Question

Define a binary operation \(\bullet\) on the set of real numbers by $$ x \bullet y=x+y+r x y $$ where \(r\) is a non-zero real number. Show that the operation \(\bullet\) is associative. Prove that \(x \bullet y=-r^{-1}\) if, and only if, \(x=-r^{-1}\) or \(y=-r^{-1}\). Hence prove that the set of all real numbers excluding \(-r^{-1}\) forms a group under the operation \(\bullet\).

Short answer

The operation \(\bullet\) is associative. \(x \bullet y = -r^{-1} \) if and only if \(x = -r^{-1} or \ y = -r^{-1} \). So \ \(\mathbb{R}\setminus \{-r^{-1}\}\ \) forms a group under \(\bullet\).

Step-by-step solution

Verify associativity

The binary operation \(\bullet\) is associative if \( (x \bullet y) \bullet z = x \bullet (y \bullet z) \). Calculate both sides of the equation:1. Compute \(x \bullet y:\)\[x \bullet y = x + y + rxy\]2. Compute \((x \bullet y) \bullet z:\)\[ (x \bullet y) \bullet z = (x + y + rxy) \bullet z = (x + y + rxy) + z + r(x + y + rxy)z \]Simplify the result:\[(x + y + rxy) \bullet z = x + y + rxy + z + r(xz + yz + rxyz)\]3. Compute \(y \bullet z:\)\[ y \bullet z = y + z + ryz\]4. Compute \((x \bullet (y \bullet z)):\)\[ x \bullet (y \bullet z) = x \bullet (y + z + ryz) = x + (y + z + ryz) + rx(y + z + ryz)\]Simplify the result:\[x \bullet (y + z + ryz) = x + y + z + ryz + rxy + rxz + r^2xyz\]After simplification, both expressions are equivalent:\[(x + y + rxy) + z + r(xz + yz + rxyz) = x + y + z + ryz + rxy + rxz + r^2xyz\]Thus, the operation \(\bullet\) is associative.

Prove \( x \bullet y = -r^{-1} \)

First, set up the equation:\[ x \bullet y = x + y + rxy = -r^{-1}\]Rearrange to:\[ x + y + rxy + r^{-1} = 0\]Factor to simplify:\[(x + r^{-1})(y + r^{-1}) = 0 \]This implies:1. \( x + r^{-1} = 0 \, \text{or} \ y + r^{-1} = 0 \)2. Therefore, \(x = -r^{-1}\ or \ y = -r^{-1}\).

Forming a group with \ \(\mathbb{R}\setminus \{-r^{-1}\}\ \) under \(\bullet \)

The set \ \(\mathbb{R}\setminus \{-r^{-1}\}\) \ \) should fulfill four properties to form a group under the operation \( \bullet \): closure, associativity, identity, and inverses.1. **Associativity**: Has been proven in Step 1.2. **Identity**: Let \( e \) be the identity element such that \( x \bullet e = x\). Solving \( x + e + rxe = x \, gives \ e = 0 \). Hence, \ 0 \ is the identity element.3. **Inverses**: For every \ x \ in \ \(\mathbb{R}\setminus \{-r^{-1}\}\ \ , find \ y \ such that \ x \bullet y = 0 \). Solving \( x + y + rxy = 0 \, gives \ y = \-x/(1+rx) \), which exists as long as \ x \eq -r^{-1} \.Therefore, \ \(\mathbb{R}\setminus \{-r^{-1}\}\ \ forms a group under \ \(\bullet\).

Key concepts

Associativity

In mathematics, a binary operation \(\bullet\) is said to be associative if it satisfies the condition \( (x \bullet y) \bullet z = x \bullet (y \bullet z) \) for all elements x, y, and z in a set.
Let's review this concept with the given operation, which is defined as \(x \bullet y = x + y + rxy\), where \(r\) is a non-zero real number.
To check if this operation is associative, we compute both sides of the equation \( (x \bullet y) \bullet z = x \bullet (y \bullet z) \):

1. Compute \(x \bullet y\):

\[ x \bullet y = x + y + rxy \]
2. Then compute \((x \bullet y) \bullet z\):

\[ (x + y + rxy) \bullet z = (x + y + rxy) + z + r(x + y + rxy)z \]\
Simplify the result:
\[ x + y + rxy + z + r(xz + yz + rxyz) \]
3. Next, compute \(y \bullet z\):

\[ y \bullet z = y + z + ryz \]
4. Finally, compute \(x \bullet (y \bullet z)\):

\[ x \bullet (y + z + ryz) = x + (y + z + ryz) + rx(y + z + ryz) \]
Simplify the outcome:
\[ x + y + z + ryz + rxy + rxz + r^2xyz \]
When we compare both sides, we find they are equal, meaning the associative property holds.
This shows \( \bullet \) is an associative operation.

Identity Element

An identity element in a set under a binary operation is an element that, when combined with any element of the set using the operation, leaves the other element unchanged.
Here, we wish to find an identity element \(e\) for the given operation \(x \bullet y = x + y + rxy\). This element \(e\) must satisfy the condition:

\[x \bullet e = x \]
To find \(e\), solve the equation:

\[ x + e + rxe = x \]
By canceling \(x\) on both sides, we have:

\[ e + rxe = 0 \]
Since \(r\) is non-zero and we need this to hold for any \(x\), the only value of \(e\) is 0.
This implies:

• The identity element for the operation \(x \bullet y\) on this set of real numbers is 0.

Group Theory

Group theory studies the algebraic structures known as groups, which are defined formally as a set equipped with an operation that combines any two of its elements to form a third element. A group needs to satisfy four main properties:

• Closure
• Associativity
• Identity element
• Inverses

Let's use these conditions to determine if the set \( \mathbb{R}\setminus \{-r^{-1}\}\ \) forms a group under our binary operation \(\bullet\):
**Closure:** For all \(x, y\) in the set, the result of \(x \bullet y \) must also be in \( \mathbb{R}\setminus \{-r^{-1}\}\ \), which is naturally satisfied as -\(r^{-1}\) is excluded.
**Associativity:** Proven in an earlier section.
**Identity:** As seen, the identity element \(e\) is 0.
**Inverses:** For each \(x\) in \( \mathbb{R}\setminus \{-r^{-1}\}\ \), there must exist a \(y\) such that\( x \bullet y = e = 0 \). Solving this, we have:

\[ x + y + rxy = 0 \]

This can be rearranged to:

\[y = \-x/(1+rx) \]
This works as long as \(x eq -r^{-1}\), providing an inverse for all elements in the set \( \mathbb{R}\setminus \{-r^{-1}\}\ \).

Hence, the set \( \mathbb{R}\setminus \{-r^{-1}\}\ \) satisfies all group properties under the operation \( \bullet \).