Thread: Prove x^n > x for all x > 1.

1. Prove x^n > x for all x > 1.

Suppose n > 1 is a positive integer. Prove $x^n > x$for all x > 1.
(hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all $x \in(0,c)$, then f(x) > g(x) for all $x \in (0,c)$.

2. Do you know how to prove something by induction? If not I am unsure how to do this for general n.

3. Originally Posted by lvleph
Do you know how to prove something by induction? If not I am unsure how to do this for general n.
we have done a couple of induction prove, not so familiar with that, can we apply the thereom in the hints that is given?

4. What you want to do is use the hint on a base case (n=2). Then you will make the assumption that the problem hypothesis holds for n=k. Use this assumption to prove that it is true for n=k+1.

5. Originally Posted by lvleph
What you want to do is use the hint on a base case (n=2). Then you will make the assumption that the problem hypothesis holds for n=k. Use this assumption to prove that it is true for n=k+1.
why do we set n = 2?

6. Because it is easy and the problem statement says n > 1.

7. Originally Posted by lvleph
Because it is easy and the problem statement says n > 1.
then how are we gonna apply the hint to the proof, I can't see the relation between them

8. Originally Posted by 450081592
Suppose n > 1 is a positive integer. Prove $x^n > x$for all x > 1.
(hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all $x \in(0,c)$, then f(x) > g(x) for all $x \in (0,c)$.
Note that if $f(x)=x^n\quad n>1$ and $g(x)=x$. That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}\ge n>1=g'(1)$ and the concluso follows.

9. Okay, I will show you the base case.

Assume n=2.
$f(x) = x^2, \; g(x) = x$
Now let's show that these are equal at $x=0$
$f(1)=1^2 = 1 = g(1)$
Now we differentiate the two
$f'(x) = 2x$
$
g'(x) = 1$

For $x>1$
$
2x > 2$

Therefore we see that $f'(x)=2x>1 =g'(x)$ for $x>1$.

10. Originally Posted by Drexel28
Note that if $f(x)=x^n\quad n>1$ and $g(x)=x$. That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}>n>1=g'(1)$ and the concluso follows.
In fact, I guess this is a much quicker and simpler proof.

11. Originally Posted by lvleph
In fact, I guess this is a much quicker and simpler proof.

so with this method, we do not need to prove by induction, right?

12. Originally Posted by Drexel28
Note that if $f(x)=x^n\quad n>1$ and $g(x)=x$. That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}\ge n>1=g'(1)$ and the concluso follows.
how do we know g(x) is x?

That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}\ge n>1=g'(1)$

and I dont get this line

13. We know $g(x)=x$ because you are trying to prove $x^n > x$ and the hint was suggesting you $f(x) > g(x)$
Wouldn't you agree $nx^{n-1} \ge n$ for $x>1$ and $n>1$? But we know that $f'(x) = nx^{n-1}$ and $g'(x) = 1$.

14. Originally Posted by lvleph
We know $g(x)=x$ because you are trying to prove $x^n > x$ and the hint was suggesting you $f(x) > g(x)$
Wouldn't you agree $nx^{n-1} \ge n$ for $x>1$ and $n>1$? But we know that $f'(x) = nx^{n-1}$ and $g'(x) = 1$.
so we are just assuming g(x) = x?

15. Yes, because the hint was to prove $f(x) > g(x)$ and we wanted to prove $x^n > x$ so it makes sense to assume $g(x)=x$.

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