Homework Help: An Easy Inequality that is Hard to Prove

1. Nov 10, 2012

bins4wins

Prove that if $m > 1$ such that there exists a $c > 1$ that satisfies
$$cm < m^c$$
then for any $k > c$
$$km < m^k$$
holds. Prove this without using logarithms or exponents or calculus. Basically using the properties of real numbers to prove this.

One attempt I have tried, but didn't seem to work, was trying to show this inequality using integers, then moving on to rationals, and ultimately real numbers.

Last edited: Nov 10, 2012
2. Nov 10, 2012

Simon Bridge

I think you need to start by listing the properties you are restricted to using.
What are "the properties of real numbers"?

3. Nov 10, 2012

Zondrina

Hmm, do you mean like if $m \in \mathbb{R}$ then any constant multiple of a real number is less than the number raised to the constant?

4. Nov 10, 2012

bins4wins

No Logarithms, no exponential function, continuity of functions cannot be used, and calculus cannot be used. Mainly algebraic manipulations.

5. Nov 10, 2012

bins4wins

No, because that statement is not true. The assumption is important.

6. Nov 10, 2012

Zondrina

It's certainly true in some cases actually. Try a few yourself. You'll notice something.

7. Nov 10, 2012

bins4wins

Yes its true in some cases, but not in all. But I'm trying to prove that its true in all cases given the assumption.

8. Nov 10, 2012

Zondrina

Well take m = 3/2 and c = 3/2. $cm = 9/4 > (3/2)^{(3/2)} = m^c$

Is that not a counter example to the claim? Did they not say anything about disproving the assumption?

9. Nov 10, 2012

bins4wins

I understand what you are saying, but please carefully read what I asked for. The assumption is something we take as true, and the following statement is what we want to prove given that the assumption is true.

10. Nov 10, 2012

Zondrina

Well if I'm reading your question correctly, there are two different questions. If that's the case, then you can prove the first one wrong by counter example. The second one is doable with the principle of mathematical induction ( As it is really over the interval where the inequality is true ).

Otherwise I don't believe I'd know, sorry.

11. Nov 10, 2012

bins4wins

Yeah I think there's a misunderstanding. I edited the statement. It was just one question.

12. Nov 10, 2012

Zondrina

Ahhh now your quantifiers make sense to me. So let m be an arbitrary real number greater than 1.

Now, since by hypothesis there exists a c>1, we choose c=2 for simplicities sake.

Then we want to show 2m < m^2 right?

EDIT : Note : It would work for any arbitrary c as a well. We could've just as easily shown cm < m^c for any real number c.

EDIT 2 : Use a very simple property of the real number system that 0 < m^2 - 2m

Last edited: Nov 10, 2012
13. Nov 10, 2012

bins4wins

But the property you suggested is a simple consequence of the assumption and doesn't prove the result I need. If you choose c = 2, then we already know 2m < m^2. If for example we chose c = 2, then we have to show that km < m^k for all k > 2. But our c must remain arbitrary.

Last edited: Nov 10, 2012
14. Nov 10, 2012

Zondrina

It's a basic law that governs the positive real numbers. There's three of these basic laws actually that govern the whole real number system :

1. If 0 < a and 0 < b, then 0 < b + a ( or b - a ).
2. If 0 < a and 0 < b, then 0< ab ( or ba ).
3. For any real number a, only one of the following three is true :
(i) 0 < a
(ii) 0 = a
(iii) 0 <-a

So obviously m^2 > 2m implies that m^2 - 2m > 0 which implies m(m-2) > 0, nes pas?

You take over from here.

15. Nov 10, 2012

Zondrina

You said there EXISTED a c, that means we can CHOOSE a c at our own discretion. Otherwise you're going to wind up having to factor something like m^c - cm > 0 which is a pain.

16. Nov 10, 2012

bins4wins

All you did was reorder terms in your proof. I'm sorry, but I really don't see where you're going with it. How does m(m^c - 1) > 0 imply that m^k > km for all k > c.

Last edited: Nov 10, 2012
17. Nov 10, 2012

Zondrina

This question is phrased quite poorly... If the statement is true in all cases, no matter what c we pick, then the inequality should hold. So I picked a c, a very conveniently picked one actually just to prove a point.

Your hypothesis is wrong then simply by counter-example, but the second hypothesis is true by induction.

I definitely would not see any other way to do this with the way the question is phrased.

EDIT : I mean induction on c > 2 in particular. Also for c < 0....

18. Nov 10, 2012

bins4wins

First off, there EXISTS a c. It is not any c. So you cannot pick any c so that the inequality should hold. I'm sorry, but I really think there's a continuous misunderstanding between you and I, I just want to see what others can offer. No disrespect, just clearly miscommunication. Also you cannot use induction because the exponents can be any reals.

19. Nov 11, 2012

SammyS

Staff Emeritus
If this is your answer to Simon's question, then it falls short.

He didn't ask what you can't use, but essentially asked what properties you can use.

20. Nov 11, 2012

Simon Bridge

Those are what you cannot use - what can you use?[*]

i.e. The following are properties of Real numbers:
1. If 0 < a and 0 < b, then 0 < b + a ( or b - a if b > a ).
2. If 0 < a and 0 < b, then 0 < ab ( or ba ).
3. For any real number a, only one of the following three is true :
(i) 0 < a
(ii) 0 = a
(iii) 0 < -a

What's wrong with starting there?

As for the actual statement to be proved: appears it is not such an easy inequality to even describe...

I'm reading this that you have an $m > 1 \in \mathbb{R}$, we find a $c >1 \in \mathbb{R}\; : \; cm < m^c$ Having found some c that makes the expression true for a given m ... we want to prove that any other real number $k > c$ will also make the expression true.

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[*] first time round you excluded "powers" but this time you didn't - which is it?
This is why a positive list is so useful - in this case, the exercise is about the definition of real numbers.