MATH.8:5 - Archetypal Grounding

MATH.8:5.1 - An even polynomial has more than one solution orbit

For real x, solve P(x)=x^4-5*x^2+4=0. The two transformations are identity and sign reversal r(x)=-x. Since P(-x)=P(x), they preserve the equation.

The known solution 1 gives the orbit {1,-1}. Applying sign reversal again returns to 1, so this orbit is complete. It is not the complete root set: 2 also solves the equation and belongs to the different orbit {2,-2}.

Factorization P(x)=(x^2-1)*(x^2-4) establishes that these two orbits cover all real roots. Symmetry generated each pair; factorization supplied the missing completeness argument.

Now change the equation to Q(x)=x^2+x-2=0. The value 1 remains a root, but Q(-1)=-2. The linear term breaks the sign symmetry. Reusing the old transformation would give a false answer to the changed problem.

MATH.8:5.2 - Binary arrangements on a cycle

Consider binary strings of length four with exactly two entries equal to 1. Positions are numbered 0 through 3 around a cycle. Let r move each entry to the next position, wrapping the last to the first. Four rotations return the original string, and rotation preserves the number of ones.

Starting with 1100, repeated rotation gives:

1100 -> 0110 -> 0011 -> 1001 -> 1100.

There are four distinct members. Only the identity rotation fixes 1100, agreeing with the count 4/1=4.

Starting instead with 1010 gives:

1010 -> 0101 -> 1010.

A rotation by two positions fixes either alternating string. Its stabilizer has two members, so the orbit count is 4/2=2.

Every two-one string either has adjacent ones around the cycle or has opposite ones. This covers the six possible strings and separates the two orbits. If the question asks for arrangements up to rotation, two representatives suffice. If it asks for every labeled string, return all six.

If position 0 receives a distinguished mark that must remain fixed, only the identity rotation preserves that data. The old orbit classification then forgets a distinction required by the new problem. The strings remain available; their identification must change.

Instead put identical marks at positions 0 and 2, with neither mark distinguished from the other. A single rotation and its inverse move the marked set to {1,3}; a half-turn returns it to {0,2}. The data stabilizer is therefore {e,r²}. To find this subgroup, examine compositions such as r² as well as the supplied generators.

MATH.8:5.3 - Permuting coefficients changes which equation was solved

Let the data be coefficients a=(2,1) and right-hand side 5. The equation is 2*x1+x2=5, with known solution x=(1,3).

The swap sends a to (1,2) and x to (3,1). Their scalar product remains 5 because both positions are exchanged. Thus the transformed solution satisfies x1+2*x2=5.

Keeping the original coefficients instead gives 2*3+1=7. The swap preserves the relation between transformed data and transformed solutions; it does not preserve this fixed original equation.

For comparison, x1+x2=5 has equal coefficients. Its data are fixed by the swap, so a solution (1,4) gives another solution (4,1) of the same equation. Neither the swap nor the equation singles out one of them. A requirement for one distinguished answer needs an additional criterion or a compatible choice method.

For a representative-data calculation, take coefficients (1,2) and its solution (3,1). The swap maps those data back to (2,1) and the solution back to (1,3), recovering the original answer. With the equal coefficients (1,1), both identity and swap return the same data, while the chosen solution (1,4) returns as either (1,4) or (4,1). Each is a valid solution, but this choice does not define a transformation-independent rule. This ambiguity concerns the chosen pair; a request for a swap-fixed solution can instead use (5/2,5/2).