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  "body": "\n## Statement\n\nThere is no total order $\\le$ on $\\mathbb{C}$ that makes $\\mathbb{C}$, with its usual addition and multiplication, an [[ordered field|010JCV9J4E]].\n\n## Remarks\n\nProof: in an ordered field, each square is nonnegative, and $1 = 1^2 \u003e 0$. Then $-1 \u003c 0$. But $-1 = i^2$ is a square, by [[i squared is minus one|07KBQCG9KV]], a contradiction.\nThere are total orders on the set $\\mathbb{C}$, for example the lexicographic order. They are not compatible with the field operations.\nThus inequalities such as $z \u003c w$ make sense only for real numbers $z$ and $w$.\n\n## Explanations\n\n### L1 human\n\nYou cannot say which of $i$ and $0$ is larger in a way that keeps the usual rules of inequalities.\nIf $i \u003e 0$, then $i \\cdot i \u003e 0$, so $-1 \u003e 0$. If $i \u003c 0$, then $-i \u003e 0$ and $(-i)(-i) \u003e 0$, so again $-1 \u003e 0$.\nBoth cases are impossible.\n",
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