## Calculus: Early Transcendentals 8th Edition

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.