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Let n $\in$ N and f: [$0, \infty$) $\to$ R be defined by $f(x) = x^n$ for all $x \ge 0$. Prove that f is an increasing function.

My attempt at a solution: Let $0 \le x \le y$. Multiply both sides by $x^{n-1}$, so

$0 \le x*x^{n-1} \le y*x^{n-1}$

So,

$0 \le x^{n} \le y*x^{n-1}$

Next, (and here's where I'm having some issues) if the function is increasing, then

$0 \le x^{n-1} \le y^{n-1}$

Multiplying both sides by $y$ yields:

$0*y \le y*x^{n-1} \le y*y^{n-1}$ so,

$0 \le y*x^{n-1} \le y^{n}$

and by transivity,

$0 \le x^{n} \le y*x^{n-1} \le y^n$

Therefore the function is increasing. Any help with the later part of the problem would be helpful.

Note that when $n = 2$ this is a very easy problem because:

for $0 \le x \le y$ multiply both sides by x and then y, so

$0 \le x^2 \le y*x$

$0 \le x*y \le y^2$

And by transivity this is true

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Option:

Let $0\le a <b$.

Want to show that $a^n<b^n$, i .e. strictly increasing.

$b^n-a^n= $

$(b-a)(b^{n-1}+b^{n-2}a+ ....+a^{n-1}).$

The second factor is positive, all positive summands, as is the first factor $(b-a)$,

hence $b^n -a^n >0$.

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The proof can be done much shorter.

Be $\epsilon > 0$. Now there must be for all $n$ a solution $x^n < (x+\epsilon)^n$.

This can be very easily proven:

  1. Take the nth root on both sides: $x<x-\epsilon$

  2. Shorten to the expression: $0<\epsilon$

As $\epsilon$ is by definition always larger than 0 this statement is for all $x$ and $n$ true.

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