# Prove that f is an increasing function

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

Option:

Let $$0\le a .

Want to show that $$a^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 $$b^n -a^n >0$$.

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

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.