Is my proof valid for this inequality problem?

[r0bHadz]

Homework Statement

Prove: if 0<a<b then a^(1/n) < b^(1/n)

Homework Equations

The Attempt at a Solution

I've already proved a^n < b^n in another problem. So I have

Assume a^n < b^n => \sqrt[n^2] a^n &lt; \sqrt[n^2] b^n => a^(1/n) < b^(1/n)

 

[StoneTemplePython]

I don't really see a proof here? How do you know you're allowed to take the $n^2$ root while preserving the inequality?

What about dividing both sides by $a$ or ($a^\frac{1}{n}$) and using the fact that nth roots are non-negative? I.e. comparing $1 \lt \frac{b}{a}$

Then you have something greater than one vs one. Should be easier to work with.

edit: my suggestion would be $c:=\frac{b}{a}$ where $1 \lt c$ but assume for a contradiction that you have $c^\frac{1}{n} \leq 1$ and see what happens

edit 2: if you want to use your existing setup, consider setting $x : =a^\frac{1}{n}$ and $y : =b^\frac{1}{n}$. Now you know that $0\lt a \lt b$. You have two cases to consider-- suppose $x \lt y$ what happens when you use your results about raising to the nth power? You also need to consider the other side of the coin: what happens if $y \leq x$? One case is consistent with desired outcome, but that isn't enough... you need to show that the 'opposite' case raises a contradiction and hence can't be true.

 

[RPinPA]

I've already proved a^n < b^n in another problem.

What exactly did you prove? That if $0 < a < b$ then $a^n < b^n$? For what $n$? Just positive integers? Did you prove the converse, that $a^n < b^n \implies a < b$? I'm guessing not, since that's what you're trying to prove here (in effect).

But then I don't understand your first step. How can you take the n-th root of both sides and then just declare it's true, since that's what you're trying to prove?

Personally, I might appeal to use of a logarithm, first establishing that the log function (with any base) is monotonically strictly increasing.

 

[WWGD]

Maybe you can also use the fact that $x^{1/n}$ is differentiable to see when it is increasing.

 

[Ray Vickson]

Homework Statement

Prove: if 0<a<b then a^(1/n) < b^(1/n)

Homework Equations

The Attempt at a Solution

I've already proved a^n < b^n in another problem. So I have

Assume a^n < b^n => \sqrt[n^2] a^n &lt; \sqrt[n^2] b^n => a^(1/n) < b^(1/n)

As others have pointed out, your use of a $\sqrt[n^2]{\cdot}$ inequality is really just the use of a $(\cdot)^{1/n^2}$ inequality, which is what you are trying to prove.

If I were doing it I would look for a proof by contradiction, using all the results available so far.

 

[jbriggs444]

How about proving a more general lemma:

If $f$ from A onto B is strictly monotone increasing then its inverse, $f^{-1}$ from B onto A exists and is strictly monotone increasing.

 

[r0bHadz]

How about proving a more general lemma:

If $f$ from A onto B is strictly monotone increasing then its inverse, $f^{-1}$ from B onto A exists and is strictly monotone increasing.

A proof by contradiction seems easier to me.

Anyways, guys I'm confused when it comes to writing proofs.

So in a previous proof I showed that, if a<b then a^n < b^n for positive n>1

I want to start this proof off as follows:

If 0<a<b, we know a^n < b^n from previous results

we also know a<a^n for n>1**

I put stars because it seems to me like I am just making an assumption there. Would the proof be valid if I said:

we also know a<a^n for n>1 because any number x times more than 1 of itself will always be bigger than x

or do I need to prove that as well?

 

[Mark44]

If 0<a<b, we know a^n < b^n from previous results
we also know a<a^n for n>1**

This conclusion is not true in general. If 0 < a < 1, then a > aⁿ for n > 1. For example, if a = 1/2, a² = 1/4

However, if a > 1, then it is true that $a < a^n$.

 

[r0bHadz]

This conclusion is not true in general. If 0 < a < 1, then a > aⁿ for n > 1. For example, if a = 1/2, a² = 1/4

However, if a > 1, then it is true that $a < a^n$.

I see. Do I have to prove the case of a<a^n for a>1, a=1, and for a>a^n for 1>a>0? or can I just state "assuming this is true" so I don't have to prove everything

 

[Mark44]

I see. Do I have to prove the case of a<a^n for a>1, a=1, and for a>a^n for 1>a>0? or can I just state "assuming this is true" so I don't have to prove everything

No, I don't think that stating "assuming this is true" will fly. You are given only that 0 < a < b, so you can't come along and assume that a is somehow also larger than 1. Depending on how you go about doing your proof, you might have to make a couple of cases: one for the interval (0, 1] and another for the interval (1, ∞).

 

[WWGD]

I see. Do I have to prove the case of a<a^n for a>1, a=1, and for a>a^n for 1>a>0? or can I just state "assuming this is true" so I don't have to prove everything

I don't see where/how you would need this. It seems all you need, if I understood correctly is whether a>b => a^k>b^k.

 

[r0bHadz]

I don't see where/how you would need this. It seems all you need, if I understood correctly is whether a>b => a^k>b^k.

hmm it seems you are right. It looks like there are a lot of ways to do this proof and I was getting too caught up in finding which is the easiest but it seems like proof by contradiction is the easiest way? Does my following proof look correct?:

I am asked to prove: if 0<a<b then a^(1/n) < b^(1/n)

I am proving by contradiction so my proposition: if 0<a<b then a(1/n) ≥b(1/n)

Let a= x^n, b=y^n

we have x^n < y^n from the proposition

Since I have proved if a<b => a^n < b^n in a previous result, I know the converse a^n < b^n => a<b is true

so from x^n < y^n => x < y

since x = a^(1/n) and y = b^(1/n), there is a contradiction