Math Is Fun Forum

Kurre

Member

Registered: 2006-07-18

Posts: 280

Inequality a^p>b^p

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

N. It also follow by axiom 1, that if a^p>b^p for p[tex]\in[/tex]N, then a>b. Now write:

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

to n decimal places.
thus we have:

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.

Last edited by Kurre (2008-05-17 00:53:43)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Inequality a^p>b^p

Your method is exactly right.  Start with the rationals, then use rationals to approximate the real numbers.  And while there is nothing wrong, the following statement is unjustified:

Certainly it holds for any finite case, but why does it hold in the limit?  You need to use the definition of a convergent sequence here.  And also, you don't need to just use the decimal expansion.  Any sequence of rationals that converges to your real number will work.

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.

The lemma, as stated, only holds in finite cases.

"Simpler" proof:

Note that since a > b, a/b > 1, so (a/b)^p > 1 which is really a^p / b^p > 1.  Multiply both sides by b^p and you get a^p > b^p.

Now here is your challenge: why can't you use this?

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Inequality a^p>b^p

Thanks for the reply Ricky!

You need to use the definition of a convergent sequence here.

Unfortunately I have never worked with definitions of convergent series before, but I can give it a try. Could you please give a short startup/link to get me going??

Note that since a > b, a/b > 1, so (a/b)^p > 1 which is really a^p / b^p > 1.  Multiply both sides by b^p and you get a^p > b^p.

Now here is your challenge: why can't you use this?

I think that is the wrong bit, since wouldnt stating that "a/b>1 implies that (a/b)^p>1^p=1", use the theorem that we are trying to prove?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Inequality a^p>b^p

I think that is the wrong bit, since wouldnt stating that "a/b>1 implies that (a/b)^p>1^p=1", use the theorem that we are trying to prove?

Right.

Is this question learning on your own, or is it part of a class?  If it's part of a class, then there should be some other way to do it.  How much have you covered in terms of real analysis (sometimes called "Advanced Calculus", which is a horrible name)?

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Inequality a^p>b^p

Is this question learning on your own, or is it part of a class?  If it's part of a class, then there should be some other way to do it.  How much have you covered in terms of real analysis (sometimes called "Advanced Calculus", which is a horrible name)?

Yes, im doing it on my own, and I dont think I have been doing a lot real analysis. Just the basic high school(?) calculus, like basic integration (substitution/parts), basic differential equations etc. In class I have not done anything close to investigating the definition of a limit or definition of a convergent serie

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Inequality a^p>b^p

In class I have not done anything close to investigating the definition of a limit or definition of a convergent series

No time like the present.  However, it will probably be simpler to work this problem another way.  As it turns out, you define a real number to be the supremum of a certain type of set, known as a Dedekind cut (look it up or ask if you want more about this).  I won't go into details, except for what a supremum is, which will be called "sup" from here on.

Given a set of real numbers, A, sup(A) = s is a real number such that:

(i) s ≥ a for all a in A
(ii) if s' ≥ a for all a in A, then s' ≥ s.

An informal definition would be that the sup of A is the smallest number that is greater than all elements in A.  It is also known as the "least upper bound", which is a more intuitive name.  Let's take a few examples:

[0,1] This sup is 1.
(0, 1) This sup is also 1.
{n : n is an integer} This set does not have a sup.

It is a theorem that every set of real numbers that is bounded above has a sup, and that this sup is unique.  So we now use this:

Let A = {r^s : r < a and s < p with both r and s being rational}
and B = {t^u : t < b and u < p with both t and u being rational}

a^p = sup A
b^p = sup B

Now it takes work to show that this definition does in fact make sense.  However, all this work is done in the construction of the real numbers with Dedekind cuts.  So using this, try to show that:

(a) a^p is not less than b^p
(b) a^p is not equal to b^p

For the first, it suffices to show that every element of B is an element of A (why?).  For the second, it suffices to show that there exists an element of A that greater than every element of B, and not in B.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."