3 ms·
There are two ways we can go about this. We can either take a closer look at the Lindemann theorem, or we can talk about the definition of “transcendental numbe
by klodolph 7y ago
There are two ways we can go about this. We can either take a closer look at the Lindemann theorem, or we can talk about the definition of “transcendental number”. I’m not going to put a full proof here.
If you look at the Lindemann theorem, you can transform the equation so that it uses π instead of e. Multiply all of the exponents by i (which is algebraic!) and then use Euler’s identity. You end up with the same formula, but with π instead of e.
However, if we already know that π is transcendental (which is proven by the Lindemann theorem using the above technique), we can rewrite any linear combination of B = {1, π, π², π³, …} as P(π) where P is a polynomial with coefficients in ℚ. Because π is transcendental, we know that P(π)=0 only if P is the zero polynomial (that is the definition of transcendental number).
In general, one of the big tricks here is that the set of polynomials is a vector space, and the powers B = {1, x, x², x³, …} span the entire vector space.