# Proving $x<y \implies n^{x}<n^{y}$, for $n>1$, $x,y \in \mathbb R$

I think I'm supposed to use the lowest upper bound property but I don't even know where to construct a set to start the problem.

• What is the definition of $n^x$? – David Jan 28 '16 at 7:04
• The usual definition i guess? $n\times n\times n\times \dots \times n$ x times. – lsy Jan 28 '16 at 7:04
• What does that mean if $x = \pi$? (Or even $x = -1$ for that matter.) – David Jan 28 '16 at 7:05
• A lecturer teaching an introductory analysis class probably wouldn't assume that all students had a very clear idea of what $n^x$ meant coming in, particularly since different approaches to defining it are possible. It may be that your teacher has given you a definition in class. – David Jan 28 '16 at 7:15
• Then the place to start with the problem is to know what the symbol is defined as - either go look at your book / course materials or ask your lecturer. – user296602 Jan 28 '16 at 7:16

Let $(x_i)_{i\in \mathbb{N}}$ and $(y_i)_{i\in \mathbb{N}}$ be given such that $x_i\to x$, $y_i\to y$, $x_i,y_i\in\mathbb{Q}$ and $x_i<y_i$, we can do this because of our initial condition and that rational numbers are dense in real numbers. We then have $$\lim_{i\to\infty}n^{x_i}=n^x$$ and $$\lim_{i\to\infty}n^{y_i}=n^y$$ We have for rational numbers that $$n^{x_i}<n^{y_i}$$ whenever $x_i<y_i$ because $x_i=\frac{p_{x_i}}{q_{x_i}}$ and clearly $$n^{p_{x_i}q_{y_i}}<n^{p_{y_i}q_{x_i}}$$ as from $x_i<y_i$ we have $p_{x_i}q_{y_i}<p_{y_i}q_{x_i}$ and as this applies for all in our sequence we have $n^x<n^y$

Setting $t=y-x$, you can easily show equivalence with

$$t>0\implies n^t>1$$

then (by logarithms)

$$t>0\implies e^t>1.$$

You can use the power series definition of the exponential,

$$e^t=1+t+\frac{t^2}2\cdots>1$$

or the limit definition with the Generalized Binomial formlua

$$\lim_{n\to\infty}\left(1+\frac tn\right)^n=\lim_{n\to\infty}1+t+n(n-1)\frac{t^2}{2n^2}\cdots>1.$$

Or, like @ZelosMalum, take $t$ to be a positive rational and

$$n^{p/q}>1\iff n^p=\left(n^{p/q}\right)^q=n\cdot n\cdots n>1^q.$$

Let $f(t)=n^t$ for $t\in\mathbb{R}$. You could show this function is differentiable over all of $\mathbb{R}$ if you like, but honestly it requires a bunch of technical hullabaloo so it will be taken by bare assertion. Since $f$ is differentiable over all of $\mathbb{R}$, it is also differentiable (and hence continuous) for any subset of the reals, namely the set $I=[x,y]$.

The mean valus theorem asserts there exists an element $c\in(x,y)$ such that \begin{align*} f'(c)=\frac{f(y)-f(x)}{y-x} \end{align*} We are interested in the derivative of $f$ with respect to $t$. \begin{align*} \frac{df}{dt}&=\frac{d}{dt}(n^t)\\ &=\frac{d}{dt}(e^{\ln(n)t})\\ &=\ln(n)e^{\ln(n)t}\\ &=\ln(n)n^t \end{align*} Now, since $n>1$, $\ln(n)>0$. From the properties of exponential functions, $n^t$ is always positive, so combining these results yields $\frac{df}{dt}>0$ for all $t\in\mathbb{R}$. Hence $\frac{f(y)-f(x)}{y-x}>0$ and thus $f(y)>f(x)$ as required.

These proofs tend to rely on a lot of the same properties (MVT, IVT, boundedness, etc. Can you see how you would prove $n^x>n^y$ for $n\in(0,1)$?

Assuming $x, y\in\mathbb{R}$.

If $x<y$ then $y$ may be defined as: $$y=x+\varepsilon,\varepsilon\in\mathbb{R}_{>0}$$

The initial equation is now: $$x<x+\varepsilon$$

The second equation is: \begin{align} n^x&<n^y\\ n^x&<n^{x+\varepsilon}\\ n^x&<n^x n^\varepsilon\\ \end{align}

Assuming $n>1$ and $x\in\mathbb{R}$ we can divide through $n^x$: \begin{align} 1&<n^\varepsilon \end{align}

For the given conditions ($\varepsilon>0$ and $n>1$) the expression $n^\varepsilon$ is always greater than $1$. $n>1$ is always true and $\varepsilon>0$ iff $x<y$.

I hope this helps.

• You seem to have circularity in your reasoning, it remains to be shown $n^x<n^y$. – user238841 Jan 29 '16 at 6:49

Prove $n^x< n^y$ is equivalent to prove that

$ln(exp(n^x))<ln(exp(n^y))$

Equivalent to:

$ln(exp(n x))= ln(exp(n y))$

If $x<y$ then $exp(nx)<exp(ny)$

Application of $ln$ to the both sides give the answer