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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 pqQ?

Solution: Yes! And the proof is cool stuff:
Consider the case where p=q=2. Let n=pq=22. If n is a rational number, then we are done. Now suppose that n is irrational, then we have n2=(22)2 or in other words n2=2 (LHS in the form of pq while RHS is a rational number), which completes our proof.


/*addition*/
A cool proof on the irrationality of 2!
Suppose that 2 is rational. Let q be the smallest positive integer such that q=(21)q is an integer. We have 0q<q. However, (21)q=(12(21))q is also an integer, which contradicts the minimality of q! Hence 2 must be irrational.

/** more addition **/
In general we have
Given f(x)=xk+ak1xk1++a1x+a0 where aiZ. 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=mn with n2 and gcd(m,n)=1. Then we have:
(pq)k+ak1(pq)k1++a1(pq)+a0=0
which simplifies to
qka0=pk+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|pk, which suggests that there exist a prime P such that P|q and P|pk. The latter means that P|p, hence contradicts the assumption that p,q are relatively prime.

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