Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I go about proving that the set $A=\{ (-2)^n : n \in \mathbb{N} \}$ is unbounded?

I started by showing (using induction) that if $n$ is even then $(-2)^n=2^n$, and since $\mathbb{N}$ is unbounded in $\mathbb{R}$ oubviously $A$ is unbounded. Am I in the right direction? Is it enought to say that for every $n \in \mathbb{N}$ there exist $k \in \mathbb{N}$ such that $(-2)^n=2^n \lt 2^k=(-2)^k$?

If I'm write then showing that $A$ is unbounded from beloew is almot similar.

Any thoughts? Thanks!

share|improve this question
    
You may want to show that $2^n\ge n$. Then you'd be done. I'm not sure what you're asking in your second to last question. –  David Mitra Mar 26 '12 at 12:53
    
One way to rephrase the statement "$A$ is unbounded" is "there is no positive real number $r$ such that $A$ is contained in the interval $[-r, r]$." That is, there is no $r>0$ such that $|x|\leq r$ for all $x\in A$. Do you see why this is true for your particular set $A$? –  Brad Mar 26 '12 at 12:53
    
I can see that, but how do I prove it? Also, it is enough to say that for every element in the set there exist a greater element? Thanks! –  yotamoo Mar 26 '12 at 12:59
    
Arguing directly, you want to show that given any integer $n$, there is an element in the set greater than $n$ (here, it would be either $(-2)^n$ or $(-2)^{n+1}$). –  David Mitra Mar 26 '12 at 13:01
add comment

1 Answer

up vote 2 down vote accepted

The argument in your first sentence is correct. But you need to justify the statement "obviously $A$ is unbounded". Why is this so? Well, of course, because $A$ contains all even powers of $2$ and the set of all even powers of 2 is unbounded from above. So, what you need to show is that:

given any positive integer $n$, there is an even positive integer $k$ so that $2^{ k}\ge n$.

The above claim would follow from a more general statement:

given any positive integer $n$, the inequality $2^n\ge n$ holds.

You can prove this statement by induction.


So, I would start of by proving the Lemma: $2^n\ge n$ for any positive integer $n$. Then to show $A$ is unbounded, argue as follows:

Let $n$ be a positive integer.

If $n$ is even, then $2^n=(-2)^n$ is in $A$ and from the lemma we have $(-2)^n=2^n\ge n$.

If $n$ is odd, we have $2^{n+1}=(-2)^{n+1}$ is in $A$ and, again from the lemma, $(-2)^{n+1}=2^{n+1}\ge n+1>n$.

So in either case, we have an element in $A$ greater than or equal to $n$. Since $n$ was an arbitrary positive integer, it follows that $A$ is not bounded from above.


Note that, to show that your $A$ is not bounded, you need only show that $A$ is not bounded from above (or show that it is not bounded from below). Of course, if you were required to show that $A$ is unbounded in both directions, you'd use a similar argument to show that $A$ is not bounded from below.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.