The difference of any two rational numbers is a rational number.
The correct proof will use 7 of the statements below.
Statements to choose from:
Your Proof: Put chosen statements in order in this column and press the Submit Answers button.
Then $r = \frac{a}{b}$, $s = \frac{c}{d}$
Thus, $a - c$ is an integer and $b - d$ is
an integer.
Thus, $r - s$ is a fraction with an
integer numerator and a nonzero
integer denominator and hence, by
definition, $r - s$ is rational.
Since $a$, $b$, $c$, and $d$ are all integers, the
product of integers is still an integer
and the difference of integers is an
integer.
By definition of a rational number,
$r = \frac{a}{b}$ and $s = \frac{c}{d}$ for integers $a$, $b$, $c$,
and $d$ with $a \neq 0$ and $c \neq 0$.
Thus, $ad - bc$ is an integer and $bd$ is an
integer.
Also, since $b \neq 0$ and $d \neq 0$ by
definition, $bd \neq 0$.
Thus, $r - s$ is a difference of two
integer and hence, by definition, $r - s$
is rational.
Suppose $r$ and $s$ are rational numbers.
By definition of a rational number,
$r = \frac{a}{b}$ and $s = \frac{c}{d}$ for integers $a$, $b$, $c$,
and $d$ with $b \neq 0$ and $d \neq 0$.
Then
$r - s = \frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}$
