If \(a > b\), is it always true that \(a^n > b^n?\)

True or False?

If $a>b$, then $a^n>b^n.$

Why some people say it's true: The more you multiply, the bigger it gets, so obviously if $a>b$, $a^n>b^n$.

Why some people say it's false: It doesn't work for all numbers.

This statement is $\color{#D61F06} {\textbf{false}}$. In particular, if $a>b>0$, then $$\begin{cases} a^n>b^n,~ \text{if } n>0; \\ a^n<b^n,~ \text{if } n<0; \\ a^n=b^n,~ \text{if } n=0.\end{cases}$$

Proof:

First, we need to understand the graph of the function $f(x)=k^x (k\neq0)$.

As shown in the figure above, we have 3 cases:
If $k>1$, the graph of $f$ is monotonically increasing; $f(x)\geqslant1$ when $x\geqslant0$, and $f(x)<1$ when $x<0$.
If $k<1$, the graph of $f$ is monotonically decreasing; $f(x)\leqslant1$ when $x\geqslant0$, and $f(x)>1$ when $x<0$.
If $k=1$, then $f(x)=1$.

Now consider 3 functions $f(x)=a^x$, $g(x)=b^x$ and $h(x)=\frac{f(x)}{g(x)},$ where $a>b>0.$ Then we have $$h(x)=\frac{a^x}{b^x}=\left(\frac{a}{b}\right)^x.$$ As $a>b>0$ and $b\neq0$, we have $\frac{a}{b}>1$.
Hence, $h(x)=\left(\frac{a}{b}\right)^x$ is in the form $k^x$ and $k=\frac{a}{b}>1$, so the graph of $h(x)$ is monotonically increasing.

When $x>0$, we have $h(x)=\frac{f(x)}{g(x)}>1$. Since $g(x)>0$, $$f(x)>g(x) \implies a^x>b^x.$$

When $x<0$, we have $h(x)=\frac{f(x)}{g(x)}<1$. Since $g(x)>0$, $$f(x)<g(x) \implies a^x<b^x.$$

When $x=0$, we have $h(x)=\frac{f(x)}{g(x)}=1$. Thus, $$f(x)=g(x) \implies a^x=b^x. \ _\square$$

Let $a=3, b=2,$ and $n=-1$. Find the relationship between $a^n$ and $b^n$.

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We have $$\begin{aligned} a^n &= 3^{-1}= 0.33\ldots\\ b^n &= 2^{-1} = 0.5. \end{aligned}$$ So, $a^n < b^n$. $_\square$

Let $a=3, b=2,$ and $n=0$. Find the relationship between $a^n$ and $b^n$.

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We have $$\begin{aligned} a^n &= 3^0 = 1\\ b^n &= 2^0 = 1. \end{aligned}$$ So, $a^n = b^n$. $_\square$

So, we see that in spite of $a$ being greater than $b$, $a^n$ is not always greater than $b^n$.