Generalizing the binary operation

Oct 13, 2025 by talynt.org 55 views

Introduction

In the realm of abstract algebra, a group is a fundamental concept that has far-reaching implications in various branches of mathematics. A group is defined as a set of elements with a binary operation that satisfies four essential properties: closure, associativity, the existence of an identity element, and the existence of inverse elements. In this article, we will explore the binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ and demonstrate that it yields a group.

The Binary Operation

The binary operation $ *: \mathbb{R}\times \mathbb{R} \backslash {(0,0)} $ is defined as follows:

$ (a,b)(c,d)=(ac-bd,ad+bc) $

This operation takes two elements $(a,b)$ and $(c,d)$ from the set $\Bbb R\times\Bbb R\backslash (0,0)$ and produces a new element $(ac-bd,ad+bc)$ .

Closure Property

To show that the binary operation satisfies the closure property, we need to demonstrate that the result of the operation is always an element of the set $\Bbb R\times\Bbb R\backslash (0,0)$ . Let's consider two elements $(a,b)$ and $(c,d)$ from the set. We can calculate the result of the operation as follows:

$ (a,b)(c,d)=(ac-bd,ad+bc) $

Since $a,b,c,d\in\Bbb R$ , we know that $ac-bd$ and $ad+bc$ are also real numbers. Therefore, the result of the operation is an element of $\Bbb R\times\Bbb R$ , but we need to show that it is not equal to $(0,0)$ . Suppose that $(ac-bd,ad+bc)=(0,0)$ . Then, we must have:

$ ac-bd=0 $ and $ ad+bc=0 $

Subtracting the second equation from the first, we get:

$ (a^2+b2)c=0 $

Since $a^2+b^2\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $c=0$ . Substituting this into the first equation, we get:

$ a^2b=0 $

Since $a^2\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $b=0$ . Therefore, $(a,b)=(0,0)$ , which is a contradiction. Hence, the result of the operation is always an element of $\Bbb R\times\Bbb R\backslash (0,0)$ , and the closure property is satisfied.

Associativity Property

To show that the binary operation satisfies the associativity property, we need to demonstrate that the result of the operation is independent of the order in which the elements are combined. Let's consider three elements $(a,b)$ , $(c,d)$ , and $(e,f)$ from the set. We can calculate the result of the operation as follows:

$ (a,b)((c,d)(e,f))=(a,b)((ce-df,cf+de))=(a(ce-df)-b(cf+de),a(cf+de)+b(ce-df)) $

$ ((a,b)(c,d))(e,f)=((ac-bd,ad+bc))(e,f)=((ac-bd)e-(ad+bc)f,(ac-bd)f+(ad+bc)e) $

Since the two expressions are equal, we can conclude that the associativity property is satisfied.

Identity Element

To show that the binary operation has an identity element, we need to find an element $(x,y)$ such that for any element $(a,b)$ , we have:

$ (a,b)(x,y)=(a,b) $

Let's consider an element $(x,y)$ from the set. We can calculate the result of the operation as follows:

$ (a,b)(x,y)=(ax-by,ay+bx) $

To satisfy the equation $(a,b)(x,y)=(a,b)$ , we must have:

$ ax-by=a $ and $ ay+bx=b $

Subtracting the second equation from the first, we get:

$ (a^2+b2)x=0 $

Since $a^2+b^2\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $x=0$ . Substituting this into the first equation, we get:

$ -by=a $

Since $a\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $y=0$ . Therefore, the identity element is $(0,0)$ .

Inverse Elements

To show that the binary operation has inverse elements, we need to find an element $(x,y)$ such that for any element $(a,b)$ , we have:

$ (a,b)(x,y)=(0,0) $

Let's consider an element $(a,b)$ from the set. We can calculate the result of the operation as follows:

$ (a,b)(x,y)=(ax-by,ay+bx) $

To satisfy the equation $(a,b)(x,y)=(0,0)$ , we must have:

$ ax-by=0 $ and $ ay+bx=0 $

Subtracting the second equation from the first, we get:

$ (a^2+b2)x=0 $

Since $a^2+b^2\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $x=0$ . Substituting this into the first equation, we get:

$ -by=a $

Since $a\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $y=0$ . Therefore, the inverse element of $(a,b)$ is $(0,0)$ .

Conclusion

In this article, we have demonstrated that the binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ satisfies the four essential properties of a group: closure, associativity, the existence of an identity element, and the existence of inverse elements. Therefore, we can conclude that the operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ yields a group.

