Monday, 1 September 2014

Analysis for Realsies: 3

I love Community. You love Community. Who doesn't love Community?

Now, if you love Community (premise), then you love Magnitude (inference). And if love Magnitude, then surely you love POP! POP! quizzes (conclusion/pun-ch line), right?

Question 1 (it's the only question): which of these four numbers do we know for certain is irrational?
A) \(\pi+e\).
B) \(e\pi\).
C) \(e^\pi\).
D) All of the above.

Strangely enough, the answer is C: we have no idea if \(\pi+e\) or \( e\pi\) are actually irrational (even though we strongly suspect that they are). Having said that, we do know that at least one of those two numbers is irrational. Assume not, then there exist integers \(p,q\in\mathbb{N}\) such that \(\pi+e=\frac{p}{q}\) and \(m,n\in\mathbb{N}\) such that \(e\pi=\frac{m}{n}\). Then, \begin{align}\frac{p}{q}\pi=\pi^2+e\pi=\pi^2+\frac{m}{n}\Rightarrow 0=\pi^2-\frac{p}{q}\pi+\frac{m}{n}.\end{align} Which is a contradiction, because \(\pi\) is transcendental and (hence) cannot be the solution to some quadratic equation with rational coefficients.

To show that \(e^\pi\) is irrational, we invoke the Gelfond-Schneider theorem. Which says that if \(a,b\) are algebraic numbers with \(a\neq0,1\) and \(b\) irrational, then \(a^b\) is transcendental (and hence necessarily irrational). The fact that \(e^\pi\) is irrational then follows from using the (kinda awesome) fact that \(e^\pi=(-1)^{-\sqrt{-1}}\).


Number theory, she is you =)

No comments:

Post a Comment