Editing Lindemann–Weierstrass theorem (section)

== {{anchor|Transcendence of ''e'' and π}} Transcendence of {{math| ''e'' }} and {{pi}} ==

{{also|e (mathematical constant)|Pi}}

The [[transcendental number|transcendence]] of {{math| [[e (mathematical constant)|''e'']] }} and {{pi}} are direct corollaries of this theorem.

Suppose {{math| α }} is a non-zero algebraic number; then {{math| {α} }} is a linearly independent set over the rationals, and therefore by the first formulation of the theorem {{math| {''e''<sup>α</sup>} }} is an algebraically independent set; or in other words {{math| ''e''<sup>α</sup> }} is transcendental. In particular, {{math| ''e''<sup>1</sup> {{=}} ''e'' }} is transcendental. (A more elementary proof that {{math| ''e'' }} is transcendental is outlined in the article on [[transcendental number]]s.)

Alternatively, by the second formulation of the theorem, if {{math| α }} is a non-zero algebraic number, then {{math| {0, α} }} is a set of distinct algebraic numbers, and so the set {{math| {''e''<sup>0</sup>,&nbsp;''e''<sup>α</sup>}&nbsp;{{=}}&nbsp;{1,&nbsp;''e''<sup>α</sup>} }} is linearly independent over the algebraic numbers and in particular {{math| ''e''<sup>α</sup> }} cannot be algebraic and so it is transcendental.

To prove that {{pi}} is transcendental, we prove that it is not algebraic. If {{pi}} were algebraic, {{pi}}''i'' would be algebraic as well, and then by the Lindemann–Weierstrass theorem {{math| ''e''<sup>{{pi}}''i''</sup> {{=}} −1 }} (see [[Euler's identity]]) would be transcendental, a contradiction. Therefore {{pi}} is not algebraic, which means that it is transcendental.

A slight variant on the same proof will show that if {{math| α }} is a non-zero algebraic number then {{math| sin(α), cos(α), tan(α) }} and their [[hyperbolic function|hyperbolic]] counterparts are also transcendental.