# Lindemann-Weierstrass Theorem Watch

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Hey, does anyone know a proof of this theorem - which makes it very easy to prove e (and pi) are transcendental. any help/ideas would be v. useful. thanks

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(Original post by

Hey, does anyone know a proof of this theorem - which makes it very easy to prove e (and pi) are transcendental. any help/ideas would be v. useful. thanks

**davidreesjones**)Hey, does anyone know a proof of this theorem - which makes it very easy to prove e (and pi) are transcendental. any help/ideas would be v. useful. thanks

the Lindemann-Weierstrass theorem states that if x_1,...,x_n are algebraic numbers which are linearly independent over the rational numbers, then are algebraically independent over the algebraic numbers.

The transcendence of e and pi are direct corollaries of this theorem. Suppose x is a nonzero algebraic number; then {0, x} is a set of distinct algebraic numbers, and so the set {e^0, e^x} = {1, e^x} is linearly indepedent over the algebraic numbers, and so e^x is immediately seen to be transcendental. In particular, e^1 = e is transcendental. Also, if b = e^ix is transcendental, so are the real and imaginary parts of b, Re(b) = (b + b−1)/2 and Im(b) = (b − b−1)/2i, and hence cos(x) = Re(b) and sin(x) = Im(b) are also. Therefore, if pi were algebraic, cos(pi) = −1 and sin(pi) = 0 would be transcendental, which proves by contradiction pi is not algebraic, and hence is transcendental.

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Euler, you haven't proven Lindemann's theorem, particularly because it is difficult to prove. Obviously, once you have proven it you can do nice things with it.

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**davidreesjones**)

Hey, does anyone know a proof of this theorem - which makes it very easy to prove e (and pi) are transcendental. any help/ideas would be v. useful. thanks

look at about 17:43. have fun with it !

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(Original post by

Euler, you haven't proven Lindemann's theorem, particularly because it is difficult to prove. Obviously, once you have proven it you can do nice things with it.

**J.F.N**)Euler, you haven't proven Lindemann's theorem, particularly because it is difficult to prove. Obviously, once you have proven it you can do nice things with it.

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thanks - i saw an alternative expression of this theory

Let {alpha1, alpha2....alpha n} be n distinct algebraic numbers

Let {beta1, beta2....beta n} be n distinct algebraic numbers, not equal to zero

(n is an integer)

then beta1 e^alpha1 + beta2 e^alpha2 + ..... + betan e^alpha n is not equal to zero.

Then prove by contradiction e is transcendental

Assume e is algebraic (clearly 0,1,-1 are algebraic

Let beta1=e

Let beta2 = 1

Let alpha1=-1

Let alpha 2 = 0

(n=2)

Then by Lindemann-Weierstrass Theorem

e*(e^-1 ) + (-1)e^0 is non-zero

but e*(e^-1 ) + (-1)e^0 = e/e - 1(0) = 1-1=0

so we have a contradiction and must be transcendental

So in order to prove e is transcendental all be have to do is to prove Lindemann-Weierstrass theorem for a specific case, n=2 (and alpha2-alpha1 = 1)

I thought this might be relatively easy to prove (perhaps using the simple continued fraction expansion of e?)

Let {alpha1, alpha2....alpha n} be n distinct algebraic numbers

Let {beta1, beta2....beta n} be n distinct algebraic numbers, not equal to zero

(n is an integer)

then beta1 e^alpha1 + beta2 e^alpha2 + ..... + betan e^alpha n is not equal to zero.

Then prove by contradiction e is transcendental

Assume e is algebraic (clearly 0,1,-1 are algebraic

Let beta1=e

Let beta2 = 1

Let alpha1=-1

Let alpha 2 = 0

(n=2)

Then by Lindemann-Weierstrass Theorem

e*(e^-1 ) + (-1)e^0 is non-zero

but e*(e^-1 ) + (-1)e^0 = e/e - 1(0) = 1-1=0

so we have a contradiction and must be transcendental

So in order to prove e is transcendental all be have to do is to prove Lindemann-Weierstrass theorem for a specific case, n=2 (and alpha2-alpha1 = 1)

I thought this might be relatively easy to prove (perhaps using the simple continued fraction expansion of e?)

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