Problem 21 - Algebra: Pure and Applied

Question

Let \(G=\\{a+b \sqrt{2} \mid a, b \in Q\\}\). Show that \(G\) is a subgroup of \(\mathbb{R}\) under addition.

Step-by-Step Solution

Answer

Set \(G\) is a subgroup of \(\mathbb{R}\) under addition because it satisfies closure, contains the additive identity, and includes inverses.

Step 1: Define the Subgroup Criteria

To show that a set is a subgroup under addition, three criteria must be satisfied: 1) Closure under addition, 2) Existence of an additive identity, 3) Existence of an additive inverse for each element.

Step 2: Verify Closure Under Addition

Take two arbitrary elements from set \(G\), i.e., \(x = a + b\sqrt{2}\) and \(y = c + d\sqrt{2}\), where \(a, b, c, d \in \mathbb{Q}\). Add these two elements to get \(x + y = (a+c) + (b+d)\sqrt{2}\). Since \(a+c\) and \(b+d\) are both in \(\mathbb{Q}\), \(x + y\) is also in \(G\). Thus, \(G\) is closed under addition.

Step 3: Identify the Additive Identity

The additive identity in \(\mathbb{R}\) is 0. In the context of \(G\), this can be written as \(0 + 0\sqrt{2}\), which is clearly in \(G\) because both coefficients are rational: \(0 \in \mathbb{Q}\). Thus, the additive identity is in \(G\).

Step 4: Verify Existence of Additive Inverses

For any element \(x = a + b\sqrt{2}\) in \(G\), the additive inverse is \(-x = -a - b\sqrt{2}\). Since \(-a\) and \(-b\) are also rational numbers, \(-x\) is an element of \(G\). Therefore, every element in \(G\) has an additive inverse.

Key Concepts

Subgroup Criteria

To determine whether a set is a subgroup under addition, we need to verify three important criteria. These criteria are:

• Closure Under Addition: If we take any two elements from the set, their sum should also be in the set.
• Additive Identity: There should be an element in the set that acts as an identity for addition, meaning any element added to this identity will remain unchanged.
• Additive Inverses: For each element in the set, there must be another element in the set such that their sum equals the additive identity.

Subgroups must satisfy all of these criteria, ensuring they are neatly contained within their parent group under addition. This structure keeps mathematical consistency intact!

Closure Under Addition

Closure under addition is the foundational step in proving a subgroup. We take two arbitrary elements from the set, say:

• x = a + b\sqrt{2}
• y = c + d\sqrt{2}

where \(a, b, c, d \in \mathbb{Q}\), the set of rational numbers. When these elements are added, they combine as follows:\[x + y = (a + c) + (b + d)\sqrt{2}\]Since \(a+c\) and \(b+d\) are also rational, \(x + y\) fits the definition and remains within the set \(G\). This verifies our closure condition, signifying that any elements combined maintain membership in the set.

Additive Identity

The additive identity is a significant concept in group theory. In the real numbers \(\mathbb{R}\), zero acts as the universal additive identity since any number plus zero remains unchanged. For our set \(G = \{a + b\sqrt{2} | a, b \in \mathbb{Q}\}\), the role of the additive identity is fulfilled by the element \(0 + 0\sqrt{2}\).

Both coefficients, 0, are rational, thus preserving the identity within the set. Consequently, adding this identity element to any number in the set \(G\) leaves it unchanged, meeting the criteria for an additive identity within \(G\).

Additive Inverses

Additive inverses ensure every element can "cancel out" another, reverting back to the identity element, zero. For any number \(x = a + b\sqrt{2}\) in our set \(G\), the inverse is:\[-x = -a - b\sqrt{2}\]Both \(-a\) and \(-b\) remain rational, confirming that \(-x\) belongs to \(G\).

By adding \(x\) and \(-x\), we achieve the identity:\[x + (-x) = (a - a) + (b - b)\sqrt{2} = 0 + 0\sqrt{2} = 0\]Thus, every element in \(G\) has an inverse also within \(G\), securing this essential subgroup property.