Calculating and Classifying \((\sqrt2 -\sqrt3)^2\)

The problem asks us to evaluate the expression \((\sqrt2 -\sqrt3)^2\) and then classify the resulting number based on the given options: whole number, natural number, rational number, or irrational number.

To evaluate \((\sqrt2 -\sqrt3)^2\), we can use the algebraic identity for squaring a binomial: \((a-b)^2 = a^2 - 2ab + b^2\).

Here, let \(a = \sqrt2\) and \(b = \sqrt3\). Applying the formula, we get:

\((\sqrt2 -\sqrt3)^2 = (\sqrt2)^2 - 2(\sqrt2)(\sqrt3) + (\sqrt3)^2\)

Now, we simplify each term:

• \((\sqrt2)^2 = 2\)
• \((\sqrt3)^2 = 3\)
• \(2(\sqrt2)(\sqrt3) = 2\sqrt{2 \times 3} = 2\sqrt{6}\)

Substituting these simplified terms back into the expression:

\((\sqrt2 -\sqrt3)^2 = 2 - 2\sqrt{6} + 3\)

Combine the rational terms (the numbers without square roots):

\((\sqrt2 -\sqrt3)^2 = (2 + 3) - 2\sqrt{6}\)

\((\sqrt2 -\sqrt3)^2 = 5 - 2\sqrt{6}\)

Now, let's analyze the resulting number \(5 - 2\sqrt{6}\) to classify it.

• The number 5 is an integer, and therefore a rational number.
• The number \(\sqrt{6}\) is an irrational number because 6 is not a perfect square.
• Multiplying a non-zero rational number (2) by an irrational number (\(\sqrt{6}\)) results in an irrational number (\(2\sqrt{6}\)).
• Subtracting an irrational number (\(2\sqrt{6}\)) from a rational number (5) results in an irrational number.

Therefore, \(5 - 2\sqrt{6}\) is an irrational number.

Based on this classification, we compare the result with the given options:

• A whole number: Whole numbers are 0, 1, 2, 3, ... \(5 - 2\sqrt{6}\) is not a whole number as it involves \(\sqrt{6}\).
• A natural number: Natural numbers are 1, 2, 3, ... \(5 - 2\sqrt{6}\) is not a natural number.
• A rational number: Rational numbers can be expressed as a fraction \(\frac{p}{q}\) where p and q are integers and q ≠ 0. Irrational numbers cannot be expressed in this form. \(5 - 2\sqrt{6}\) is not rational.
• An irrational number: Irrational numbers are real numbers that are not rational. \(5 - 2\sqrt{6}\) fits this description.

Thus, the expression \((\sqrt2 -\sqrt3)^2\) simplifies to an irrational number.

Revision Table: Real Number System Classification

Additional Information: Operations with Irrational Numbers

Understanding how operations affect the type of number is crucial for classifying results.

• The sum or difference of a rational number and an irrational number is always irrational. (e.g., \(5 + \sqrt{2}\), \(5 - \sqrt{2}\), \(\sqrt{2} - 5\))
• The product or quotient of a non-zero rational number and an irrational number is always irrational. (e.g., \(2\sqrt{6}\), \(\frac{\sqrt{2}}{3}\))
• The sum, difference, product, or quotient of two irrational numbers can be either rational or irrational.
  • Example (Rational Result): \(\sqrt{2} + (-\sqrt{2}) = 0\) (Rational), \(\sqrt{2} \times \sqrt{2} = 2\) (Rational), \(\frac{\sqrt{8}}{\sqrt{2}} = \sqrt{4} = 2\) (Rational).
  • Example (Irrational Result): \(\sqrt{2} + \sqrt{3}\) (Irrational), \(\sqrt{2} \times \sqrt{3} = \sqrt{6}\) (Irrational).

In our case, \(5 - 2\sqrt{6}\) is the difference between a rational number (5) and an irrational number (\(2\sqrt{6}\)), which results in an irrational number.