Properties of Inequalities I am unable to understand the following statements. If someone can explain with the help of numerical examples it would be great:


*

*If sides of an inequality are both positive or both negative, taking the reciprocal reverses the inequality

*If $0<x<1$ and $m$ and $n$ are integers with $m>n$, then $x^m<x^n<x$

*If $0<x<1$, then $\sqrt{x}>x$

*If $0<x<1$, then $\dfrac{1}{x}>x$ and $\dfrac{1}{x}>1$
 A: Below are some numerical examples of each of these inequalities.


*

*You know that $2<4$. Now, $0.5=\dfrac{1}{2} > \dfrac{1}{4}=0.25$. Similarly, $-2>-4$. now, $-0.5=\dfrac{-1}{2}<\dfrac{-1}{4}=0.25$.

*Consider $a=4$ and $x=0.5$. Everything meets the stipulations. Then $xa=0.5(4)=2<4=a$.

*Suppose $x=\dfrac{1}{2}$ and $m=3$ and $n=2$. Then we have that $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$. Furthermore, $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$. We know that $\dfrac{1}{8}<\dfrac{1}{4}$.

*Suppose that $x=\dfrac{1}{4}$. Then we have that $\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$. Furthermore, we know that $\dfrac{1}{2}>\dfrac{1}{4}$.

*Suppose that $x=\dfrac{1}{2}$. Then we know that $\dfrac{1}{\left(\frac{1}{2}\right)}=2$. Furthermore, $2>\dfrac{1}{2}$. Also, $\dfrac{2}>1$.

A: These statements can all be shown using basic rules of algebra.
The first statement "If both sides of an inequality are both positive and both negative, taking the reciprocal reverses the inequality", can be easily understood by considering $a>b$ for some real-numbers $a$ and $b$. Clearly $\frac{1}{a}<\frac{1}{b}$ for the bigger the number the smaller it's reciprocal is. For instance if we take $a=2, b=1$, we have:
$$2>1 \text{ and } \frac{1}{2}<1$$
For the second statement, if we have $0<x<1$ then clearly multiplying a number by it will make it smaller because it is the same as dividing by some number greater than $1$. For instance, if we take $x=0.5=\frac{1}{2}$ then we have $ax=\frac{a}{2}<a$.
For the third statement we can remember that raising a base $b$ to a power $m$ can be viewed as multiplying $b$ by itself $m$ times, i.e.:
$$b^{m}=\underbrace{b\times \cdots \times b}_{m}$$
As we have $0<x<1$ multiplying $x$ by itself will give a smaller number still between $0$ and $1$. The more times we multiply itself by itself the smaller the result will be. For instance, if we take $x=0.5=\frac{1}{2}$ again and $n=2,m=3$ then we have: 
$$x^n=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \text{ and } x^{m}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$$
And so clearly as $\frac{1}{8}<\frac{1}{4}<\frac{1}{2}$ we have $x^{m}<x^{n}<x$.
For your fourth statement, a nice way to view a square root for positive real numbers is bringing the number closer to $1$, so for some $x$ between $0$ and $1$ we have $\sqrt{x}>x$. This is obvious by squaring both sides to give $x>x^{2}$, which is clearly true as we showed in your third statement.
For your last statement it can be clearly seen to be true as dividing a number by some small $x$ is similar to asking how many times can $x$ fit into that number, therefore if $x$ is really small, it can clearly fit into $1$ many times. For a numerical example we can again take $x=\frac{1}{2}$, therefore we have $\frac{1}{x}=\frac{1}{\frac{1}{2}}=2>1>x$.
