How does $a^n$ compare to $b^n$, in case $a>b$ and $n>0$? ( On an inequality rule regarding exponentiation).

Question

Suppose $a>b$ and that $n > 0$.

How does $a^n$ compare to $b^n$?

By considering various cases , I arrived at the following rule :

"In case $a> b$ , and $n > 0$ , then $n$-powers conserve the order of $a$ and $b$ , that is ,$a^n > b^n$ , except when n is even and

• either (1) $a$ and $b$ are both negative ,

• or (2) $a$ and $b$ have different signs and $a$ is smaller than $b$ in absolute value."

Is this rule corrrect? In case it is, is there a simpler way to phrase it?

Answer 1

I assume $n$ is an integer here.

Yes, it's correct. Note that your "(1) or (2)" can be simplified to "$a$ is smaller than $b$ in absolute value". (It should actually be "smaller than or equal to" here, since if $b=-a$ then $a>b$ but $a^n\not>b^n$.)

An alternative way to look at it: if $n$ is odd then raising to the $n$th power preserves order relations; if $n$ is even then it reorders based on absolute value.

Answer 2

Yes this rule is true :

$f(x)=x^n$ then $f'(x)=nx^{n-1}$ for positive value the derivative is positive and therefore the function is increasing.

for negative value :

if n is even n-1 is odd and therefore for negative value the function is decreasing

if n is odd n-1 is even and therefore the function is increasing.