• Use the positive cone axioms (P1–P3) to derive \(1>0\) and that squares are \(\ge 0\), with equality only at \(0\).
• Conclude that \(x^2+1=0\) has no solution in any ordered field, in particular in \(\mathbb{R}\).
• Construct \(\mathbb{C}\) concretely as \(\mathbb{R}^2\) with the twisted product; prove the inverse formula \(z^{-1}=\overline z/|z|^2\).
• Compute with \(\mathbb{C}\) via the \(\mathbb{R}^2\) model and verify basic identities; visualize multiplication by \(i\) as a \(90^\circ\) rotation.

1. State the positive cone axioms (P1–P3) and how they induce \(>\).
2. Explain why squares are \(\ge 0\) (and equal \(0\) only at \(0\)).
3. Describe the construction \(\mathbb{C}\cong\mathbb{R}^2\) with the twisted product and identify \(i=(0,1)\).

End with one precise question you still have.

Option 1 (ordered field argument). Prove, tagging explicit uses of (P1)–(P3): for all \(x\in\mathbb{R}\), \(x^2\in P\cup\{0\}\) and \(x^2=0\Rightarrow x=0\). Conclude that \(x^2+1=0\) has no solution in \(\mathbb{R}\).

Option 2 (non-orderability of \(\mathbb{C}\)). In any ordered field, squares are \(\ge 0\). Since \(i^2=-1\), show this forces a contradiction (\(-1\ge 0\Rightarrow 0\ge 1\)).

Implement complex arithmetic using the \(\mathbb{R}^2\) model (do not rely on complex built-ins for these core operations):

CAdd[{a_,b_},{c_,d_}]:={a+c, b+d}
CMult[{a_,b_},{c_,d_}]:={a c - b d, a d + b c}
CConj[{a_,b_}]:={a, -b}
CAbs[{a_,b_}]:=Sqrt[a^2 + b^2]
CInv[{a_,b_}]:=If[a==0 && b==0, "undefined", CConj[{a,b}]/(a^2 + b^2)]

Numerical checks (state a tolerance):

1. With \(z=(2,-3)\), verify \(CMult[z, CInv[z]]=(1,0)\) up to tolerance.
2. With \(z=(2,-3)\), \(w=(1,4)\), verify \(|zw|=|z||w|\) by comparing CAbs[CMult[z,w]] and CAbs[z]*CAbs[w].

1. Pick any nonzero \(z=(a,b)\). Compute \(i z = CMult[{0,1}, z]\).
2. Report the scalars \(CAbs[z]\) and \(CAbs[iz]\) and confirm they match.
3. Using real-vector arithmetic, compute the dot product \(z\cdot (iz)=a(-b)+b(a)\) and confirm it is \(0\).
4. Produce one labeled figure with arrows from the origin to \(z\) and to \(iz\); include axes and a short caption (“Multiplication by \(i\) is a \(90^\circ\) rotation, norm-preserving”).

With[
  {z={2,-1}, iz=CMult[{0,1}, {2,-1}]},
  Show[
    Graphics[{
      Thick, Arrow[{{0,0}, z}],
      Arrow[{{0,0}, iz}]
    }, Axes->True, PlotRange->{{-3,3},{-3,3}}],
    ImageSize->360
  ]
]

Why is proving “no real root of \(x^2+1\)” from the order axioms an appropriate prelude to adjoining \(i\)? Relate in one sentence to your rotation plot from Problem D.