Here is an adaptation of Lang’s proof of Gelfond’s theorem to prove that is not rational – needed for http://mathoverflow.net/questions/138247/prove-that-sqrt2-sqrt2-is-an-irrational-number-without-using-a-theorem/138291#138291. Consider the functions . Let , where is big enough (we shall see how large should be below). Suppose that is rational. Then is a rational number for every .

Lemma 1. Let us have a homogeneous system of linear equations with variables with integer coefficients, the abs. value of each coefficient is at most . Then there exists a non-trivial integer solution with provided .

Proof (half a page) see in Lang “Algebra”.

Consider the function with rational coefficients. Set (choose so that it is an integer). By Lemma 1, we can find integers such that all derivatives , , moreover absolute values of are bounded by something like for very large by Lemma 1. Since are algebraically independent (here we use that is irrational), the function is not identically 0. Let be the smallest integer such that all derivatives up to the order of are 0 at each , but the derivative for some (for simplicity let , the proof is the same for any ). Then belongs to the field , that is where are rational numbers. Moreover the least common denominator of is bounded by for some constant . The conjugate of (i.e.,) is bounded by for some constant . Then the norm of in , i.e. the product of and its conjugate, which is an integer , is bounded by . Now consider the function . Then differs from by some factor bounded by (which is “small”). By the maximum modulus principle, the value of is bounded above by the value of where is a complex number on a circle of large enough radius (we need only that is bigger than each ). Taking large enough, we get that is bounded from above by a number of the form . For ( is an appropriate constant) this gives a contradiction with the previous inequality for (the denominator grows much faster than the numerator, so the fraction cannot be always greater than 1).

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