# Inequality a^p>b^p

1. May 17, 2008

### Kurret

It struck me when reading about the axioms of inequalities, that this inequality wasnt that easy to straight ahead deduce from the axioms.
the inequality to be proven is given p>0, a>b>0, prove that a^p>b^p.
the axioms are:
1. for two real numbers x and y exactly one of the following cases i true:
a) x=y b) x<y c)x>y
2. if x<y, then for all z x+z<y+z
3. x<y, z>0, then xz<yz
4. if x<y and y<z, then x<z
proof:
lemma: for x,y,z,w>0, if x<y, w<z, then xw<yz.
By axiom 2 we have xw<yw, and again by axiom 2 wy<yz, and by axiom 4 xw<yz. Now assume p is natural, then for p=1 the inequaliy holds. Assume it holds for p=n, ie a^n>b^n. then by the lemma we can multiply left side with a, and right side with b and establishing a^(n+1)>b^(n+1), thus by the induction principle it holds for all p $$\in$$N. It also follow by axiom 1, that if a^p>b^p for p$$\in$$N, then a>b. Now write:
$$a^n=(a^{\frac{n}{m}})^m>(a^{\frac{n}{m}})^m=b^n$$
ad by earlier result, we get that a^p>b^p for rational p.
Now, let the sequence {p_k} of rational numbers be the numbers of the decimal expansion of the real number p, ie $$\sum_{k=1}^n p_k = p$$ to n decimal places.
thus we have:
$$a^p=\lim_{n\to \infty}a^{\sum_{k=1}^n p_k}>\lim_{n\to \infty}b^{\sum_{k=1}^n p_k}=b^p$$
thus a>b implies that a^p>b^p for all and p,a,b>0.

Now my questions are:
Is this proof correct? Im mostly concerned about the last step when introducing a limit. Do I need to state anything about continuity of the function a^p? Or is it sufficient to use the proved lemma, thus I can continue to multiply the inequality with a^(pk)>b^(pk) and continue to infinity to establish the inequality.
Also, thus the introduction of a limit need extra proofs of the limit, if I want to deduce the inequality straight down from the axioms?
I know that I can just state the the function a^x is strictly increasing for a,x>0 and thus a>b -> a^x>b^x, but that does need extra proofs conecrning derivatives etc, right? (and its not so much fun )
Or have I missed a really obvious way to prove this kind of intuitive inequality?
and last, a challenge for you, find a simplier proof.