Answer
The sum, difference, and product of rational numbers are rational numbers.
Work Step by Step
Show that the sum, difference, and product of rational numbers are rational numbers.
1. Take $a, b$ two rational numbers.
By the definition of rational numbers $a=\frac{m_{1}}{n_{1}}$
Here, $m_{1}$ and $n_{1}$ are integers and $n_{1}\ne 0$.
Also, $b=\frac{m_{2}}{n_{2}}$
Here, $m_{2}$ and $n_{2}$ are integers and $n_{2}\ne 0$
2. Take the sum of two rational numbers.
$a+b=\frac{m_{1}}{n_{1}}+\frac{m_{2}}{n_{2}}=\frac{m_{1}n_{2}+m_{2}n_{1}}{n_{1}n_{2}}$
Here, $({m_{1}n_{2}+m_{2}n_{1}})$ are integers and ${n_{1}n_{2}}\ne 0$
Because $m_{1},m_{2},n_{1}$and $n_{2}$ are integers and $n_{1}\ne 0$ , $n_{2}\ne 0$, the sum $a+b$ is also a rational number.
3. Take the product of two rational numbers.
$ab=\frac{m_{1}}{n_{1}}\frac{m_{2}}{n_{2}}=\frac{m_{1}m_{2}}{n_{1}n_{2}}$
Here, ${m_{1}m_{2}}$ are integers and ${n_{1}n_{2}}\ne 0$
Because $m_{1},m_{2},n_{1}$and $n_{2}$ are integers and $n_{1}\ne 0$ , $n_{2}\ne 0$, the sum $ab$ is also a rational number.
4. Take the differnce of two rational numbers.
$a-b=\frac{m_{1}}{n_{1}}-\frac{m_{2}}{n_{2}}=\frac{m_{1}n_{2}-m_{2}n_{1}}{n_{1}n_{2}}$
Here, $({m_{1}n_{2}-m_{2}n_{1}})$ are integers and ${n_{1}n_{2}}\ne 0$
Because $m_{1},m_{2},n_{1}$and $n_{2}$ are integers and $n_{1}\ne 0$ , $n_{2}\ne 0$, the sum $a-b$ is also a rational number.
