a product of two elements). As Tobias and co. poin…

[Division by zero] binary zero term algebra that satisfy the composition table property in general have a more restrictive form of the distributive law. In general, given two zero terms A and B where a partial ordering had not been specified, either A+B=B or A+B=A. In the cayley table, this will translate to a row will either be dominated by all rows or dominates all rows, on a row by row basis
(that is, if there are n rows, then there are a total of n distributive laws phrased in terms of rows need to be specified to completely determine the + composition table)