A more elementary proof that e is transcendental is outlined in the article on transcendental numbers. Modular conjecture An analogue of the theorem involving the modular function j was conjectured by Daniel Bertrand in , and remains an open problem. Let q1, If a1,
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So we are in the situation of Lemma A. To reach a contradiction it suffices to see that at least one of the coefficients is non-zero. This is seen by equipping with the lexicographic order and by choosing for each factor in the product the term with non-zero coefficient which has maximum exponent according to this ordering: the product of these terms has non-zero coefficient in the expansion and does not get simplified by any other term.
This proves Lemma B. We will show that this leads to contradiction and thus prove the theorem. Let us denote the distinct roots of this polynomial a i 1, We form the polynomial in the variables x11, Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in a i i, By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b 1 , Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof.
Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0,
In order to complete the proof we need to reach a contradiction. This proves Lemma A. Lemma B. Thus after having grouped the terms with the same exponent we see that the set of the exponents form a complete set of conjugates and, if two terms have conjugate exponents, they are multiplied by the same coefficient. So we are in the situation of Lemma A. To reach a contradiction it suffices to see that at least one of the coefficients is non-zero.
Lindemann–Weierstrass theorem explained