# Inversion of the inequality sign when raising to a negative power

How come since $e>1 \implies e^{-1/2} < 1^{-1/2}$. I know that one reverses the inequality signs when we take reciprocal of both sides or multiplies by a negative number. I have never seen inequality sign inverted because we raise both sides to the negative power.

What would in this case then, given:

$a > -\ln (2e^{-1/2}-1)$

and I wanted to take exponent and as a power have each side of the inequality:

$e^a \ \ ? \ \ e^{\ln(2e^{-1/2}-1)^{-1}}$

• "when we divide both sides by $1$"??? Raising to a negative power is sort of equivalent to what you probably meant by "when we divide both sides by $1$". May 4, 2016 at 8:20
• $$a>b \Rightarrow a^{-1}< b^{-1}$$ because $a^{-1}=1/a$. Now raise to any power you want. May 4, 2016 at 8:20
• Because $x\mapsto x^{-1/2}$ is decreasing on its domain $x>0$. Or, because $x\mapsto 1/x$ is decreasing on $x>0$ and $x\mapsto x^{1/2}$ is increasing on its domain $x>0$. On the other hand, $\exp$ is increasing hence $a>b$ implies $e^a>e^b$ for every $a$ and $b$ real numbers.
– Did
May 4, 2016 at 8:22

Taking the reciprocal of each side (which is the same thing as raising to the negative first power) only flips the inequality if $a\times b$ is positive. That is, it only holds if $a,b$ are of the same sign.

Consider $a>b$.

If $ab>0$, then dividing both sides above by $ab$ gets $\frac1b>\frac1a$ as per the rule. The inequality appears to have flipped, $a^{-1} < b^{-1}$.

However, if $ab<0$, then dividing both sides above by $ab$ gets $\frac1b<\frac1a$, or in other phrasing, $a^{-1} > b^{-1}$.

$$e>1\implies\frac1{e}<\frac11\implies e^{-\frac12}=\left(\frac1{e}\right)^{\frac12}<\left(\frac11\right)^{\frac12}=1^{-\frac12}$$

• Alright. I think I can answer that second part of my question myself
– Naz
May 4, 2016 at 8:24
• $a > -\ln (2e^{-1/2}-1) \implies (1/a) < \ln(2e^{-1/2}-1) \implies e^{1/a} < 2e^{-1/2}-1$
– Naz
May 4, 2016 at 8:25
• Your first implication is wrong. $a>-\ln(z)\implies-a<\ln(z)\implies e^{-a}<z$ or $a>-\ln(z)=\ln(z^{-1})\implies e^a>z^{-1}$. May 4, 2016 at 8:29