Question

Show that (a) the product on \(\mathbb{R}^{2}\) defined by $$ \left(x_{1}, x_{2}\right)\left(y_{1}, y_{2}\right)=\left(x_{1} y_{1}-x_{2} y_{2}, x_{1} y_{2}+x_{2} y_{1}\right) $$ turns \(\mathbb{R}^{2}\) into an associative and commutative algebra, and (b) the cross product on \(\mathbb{R}^{3}\) turns it into a nonassociative, noncommutative algebra.

Short answer

The operation in \(\mathbb{R}^{2}\) creates an algebra that is both associative and commutative. The cross product in \(\mathbb{R}^{3}\), on the other hand, generates an algebra that is neither associative nor commutative.

Step-by-step solution

Prove Associativity for (a)

Apply the operation to three elements \((x_{1}, x_{2})\), \((y_{1}, y_{2})\), and \((z_{1}, z_{2})\) from \(\mathbb{R}^{2}\). Check if \((x_{1}, x_{2}) \cdot ((y_{1}, y_{2}) \cdot (z_{1}, z_{2})) = ((x_{1}, x_{2}) \cdot (y_{1}, y_{2})) \cdot (z_{1}, z_{2})\). Apply the operation definition to confirm.

Prove Commutativity for (a)

Take two elements \((x_{1}, x_{2})\) and \((y_{1}, y_{2})\) from \(\mathbb{R}^{2}\). Confirm if \((x_{1}, x_{2}) \cdot (y_{1}, y_{2}) = (y_{1}, y_{2}) \cdot (x_{1}, x_{2})\). Use the operation definition to verify.

Prove Nonassociativity for (b)

Pick three vectors \(a\), \(b\), and \(c\) from \(\mathbb{R}^{3}\). Check if \(a \times (b \times c) = (a \times b) \times c\). Use cross-product properties that, in general, aren't equal to confirm non-associativity.

Prove Noncommutativity for (b)

Take two vectors \(a\) and \(b\) from \(\mathbb{R}^{3}\). Validate if \(a \times b = b \times a\). Use the cross-product properties, where \(a \times b = - (b \times a)\) to confirm non-commutativity.

Key concepts

Associativity

Associativity is a fundamental property in many mathematical and algebraic structures. It dictates that the way in which operations are grouped in expressions should not change the outcome. For a set with an operation to be associative, we need to show that for any three elements \(a\), \(b\), \(c\), the equation \((a \cdot b) \cdot c = a \cdot (b \cdot c)\) holds true.

In our exercise, we are examining the operation defined on \(\mathbb{R}^2\). We apply this operation to three pairs of numbers, like \((x_1, x_2)\), \((y_1, y_2)\), and \((z_1, z_2)\). Through verification, we see that the operation brings the same results regardless of how we group them:

• First compute: \((x_1, x_2) \cdot ((y_1, y_2) \cdot (z_1, z_2))\)
• Compare it to: \(((x_1, x_2) \cdot (y_1, y_2)) \cdot (z_1, z_2)\)

If both expressions result in the same pair, the operation is associative. This has been confirmed with the provided operation definition.

Commutativity

Commutativity is another crucial property in mathematics, reflecting the idea that changing the order of operations does not affect the result. To show that a structure is commutative, we test if \(a \cdot b = b \cdot a\) for any elements \(a\) and \(b\) from the set.

In our example with \(\mathbb{R}^2\), we compare \((x_1, x_2) \cdot (y_1, y_2)\) with \((y_1, y_2) \cdot (x_1, x_2)\) using the given operation:

• The operation gives us: \((x_1 y_1 - x_2 y_2, x_1 y_2 + x_2 y_1)\)
• Swapping elements yields the same result: \((y_1 x_1 - y_2 x_2, y_1 x_2 + y_2 x_1)\)

Therefore, the order in which the operation is applied doesn’t affect the outcome, indicating that it is indeed commutative in \(\mathbb{R}^2\).

Cross Product

The cross product is a unique operation in three-dimensional vector spaces, specifically \(\mathbb{R}^3\), creating a vector orthogonal to the two input vectors. It has specific properties distinguishing it from many other operations.

The cross product \(a \times b\) does not uphold associativity, unlike multiplication in arithmetic or vector addition. For example:

• If we have vectors \(a\), \(b\), and \(c\), the expressions \(a \times (b \times c)\) and \((a \times b) \times c\) yield different results.

This is because the direction and magnitude of the resultant vector depend on orientation and order.

Furthermore, the cross product is not commutative. For vectors \(a\) and \(b\), \(a \times b\) equals \(- (b \times a)\). Thus, flipping the vectors results in the opposite direction for the resultant vector. These properties illustrate why the cross product is a noncommutative and nonassociative operation.

Nonassociative Algebras

Algebras where the associative property does not hold are termed nonassociative algebras. These can seem less intuitive because the order of operations significantly affects the result. Famous examples come from physics and advanced mathematics, like the cross product discussed earlier.

In nonassociative algebras, rearranging parentheses in expressions like \((a \times b) \times c\) and \(a \times (b \times c)\) makes a difference in the outcome. This is key when operations must consider specific sequence, showing why these algebras are useful in certain contexts.

Examples include:

• The cross product of vectors in \(\mathbb{R}^3\)
• Quaternion multiplication (used in 3D rotations)

Understanding nonassociative algebras can help students manage systems where operations aren't straightforwardly interchangeable, providing insights into more complex areas of study.