Monday, September 1, 2014

a bit of something: The power of irrational number

I feel that the title is misleading in so many level :P. Anyway here I'm not intending to praise irrational number (as the term "power" suggests), but rather there is a discussion that I came across about irrational numbers that is quite interesting!

Lemma: Let \(p,q\) be irrational numbers. If we raise an irrational number to an irrational power, can the result be rational? I.e. Are there \(p,q\) such that \(p^q \in \mathbb{Q}\)?

Solution: Yes! And the proof is cool stuff:
Consider the case where \(p=q= \sqrt{2}\). Let \(n = p^q = \sqrt{2}^{\sqrt{2}}\). If \(n\) is a rational number, then we are done. Now suppose that \(n\) is irrational, then we have \(n^{\sqrt{2}} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} \) or in other words \(n^{\sqrt{2}} = 2 \) (LHS in the form of \(p^q\) while RHS is a rational number), which completes our proof. \( \blacksquare\)


/*addition*/
A cool proof on the irrationality of \( \sqrt{2} \)!
Suppose that \(\sqrt{2}\) is rational. Let \(q\) be the smallest positive integer such that \(q' = (\sqrt{2} - 1)q \) is an integer. We have \(0 \leq q' < q\). However, \((\sqrt{2} - 1)q' = (1-2(\sqrt{2}-1))q\) is also an integer, which contradicts the minimality of \(q\)! Hence \(\sqrt{2}\) must be irrational. \(\blacksquare\)

/** more addition **/
In general we have
Given \(f(x) = x^k + a_{k-1}x^{k-1} + \ldots + a_1x+a_0\) where \(a_i \in \mathbb{Z}\). Then if \(f(x)\) has real roots, they must be of integer or irrational numbers.

Proof:
Let there be (how poetic) \(f(x)\) such that there exists \(x = \frac{m}{n} \) with \(n \not=2 \) and \(\text{gcd}(m,n) = 1\). Then we have:
\( (\frac{p}{q})^k + a_{k-1}(\frac{p}{q})^{k-1} + \ldots + a_1(\frac{p}{q}) + a_0 = 0 \)
which simplifies to
\( -q^ka_0 = p^k + qA \) where A is the remaining terms, the black box, the pandora box
We know that LHS is divisible by \(q\), hence RHS must also be divisible by q. That means \(q | p^k\), which suggests that there exist a prime \(P\) such that \(P | q\) and \(P | p^k\). The latter means that \(P | p\), hence contradicts the assumption that \(p,q\) are relatively prime. \(\blacksquare\)