91Ó°ÊÓ

When dealing with the concept of closure in mathematics, particularly in fields such as algebra, the term refers to a set’s stability under certain operations. For a set to be considered closed under an operation, applying that operation to any members within the set should result in a member that is also within the set. In the context of the set \( F = \{ x + \sqrt{2} y \mid x, y \in \mathbb{Q} \} \), closure under addition and multiplication is demonstrated as follows.

1. **Addition:** Consider two elements of the set, say \( a = x_1 + \sqrt{2}y_1 \) and \( b = x_2 + \sqrt{2}y_2 \). Adding these results in \( a + b = (x_1 + x_2) + \sqrt{2}(y_1 + y_2) \). Given that sums of rational numbers are rational, \( x_1 + x_2 \) and \( y_1 + y_2 \) remain rational, thus ensuring that \( a + b \) is still in the set \( F \).

2. **Multiplication:** Similarly, considering the product of the same elements \( a \times b = (x_1x_2 + 2y_1y_2) + \sqrt{2}(x_1y_2 + x_2y_1) \) reveals that both expressions \( x_1x_2 + 2y_1y_2 \) and \( x_1y_2 + x_2y_1 \) are rational. Therefore, the set \( F \) is closed under multiplication as well.

Ultimately, closure under addition and multiplication confirms that any basic arithmetic operation on members of \( F \) will result in a value that also resides in \( F \), thereby maintaining the set's integrity.

Completeness is a concept often associated with real numbers and the ability to fill out any gaps that might exist within a number set, particularly when considering least upper bounds or supremum. In other words, a set is complete if every non-empty subset that has an upper bound also has a least upper bound within the set. Let's explore why the set \( F = \{ x + \sqrt{2} y \mid x, y \in \mathbb{Q} \} \) lacks this property.

Consider the subset \( \{ x \in \mathbb{Q} \mid x^2 < 2 \} \). The elements of this subset are rational numbers whose squares are less than 2. In the real numbers, we know that the least upper bound for this set is \( \sqrt{2} \), a number that is not rational. Since \( \sqrt{2} \) cannot be expressed as any member of \( F \) (where elements are formed of rational components), \( F \) does not contain the least upper bound required for completeness.

By illustrating this scenario, it becomes apparent that \( F \), while behaving well under arithmetic operations, fails to exhibit the completeness property inherent to the real numbers, reaffirming that \( F \) is incomplete.

The concept of multiplicative inverses involves finding a value that, when multiplied by a given non-zero element, results in the multiplicative identity, usually the number 1 in the context of standard numbers. For the set \( F = \{ x + \sqrt{2} y \mid x, y \in \mathbb{Q} \} \), this means confirming that for every non-zero element of \( F \), there also exists an inverse within \( F \).

Consider the element \( a = x + \sqrt{2}y \). Its inverse is formulated as \( a^{-1} = \frac{x - \sqrt{2}y}{x^2 - 2y^2} \), provided \( x^2 eq 2y^2 \). First, the denominator \( x^2 - 2y^2 \) must be non-zero to avoid division by zero, which would otherwise invalidate the expression. As long as this condition is satisfied, the inverse expression falls within \( F \) because \( x - \sqrt{2}y \) involves rational numbers, as does the ratio as a whole.

This existence of a multiplicative inverse for every non-zero element showcases one of the essential features of a field, highlighting the algebraic robustness of \( F \), aside from its incompleteness.