{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe check the axioms of a [[field|0QA8CHH8MG]] one by one, using the rules for the [[complex numbers|09FTAET5Z4]].\n\n1. Let $z = a + b i$, $w = c + d i$ and $u = e + f i$ be complex numbers, with $a, b, c, d, e, f$ real.\nThe operations are $z + w = (a + c) + (b + d) i$ and $z w = (a c - b d) + (a d + b c) i$.\n2. All computations with $a, \\dots, f$ below take place in $\\R$. They use only the field axioms of $\\R$, by [[the reals form a field|0FGENPMPJX]].\nTwo complex numbers are compared through their real and imaginary parts, by [[equality of complex numbers|05V7J4WD1C]].\n3. Addition is commutative and associative. Indeed, both properties hold for the real parts and for the imaginary parts separately, since they hold in $\\R$.\n4. Let $0 = 0 + 0 i$. Then $z + 0 = (a + 0) + (b + 0) i = z$.\n5. Let $-z = (-a) + (-b) i$. Then $z + (-z) = 0 + 0 i = 0$.\nSo $(\\C, +)$ is an abelian group with identity $0$.\n6. Multiplication is commutative. Indeed, $w z = (c a - d b) + (c b + d a) i$, and this equals $z w$ because real multiplication is commutative.\n7. Multiplication is associative. We have $z w = (a c - b d) + (a d + b c) i$. Multiplying by $u$ gives\n$$(z w) u = (a c e - b d e - a d f - b c f) + (a c f - b d f + a d e + b c e) i.$$\n8. Also $w u = (c e - d f) + (c f + d e) i$. Multiplying $z$ by it gives\n$$z (w u) = (a c e - a d f - b c f - b d e) + (a c f + a d e + b c e - b d f) i.$$\nThe real parts in steps 7 and 8 agree, and so do the imaginary parts. So $(z w) u = z (w u)$.\n9. Let $1 = 1 + 0 i$. Then $1 \\cdot z = (1 a - 0 b) + (1 b + 0 a) i = z$. By step 6, also $z \\cdot 1 = z$.\n10. Multiplication distributes over addition. We have $w + u = (c + e) + (d + f) i$, so\n$$z (w + u) = (a c + a e - b d - b f) + (a d + a f + b c + b e) i.$$\n11. Also $z w + z u = \\bigl((a c - b d) + (a e - b f)\\bigr) + \\bigl((a d + b c) + (a f + b e)\\bigr) i$. This has the same parts as step 10.\nSo $z (w + u) = z w + z u$. By step 6, also $(w + u) z = w z + u z$.\n12. Steps 3 to 11 show that $\\C$ is a [[ring|0894J980RK]], and step 6 shows that it is a [[commutative ring|0GYGHKX834]].\n13. We have $0 \\neq 1$, because $\\operatorname{Re} 0 = 0$ and $\\operatorname{Re} 1 = 1$ differ.\n14. Now let $z = a + b i \\neq 0$. Then $a \\neq 0$ or $b \\neq 0$, by equality of complex numbers.\nSquares of real numbers are nonnegative, and the square of a nonzero real number is positive. So $s = a^2 + b^2 \u003e 0$.\n15. Let $v = \\dfrac{a}{s} + \\dfrac{-b}{s} i$, which makes sense because $s \\neq 0$. Then\n$$z v = \\left( a \\cdot \\frac{a}{s} - b \\cdot \\frac{-b}{s} \\right) + \\left( a \\cdot \\frac{-b}{s} + b \\cdot \\frac{a}{s} \\right) i = \\frac{a^2 + b^2}{s} + 0 i = 1.$$\n16. So every nonzero $z$ has a multiplicative inverse $z^{-1} = v$.\nTogether with steps 12 and 13, this proves that $\\C$ is a field.\n\n## Idea\n\nEvery field axiom reduces to identities among real numbers, by comparing real and imaginary parts.\nThe only axiom that needs an idea is the inverse.\nThe number $a^2 + b^2$ is a positive real, so we can divide by it, and $(a + b i)(a - b i) = a^2 + b^2$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "12D99"
    ],
    "name": "Proof that the complex numbers form a field",
    "proves": "0FRSR10DWF",
    "sources": [
      {
        "locator": "Abstract and algebraic definitions",
        "url": "https://en.wikipedia.org/wiki/Complex_number"
      }
    ],
    "tag": "P4N3NXZ6XS",
    "uses": [
      "09FTAET5Z4",
      "0FGENPMPJX",
      "05V7J4WD1C",
      "0QA8CHH8MG",
      "0GYGHKX834",
      "0894J980RK"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/P4N3NXZ6XS/",
  "links": [
    {
      "field": "proves",
      "target": "0FRSR10DWF"
    },
    {
      "field": "uses",
      "target": "09FTAET5Z4"
    },
    {
      "field": "uses",
      "target": "0FGENPMPJX"
    },
    {
      "field": "uses",
      "target": "05V7J4WD1C"
    },
    {
      "field": "uses",
      "target": "0QA8CHH8MG"
    },
    {
      "field": "uses",
      "target": "0GYGHKX834"
    },
    {
      "field": "uses",
      "target": "0894J980RK"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/P4N3NXZ6XS.md",
  "msc": [
    "12D99"
  ],
  "name": "Proof that the complex numbers form a field",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/P4N/P4N3NXZ6XS.md",
  "status": null,
  "tag": "P4N3NXZ6XS",
  "uri": "m2://P4N3NXZ6XS",
  "url": "https://math2.tamnd.com/wiki/proof-that-the-complex-numbers-form-a-field/",
  "writing": {
    "tag": "P4N3NXZ6XS",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}