Beginning Algebra With Applications, Chapter 1, 1.4, Section 1.4, Problem 70

Find a rational number, $r$, that satisfies the condition.

a.) $r^2 < r$
For $r^2 < r$
If we substitute a rational number for $r < 0$, let's say $-2$, we get

$
\begin{equation}
\begin{aligned}
r^2 &< r\\
\\
(-2)^2 &< -2\\
\\
4 &< -2 && \text{Which is false}
\end{aligned}
\end{equation}
$


Then if we substitute a rational number for $r > 1$, let's say $2$, we get

$
\begin{equation}
\begin{aligned}
r^2 &< r\\
\\
2^2 &< 2\\
\\
4 &< 2 && \text{Which is also incorrect}
\end{aligned}
\end{equation}
$


But if we substitute a rational number for $0 < r < 1$, let's say $\displaystyle \frac{1}{2}$, we get

$
\begin{equation}
\begin{aligned}
r^2 &< r \\
\\
\left( \frac{1}{2} \right)^2 &< \frac{1}{2}\\
\\
\frac{1}{4} &< \frac{1}{2} && \text{Which is true}
\end{aligned}
\end{equation}
$


Thus, the rational numbers that will satisfy the inequality $r^2 < r$ are from $0 < r < 1$.

b.) $r^2 = r$
For $r^2 = r$

$
\begin{equation}
\begin{aligned}
r^2 - r &= 0 && \text{Subtract $r$ each side}\\
\\
r (r - 1) &= 0 && \text{Factor } r
\end{aligned}
\end{equation}
$

We have,

$
\begin{equation}
\begin{aligned}
r &=0 &&\text{and}& r - 1 &= 0 \\
\\
r &= 0 &&\text{and}& r &= 1
\end{aligned}
\end{equation}
$

Thus, the rational numbers that will satisfy $r^2 = r$ are and $1$.

c.) $r^2 > r$
For $r^2 > r$
If we substitute a rational number for $r < 0$, let's say $-2$, we get


$
\begin{equation}
\begin{aligned}
r^2 &> r\\
\\
(-2)^2 &> -2\\
\\
4 &> -2 && \text{Which is true}
\end{aligned}
\end{equation}
$


Then if we substitute a rational number for $r > 1$, let's say $2$, we get

$
\begin{equation}
\begin{aligned}
r^2 &> r \\
\\
2^2 &> 2\\
\\
4 &> 2 && \text{Which is also true}
\end{aligned}
\end{equation}
$


But if we substitute a rational number for $0 < r < 1$, let's say $\displaystyle \frac{1}{2}$, we get

$
\begin{equation}
\begin{aligned}
r^2 &> r\\
\\
\left( \frac{1}{2} \right)^2 &> \frac{1}{2}\\
\\
\frac{1}{4} &> \frac{1}{2} && \text{Which is false}
\end{aligned}
\end{equation}
$

Thus, the rational numbers that will satisfy the inequality $r^2 > r$ are from $r < 0$ and $r > 1$