## 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. $$\blacksquare$$