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  "body": "\n## Statement\n\nLet $a$ and $b$ be complex [[algebraic numbers|0MNF0BYFQC]] with $a \\ne 0$, $a \\ne 1$, and $b \\notin \\Q$.\nLet $\\log a$ be any value of the [[complex logarithm|02QGGD5G87]] of $a$.\nThen the number $a^b = e^{b \\log a}$ is [[transcendental|0VTQXTCR2G]].\n\n## Remarks\n\nThe proof builds an auxiliary function $F(z) = \\sum_{i,j} c_{ij} e^{(i + j b) z \\log a}$ with integer coefficients $c_{ij}$, not all zero, chosen so that $F$ vanishes at many integer points.\nIf $a^b$ were algebraic, the values of $F$ and its derivatives at integers would be algebraic numbers of controlled size.\nAn analytic estimate and a lower bound for nonzero algebraic numbers then force too many zeros, which is a contradiction.\nExamples: $2^{\\sqrt{2}}$ is transcendental, and so is $e^{\\pi}$, since $e^{\\pi} = (-1)^{-i}$ with $\\log(-1) = i\\pi$.\nAnother consequence: $\\log 3 / \\log 2$ is transcendental, since it is irrational and $2^{\\log 3 / \\log 2} = 3$.\nThe hypothesis $b \\notin \\Q$ is necessary: $2^{1/2} = \\sqrt{2}$ is algebraic.\nThe theorem answers Hilbert's seventh problem.\n[[Baker's theorem|WMBJSNG57F]] generalizes it to linear forms in several logarithms.\n\n## Explanations\n\n### L1 human\n\nRaise an algebraic number to an irrational algebraic power, for example $2^{\\sqrt{2}}$.\nThe theorem says that the result is never a root of a polynomial with integer coefficients.\nSo numbers such as $2^{\\sqrt{2}}$ and $e^{\\pi}$ are transcendental, like $\\pi$ and $e$.\n\n### L2 human\n\nAn equivalent form: if $\\alpha$ and $\\beta$ are nonzero algebraic numbers and $\\log \\alpha$ and $\\log \\beta$ are linearly independent over $\\Q$, then they are linearly independent over the algebraic numbers.\nThis is the case of two logarithms in Baker's theorem.\nTogether with the [[Lindemann–Weierstrass theorem|0JNCBFMPXR]], it is one of the classical results of transcendence theory.\nThe question whether $e + \\pi$ or $e \\pi$ is transcendental is still open.\n",
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  "name": "Gelfond–Schneider theorem",
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