The **aim of this question** is to understand** deductive logic** and the concept of **irrational and rational numbers.**

A number (N) is said to be **rational** if it can be written **in the form of a fraction** such that the numerator and denominator both belong to a set of **integers**. Also it’s a necessary condition that the **denominator has to be non-zero.** This definition can be written in the **mathematical form** as follows:

\[ N \ = \ \dfrac{ P }{ Q } \text{ where } P, \ Q \ \in Z \text{ and } Q \neq 0 \]

Where $ N $ is the **rational number** while $ P $ and $ Q $ are the **integers** belonging to the set of integers $ Z $. On similar lines, we can conclude that **any number** that **can’t be written in the form of a fraction** (with numerator and denominator being integers) is called an **irrational number**.

An **integer** is such a number that does not have** any fractional part** or doesn’t have **any decimal**. An integer can be both** positive and negative**. Zero is also included in the set of integers.

\[ Z \ = \ \{ \ …, \ -3, \ -2, \ -1, \ 0, \ +1, \ +2, \ +3, \ … \ \} \]

## Expert Answer

Now **to prove the given statement,** we can prove the **contraposition.** The contraposition statement of the given statement can be written as follows:

**“A product of two rational numbers is also a rational number.”**

Let us say that:

\[ \text{ 1st rational number } \ = \ A \]

\[ \text{ 2nd rational number } \ = \ B \]

\[ \text{ Product of two rational numbers } \ = \ C \ = \ A \times B \]

**By definition of rational numbers** as described above, $ C $ can be written as:

\[ \text{ A rational number } \ = \ C \]

\[ \text{ A rational number } \ = \ A \times \ B \]

\[ \text{ A rational number } \ = \ \dfrac{ A }{ 1 } \times \dfrac{ 1 }{ B } \]

\[ \text{ A rational number } \ = \ \text{ Product of two rational numbers } \]

Now we know that $ \dfrac{ A }{ 1 } $ and $ \dfrac{ 1 }{ B } $ **are rational numbers**. Hence proved that a **product of two rational numbers** $ A $ and $ B $ is also a rational number $ C $.

So the **contrapositive statement must also be true**, that is, the product of two irrational numbers must be an irrational number.

## Numerical Result

The product of two irrational numbers must be an irrational number.

## Example

Is there a condition **where the above statement does not hold true**. Explain with the help of **example**.

Let’s **consider an irrational number** $ \sqrt{ 2 } $. Now if we **multiply this number with itself**:

\[ \text{ Product of two irrational numbers } \ = \ \sqrt{ 2 } \ \times \ \sqrt{ 2 } \]

\[ \text{ Product of two irrational numbers } \ = \ ( \sqrt{ 2 } )^2 \]

\[ \text{ Product of two irrational numbers } \ = \ 2 \]

\[ \text{ Product of two irrational numbers } \ = \text{ a rational number } \]

Hence, the **statement does not hold true when we multiply an irrational number with itself.**