Finding the Product of √2 and √3

The question asks for the product of two numbers, \(\sqrt{2}\) and \(\sqrt{3}\). To find the product, we multiply these two numbers together.

The product is calculated as follows:

\(\sqrt{2} \times \sqrt{3}\)

There is a property of square roots that states the product of the square roots of two non-negative numbers is equal to the square root of their product. Mathematically, this property is expressed as:

\(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\), where \(a \ge 0\) and \(b \ge 0\).

Using this property, we can simplify the product of \(\sqrt{2}\) and \(\sqrt{3}\):

\(\sqrt{2} \times \sqrt{3} = \sqrt{2 \times 3} = \sqrt{6}\)

So, the product of \(\sqrt{2}\) and \(\sqrt{3}\) is \(\sqrt{6}\).

Identifying the Nature of the Product √6

Now, we need to determine if \(\sqrt{6}\) is a rational number or an irrational number.

• Rational Numbers: A rational number is any number that can be expressed as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Examples include \(\frac{1}{2}\), \(3\) (which can be written as \(\frac{3}{1}\)), \(0.75\) (which is \(\frac{3}{4}\)), and \(-5\) (which is \(\frac{-5}{1}\)). The decimal expansion of a rational number is either terminating or repeating.
• Irrational Numbers: An irrational number is a number that cannot be expressed as a simple fraction \(\frac{p}{q}\). Their decimal expansions are non-terminating and non-repeating. Examples include \(\pi\) (pi), \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{5}\).

To determine if \(\sqrt{6}\) is rational or irrational, we consider if 6 is a perfect square. A perfect square is an integer that is the square of an integer (e.g., \(4 = 2^2\), \(9 = 3^2\), \(16 = 4^2\)).

The number 6 is not a perfect square because there is no integer whose square is exactly 6. The square root of a positive integer that is not a perfect square is always an irrational number.

Since 6 is not a perfect square, \(\sqrt{6}\) is an irrational number.

Analyzing the Given Options

Based on our finding that the product \(\sqrt{2} \times \sqrt{3}\) is \(\sqrt{6}\), which is an irrational number, let's look at the options:

1. an irrational number: This matches our result.
2. sometimes a rational number and sometimes an irrational number: A specific product like \(\sqrt{2} \times \sqrt{3}\) has a fixed value and fixed nature (either rational or irrational, not both). This option is incorrect.
3. equal to 4: We calculated the product as \(\sqrt{6}\). The value of \(\sqrt{6}\) is between \(\sqrt{4}=2\) and \(\sqrt{9}=3\), so it is not equal to 4. This option is incorrect.
4. a rational number: We determined that \(\sqrt{6}\) is an irrational number. This option is incorrect.

Therefore, the product of \(\sqrt{2}\) and \(\sqrt{3}\) is an irrational number.

Revision Table: Understanding Number Types and Products

Additional Information: Properties of Irrational Numbers

Irrational numbers are part of the real number system. When you perform operations with irrational numbers, the result can be rational or irrational, depending on the specific numbers and the operation.

• The sum or difference of a rational number and an irrational number is always irrational.
• The product or quotient of a non-zero rational number and an irrational number is always irrational.
• The sum, difference, product, or quotient of two irrational numbers can be either rational or irrational. As seen in our problem, the product of two irrational numbers (\(\sqrt{2}\) and \(\sqrt{3}\)) resulted in another irrational number (\(\sqrt{6}\)). However, the product of \(\sqrt{2}\) and \(\sqrt{2}\) (both irrational) is 2 (rational).
• Square roots of positive integers that are not perfect squares are irrational.

Understanding the properties of rational and irrational numbers is crucial for working with various mathematical expressions.