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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Suppose that $\\pi$ is not [[transcendental|00SW8B91E7]] over $\\Q$. Then $\\pi$ is an [[algebraic number|0MB4TGMXG6]].\n2. The number $i$ is algebraic, since it is a root of $x^2 + 1$.\n3. The [[algebraic numbers form a field|0QFTETCHKB]], so $\\alpha_1 = i\\pi$ is algebraic. Also $\\alpha_2 = 0$ is algebraic, and $\\alpha_1 \\neq \\alpha_2$ because $\\pi \\neq 0$.\n4. By [[Euler's formula|04R3CY41VJ]], $e^{i\\pi} = \\cos \\pi + i \\sin \\pi = -1$. Also $e^0 = 1$, from the series of the exponential.\n5. With $\\beta_1 = \\beta_2 = 1$, which are algebraic, we get $\\beta_1 e^{\\alpha_1} + \\beta_2 e^{\\alpha_2} = -1 + 1 = 0$.\n6. By the [[Lindemann–Weierstrass theorem|0JNCBFMPXR]], this forces $\\beta_1 = \\beta_2 = 0$. This contradicts $\\beta_1 = 1$.\n7. So $\\pi$ is transcendental over $\\Q$.\n\n## Idea\n\nEuler's identity $e^{i\\pi} + 1 = 0$ is a linear relation, with algebraic coefficients, between exponentials of $i\\pi$ and $0$. The Lindemann–Weierstrass theorem forbids such a relation when the exponents are distinct algebraic numbers. So $i\\pi$, and therefore $\\pi$, cannot be algebraic.\n",
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