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 pq∈Q?
Solution: Yes! And the proof is cool stuff:
Consider the case where p=q=√2. Let n=pq=√2√2. If n is a rational number, then we are done. Now suppose that n is irrational, then we have n√2=(√2√2)√2 or in other words n√2=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′=(√2−1)q is an integer. We have 0≤q′<q. However, (√2−1)q′=(1−2(√2−1))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+ak−1xk−1+…+a1x+a0 where ai∈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=mn with n≠2 and gcd(m,n)=1. Then we have:
(pq)k+ak−1(pq)k−1+…+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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