Chapter 1.1, Problem 81E

To determine

To explain: the sum, the difference, and the product of two rational numbers are rational numbers, whether the product of two irrational numbers necessarily irrational, to find about the sum.

Explanation of Solution

Suppose that

  $\frac{p}{q},\frac{r}{s}$ are two rational numbers. Then $p,q,r,s$ are integers and

  $q\ne 0,s\ne 0$

  $\therefore ps,qr,qs$ are integers and $qs\ne 0$

Since the sum of integers is an integer, then $ps+qr$ is an integer

  $\therefore \frac{ps+qr}{qs}$ is a rational number

  $\therefore$ The sum of rational numbers is rational

Similarly, $ps-qr$ is an integer, so $\frac{ps-qr}{qs}$ is rational, I.e., $\frac{p}{q}-\frac{r}{s}$ is rational

  $\therefore$ The product of rational numbers is a rational number

  $\sqrt{2},\sqrt{18}$ are irrational

But

  $\begin{array}{c}\sqrt{2}\ast \sqrt{18}=\sqrt{36}\\ =6\text{is}\text{a}\text{rational}\text{number}\end{array}$

The sum of irrational numbers is irrational

The product of irrational numbers is not necessarily irrational.