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Prove or disprove that the product of two irrational numbers is irrational.

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.

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