An Easy Inequality that is Hard to Prove

[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.

 

[Simon Bridge]

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

 

[STEMucator]

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?

 

[bins4wins]

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

 

[bins4wins]

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?

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

 

[STEMucator]

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

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

 

[bins4wins]

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

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.

 

[STEMucator]

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.

Well take m = 3/2 and c = 3/2. cm = 9/4 &gt; (3/2)^{(3/2)} = m^c

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

 

[bins4wins]

Well take m = 3/2 and c = 3/2. cm = 9/4 &gt; (3/2)^{(3/2)} = m^c

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

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.

 

[STEMucator]

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.

 

[bins4wins]

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

 

[STEMucator]

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

 

[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.

 

[STEMucator]

Can you prove to me this property?

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.

 

[STEMucator]

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.

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.

 

[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.

 

[STEMucator]

Let me clarify. I was just saying that you gave c a value, which we should not do in the proof.

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...

 

[bins4wins]

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.

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.

 

[SammyS]

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

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

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.

 

[Simon Bridge]

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

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

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.

 

[MarneMath]

I'm assuming by basic properties of real numbers, I assume he is really talking about the definition of a field. Thus closure, communutativity, associativity, etc

Anyway, I think the OP should consider what Z is trying to get you to realize and I think you are struggling very hard to understand the very intent of your own question. It would be easy to prove this by complete induction if you'are allowed too, since from my understanding is, if there is a exist an m and c that statisfy the first inequality (which z has shown), then it's possible to to find a k greater than c that statisfies the second. Ponder that for a bit.

 

[bins4wins]

Thanks. I would use induction but I have too prove this for all real numbers k > c. Also, to clarify what I can use, I can use, like MarneMath said, the properties of fields, the least upper bound property of real numbers, the properties of metric spaces regarding the reals, and sequences/convergence. If you are familiar with Rudin's text, I am free to use any of the material in the first two chapters and the sequence portion of the third chapter. It's just important that I avoid logarithms, exponentials, and derivatives. I hope this clarifies.