When is the function $x^{1/n}$ a monotonic function? I'm trying to prove that for every $x,y>0$, $x>y$, and $n \in 
\mathbb N$, $x^{1/n}>y^{1/n}$. Just some direction will be fine. We haven't learned derivatives yet. Thanks in advance.
 A: Since $x > y > 0$, for any $a > 0$, you have $a \cdot x > a \cdot y > 0$. Therefore 
$$
x^2= x \cdot x > x \cdot y  = y \cdot x > y \cdot y.
$$
(In the first inequality, $a=x$, but in the second, $a=y$.) You can repeat this trick for $3$ as the exponent instead of $2$ : 
$$
x^3 = x \cdot x^2 > x \cdot y^2 = y^2 \cdot x > y^2 \cdot y = y^3. 
$$
You could repeat this argument using induction (! I don't know if you have seen that, but it is a very useful thing to know.). It goes like this : assuming $x > y$ implies $x^{n-1} > y^{n-1}$ for some $n \ge 1$, 
$$
x^n = x\cdot x^{n-1} > x \cdot y^{n-1} = y^{n-1} \cdot x > y^{n-1} \cdot y = y^n.
$$
Now if $x > y$ and $x^{1/n} \le y^{1/n}$, using the same trick you conclude 
$$
x = (x^{1/n})^n \le (y^{1/n})^n = y < x,
$$
a contradiction. Therefore $x^{1/n} > y^{1/n}$. 
Hope that helps,
A: I don't know how much tools you have access to, but if you can use derivatives, then $\frac{d}{dx} x^n = nx^{n-1} $. Thus if n>0 and x>0, then the derivative of $x^n$ is positive, and thus the function is monotonic. 