References

[1] Abstract Algebra by David S. Dummit and Richard M. Foote
[2] Group Theory by Joseph J. Rotman
[3] Functions by Michael Spivak
[4] Lie Groups by Anthony W. Knapp
[5] Binary Operations by John L. Kelley
Q&A: The Operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ Yields a Group ===========================================================

Q: What is the binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ ?

A: The binary operation $ *: \mathbb{R}\times \mathbb{R} \backslash {(0,0)} $ is defined as follows:

$ (a,b)(c,d)=(ac-bd,ad+bc) $

This operation takes two elements $(a,b)$ and $(c,d)$ from the set $\Bbb R\times\Bbb R\backslash (0,0)$ and produces a new element $(ac-bd,ad+bc)$ .

Q: Why is the set $\Bbb R\times\Bbb R\backslash (0,0)$ used in the binary operation?

A: The set $\Bbb R\times\Bbb R\backslash (0,0)$ is used to exclude the element $(0,0)$ , which would otherwise cause problems with the operation. Specifically, if $(a,b)=(0,0)$ , then the operation would result in $(0,0)$ for any $(c,d)$ , which would violate the properties of a group.

Q: How does the binary operation satisfy the closure property?

A: To show that the binary operation satisfies the closure property, we need to demonstrate that the result of the operation is always an element of the set $\Bbb R\times\Bbb R\backslash (0,0)$ . Let's consider two elements $(a,b)$ and $(c,d)$ from the set. We can calculate the result of the operation as follows:

$ (a,b)(c,d)=(ac-bd,ad+bc) $

$ ac-bd=0 $ and $ ad+bc=0 $

Subtracting the second equation from the first, we get:

$ (a^2+b2)c=0 $

Since $a^2+b^2\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $c=0$ . Substituting this into the first equation, we get:

$ a^2b=0 $

Q: How does the binary operation satisfy the associativity property?

A: To show that the binary operation satisfies the associativity property, we need to demonstrate that the result of the operation is independent of the order in which the elements are combined. Let's consider three elements $(a,b)$ , $(c,d)$ , and $(e,f)$ from the set. We can calculate the result of the operation as follows:

$ (a,b)((c,d)(e,f))=(a,b)((ce-df,cf+de))=(a(ce-df)-b(cf+de),a(cf+de)+b(ce-df)) $

$ ((a,b)(c,d))(e,f)=((ac-bd,ad+bc))(e,f)=((ac-bd)e-(ad+bc)f,(ac-bd)f+(ad+bc)e) $

Since the two expressions are equal, we can conclude that the associativity property is satisfied.

Q: What is the identity element of the binary operation?

A: The identity element of the binary operation is $(0,0)$ .

Q: How do we find the inverse element of a given element?

A: To find the inverse element of a given element $(a,b)$ , we need to find an element $(x,y)$ such that:

$ (a,b)(x,y)=(0,0) $

Let's consider an element $(a,b)$ from the set. We can calculate the result of the operation as follows:

$ (a,b)(x,y)=(ax-by,ay+bx) $

To satisfy the equation $(a,b)(x,y)=(0,0)$ , we must have:

$ ax-by=0 $ and $ ay+bx=0 $

Subtracting the second equation from the first, we get:

$ (a^2+b2)x=0 $

Since $a^2+b^2\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $x=0$ . Substituting this into the first equation, we get:

$ -by=a $

Since $a\neq 0$ (otherwise, $(a,b)=(0,0)$ , which is excluded from the set), we can conclude that $y=0$ . Therefore, the inverse element of $(a,b)$ is $(0,0)$ .

Q: What are some real-world applications of the binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ ?

A: The binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ has several real-world applications, including:

Linear Algebra: The binary operation can be used to represent linear transformations between vector spaces.
Differential Equations: The binary operation can be used to solve systems of differential equations.
Computer Science: The binary operation can be used in computer graphics and game development to represent transformations between 2D and 3D spaces.

Q: What are some limitations of the binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ ?

A: The binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ has several limitations, including:

Non-commutativity: The binary operation is not commutative, meaning that the order of the elements matters.
Non-associativity: The binary operation is not associative, meaning that the order in which the elements are combined matters.
Lack of identity element: The binary operation does not have an identity element, meaning that there is no element that leaves other elements unchanged.

Q: What are some future directions for research on the binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ ?

A: Some future directions for research on the binary operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\backslash (0,0)$ include:

Generalizing the binary operation: Researchers could explore generalizing the binary operation to other sets, such as $\Bbb C\times\Bbb C$ or $\Bbb H\times\Bbb H$ .
Developing new applications: Researchers could explore developing new applications for the binary operation, such as in computer science or physics.
Investigating the properties of the binary operation: Researchers could explore investigating the properties of the binary operation, such as its commutativity, associativity, and distributivity.