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  "body": "\n## Statement\n\nLet $D$ be a [[division ring|0WZXK05ME7]], and let $a, b \\in D$ with $a \\ne 0$, $b \\ne 0$ and $ab \\ne 1$.\nThen $a^{-1} + (b^{-1} - a)^{-1}$ is invertible, and\n$$a - \\left(a^{-1} + (b^{-1} - a)^{-1}\\right)^{-1} = aba.$$\n\n## Remarks\n\nThe element $b^{-1} - a$ is nonzero because $ab \\ne 1$.\nThe proof is a direct factorization: $a^{-1} + (b^{-1} - a)^{-1} = a^{-1}\\left((b^{-1} - a) + a\\right)(b^{-1} - a)^{-1} = a^{-1} b^{-1} (b^{-1} - a)^{-1}$.\nIts inverse is $(b^{-1} - a) b a = a - aba$, and subtracting it from $a$ gives $aba$.\nFor example, in $\\Q$ with $a = 2$ and $b = 3$: $1/2 + (1/3 - 2)^{-1} = 1/2 - 3/5 = -1/10$, and $2 - (-10) = 12 = 2 \\cdot 3 \\cdot 2$.\nThe hypothesis $ab \\ne 1$ is needed, since for $b = a^{-1}$ the term $(b^{-1} - a)^{-1}$ is not defined.\n\n## Explanations\n\n### L1 human\n\nThe identity writes the product $aba$ using only addition, subtraction and inverses.\nSo a map that respects sums, inverses and $1$ automatically respects these \"sandwich\" products.\n\n### L2 human\n\nHua used the identity to show that a map $\\sigma$ between division rings that is additive, sends $1$ to $1$ and satisfies $\\sigma(x^{-1}) = \\sigma(x)^{-1}$ preserves $aba$.\nFrom this one deduces Hua's theorem: such a map is a ring homomorphism or an antihomomorphism.\nThe identity also underlies the use of the product $aba$ in Jordan algebras, where $aba$ plays the role of the quadratic operator.\n",
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