{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Assume that $\\le$ is a [[total order|0316AYH1SC]] on $\\C$ that makes $\\C$ an [[ordered field|010JCV9J4E]].\nIts two axioms are: $a \\le b$ implies $a + c \\le b + c$, and $0 \\le a$, $0 \\le b$ imply $0 \\le a b$.\n2. Claim: for each $x \\in \\C$, we have $(-x)(-x) = x x$.\nWrite $x = p + q i$ with $p, q$ real. Then $-x = (-p) + (-q) i$.\nThe product rule for [[complex numbers|09FTAET5Z4]] gives $(-x)(-x) = (p^2 - q^2) + 2 p q \\, i = x x$.\n3. Claim: $0 \\le x x$ for each $x \\in \\C$. Since the order is total, $0 \\le x$ or $x \\le 0$.\n4. If $0 \\le x$, then the second axiom with $a = b = x$ gives $0 \\le x x$.\n5. If $x \\le 0$, the first axiom with $c = -x$ gives $0 \\le -x$. The second axiom gives $0 \\le (-x)(-x)$, which is $x x$ by step 2.\n6. By step 3 with $x = 1$, we get $0 \\le 1$.\n7. By step 3 with $x = i$, we get $0 \\le i^2$. By [[i squared is minus one|07KBQCG9KV]], this says $0 \\le -1$.\n8. Adding $1$ to both sides of $0 \\le -1$ gives $1 \\le 0$, by the first axiom.\n9. A total order is a [[partial order|011C29Q1S1]], so it is antisymmetric. Steps 6 and 8 then give $0 = 1$.\n10. But $0 \\neq 1$ in $\\C$, because $\\C$ is a [[field|0FRSR10DWF]]. This contradiction shows that no such order exists.\n\n## Idea\n\nIn any ordered field every square is nonnegative, so $1 = 1^2 \\ge 0$.\nIn $\\C$ the number $-1$ is also a square, namely $i^2$, so $-1 \\ge 0$ as well.\nThese two facts force $0 = 1$, which is impossible.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "12J15"
    ],
    "name": "Proof that the complex numbers cannot be ordered",
    "proves": "01CJRJKDAA",
    "sources": [
      {
        "locator": "Orderability of fields",
        "url": "https://en.wikipedia.org/wiki/Ordered_field"
      }
    ],
    "tag": "9DM7JGR0K4",
    "uses": [
      "010JCV9J4E",
      "07KBQCG9KV",
      "0316AYH1SC",
      "011C29Q1S1",
      "0FRSR10DWF",
      "09FTAET5Z4"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/9DM7JGR0K4/",
  "links": [
    {
      "field": "proves",
      "target": "01CJRJKDAA"
    },
    {
      "field": "uses",
      "target": "010JCV9J4E"
    },
    {
      "field": "uses",
      "target": "07KBQCG9KV"
    },
    {
      "field": "uses",
      "target": "0316AYH1SC"
    },
    {
      "field": "uses",
      "target": "011C29Q1S1"
    },
    {
      "field": "uses",
      "target": "0FRSR10DWF"
    },
    {
      "field": "uses",
      "target": "09FTAET5Z4"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/9DM7JGR0K4.md",
  "msc": [
    "12J15"
  ],
  "name": "Proof that the complex numbers cannot be ordered",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/9DM/9DM7JGR0K4.md",
  "status": null,
  "tag": "9DM7JGR0K4",
  "uri": "m2://9DM7JGR0K4",
  "url": "https://math2.tamnd.com/wiki/proof-that-the-complex-numbers-cannot-be-ordered/",
  "writing": {
    "tag": "9DM7JGR0K4",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}