Chapter 6.6, Problem 39WE

(a)

To determine

To Find:

If x and y are rational numbers , what can you conclude about their product.

Answer to Problem 39WE

The product of two rational number is also a rational number.

Explanation of Solution

Concept Used:

Rational number is of the form $\frac{p}{q}\text{ where p and q are integers and q}\ne \text{0}$ .

Calculation:

x and y are rational numbers.

Let

  $\begin{array}{l}x=\frac{p}{q};\text{ p and q are integers and q}\ne \text{0}\\ y=\frac{r}{s};\text{r and s are integers and s}\ne \text{0}\\ x\cdot y=\frac{p}{q}\cdot \frac{r}{s}=\frac{pr}{qs}\\ \text{Since p,q,r,s are integers and integers are closed under}\\ \text{multiplication }\text{. So, }pr\text{ and }qs\text{ are integers }\text{.}\\ \text{q}\ne \text{0 and s}\ne \text{0}\Rightarrow qs\ne 0.\\ So,\frac{pr}{qs}\text{ is a rational number}\text{.}\end{array}$

Hence, the product of two rational numbers is also a rational number.

(b)

To determine

To Find:

If x and y are irrational numbers , what can you conclude about their product.

Answer to Problem 39WE

Depending on the irrational numbers , the product can be rational or irrational.

Explanation of Solution

Concept Used:

Rational number is of the form $\frac{p}{q}\text{ where p and q are integers and q}\ne \text{0}$ .

Calculation:

x and y are irrational numbers.

Let

  $\begin{array}{l}x=\sqrt{2}\\ y=\sqrt{3}\\ \text{Both x and y are irrational numbers}\text{.}\\ \text{Product = }\sqrt{2}\cdot \sqrt{3}=\sqrt{2\cdot 3}=\sqrt{6}\text{ , which is an irrational number}\text{.}\\ \text{Let }\\ x=\sqrt{2}\\ y=\sqrt{2}\\ \mathrm{Pr}oduct=\sqrt{2}\cdot \sqrt{2}=2,\text{which is a rational number}\text{.}\end{array}$

Hence, the paroduct of two irrational numbers can be rational or irrational.