Show that if $x > 1$ is a real number and if $a < b$ are rational numbers, then $0\le x^a \le x^b$.

My professor told me that I'm supposed to use some $x^c$, such that $c$ $\epsilon$ $Q$ > $0$. I have no idea how to use this fact and I wouldn't have thought of it because there is no $c$ in the problem. Any suggestions on what he meant/or an alternative way to solve this? Thanks!

  • $\begingroup$ How do you define $x^a$ when $a$ is rational? $\endgroup$ – Thomas Andrews Dec 15 '14 at 4:08
  • $\begingroup$ you can define it as a= p/q right? with p&q belonging to the integers $\endgroup$ – Zoë Soriano Dec 15 '14 at 4:10
  • $\begingroup$ That wasn't what I asked. How do you define $x^{p/q}$? $\endgroup$ – Thomas Andrews Dec 15 '14 at 4:42
  • $\begingroup$ not sure what you mean by that. $\endgroup$ – Zoë Soriano Dec 15 '14 at 4:59
  • $\begingroup$ You can define $x^{p/q}$ as $e^{p\ln x /q}$ or as the unique positive real number such $y$ such that $y^q=x^p$, or... There are several ways to define $x^{p/q}$, and so, how you prove something about $x^{p/q}$ depends on the definitions. $\endgroup$ – Thomas Andrews Dec 15 '14 at 5:04

Since $a$ and $b$ are rational numbers you can write them as $$a=\frac{m}{n}, b=\frac{c}{d}, m,n,c,d\in \mathbb{Z},$$ or,

$$a=\frac{md}{nd}, b=\frac{cn}{nd}. $$

Now, assuming $n,d>0,$ we have that

$$x>1\implies 1<\sqrt[nd]{x}\underbrace{\implies}_{md<cn} (\sqrt[nd]{x})^{md}<(\sqrt[nd]{x})^{nc} \iff x^{md/nd}< x^{nc/nd}\iff x^{m/n}<x^{c/d}.$$


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