All Exams Test series for 1 year @ ₹349 only
Question

If the group (z, ∗) of all integers, where a ∗ b = a + b + 1 for all a, b ∈ z, the inverse of -2 is

The correct answer is

0

Understanding the Group and Binary Operation

The problem asks us to find the inverse of the element -2 within a specific algebraic structure. This structure is the set of all integers, denoted by \( \mathbb{Z} \), together with a special binary operation denoted by \( \ast \). The binary operation \( \ast \) is defined for any two integers \( a \) and \( b \) as \( a \ast b = a + b + 1 \).

We are told that \( (\mathbb{Z}, \ast) \) forms a group. In any group, each element must have an inverse element. The existence of an **inverse element** is a fundamental property of a group.

Finding the Identity Element

To find the inverse of an element, we first need to know the identity element for this binary operation \( \ast \). The identity element, often denoted by \( e \), is an element in the group such that for any element \( a \in \mathbb{Z} \), \( a \ast e = a \) and \( e \ast a = a \).

Let's use the definition of the binary operation to find \( e \):

Using \( a \ast e = a \):

\( a + e + 1 = a \)

Subtracting \( a \) from both sides:

\( e + 1 = 0 \)

Subtracting 1 from both sides:

\( e = -1 \)

Let's verify this using \( e \ast a = a \):

\( e \ast a = -1 \ast a = -1 + a + 1 = a \)

Since both conditions are satisfied, the identity element \( e \) for the group \( (\mathbb{Z}, \ast) \) is -1. Finding the **identity element** is the necessary first step to compute any **inverse element** in a group.

Calculating the Inverse Element of -2

Now we need to find the **inverse element** of -2. Let's call the inverse of -2 as \( x \). By the definition of an inverse element in a group, the inverse \( x \) of -2 must satisfy:

\( -2 \ast x = e \)

and

\( x \ast (-2) = e \)

where \( e = -1 \) is the identity element we just found.

Let's use the first equation and the definition of the binary operation \( \ast \):

\( -2 \ast x = -1 \)

\( -2 + x + 1 = -1 \)

Simplify the left side:

\( x - 1 = -1 \)

Add 1 to both sides to solve for \( x \):

\( x = -1 + 1 \)

\( x = 0 \)

So, the calculated inverse element is 0. Let's verify this using the second equation, \( x \ast (-2) = e \):

\( 0 \ast (-2) = 0 + (-2) + 1 = -2 + 1 = -1 \)

Since \( 0 \ast (-2) = -1 \), and \( -2 \ast 0 = -1 \), where -1 is the identity element, 0 is indeed the inverse of -2 in this group \( (\mathbb{Z}, \ast) \).

This process demonstrates how the binary operation and the **identity element** work together to determine the **inverse element** for any given element in the group of integers.

Conclusion

For the group \( (\mathbb{Z}, \ast) \) where \( a \ast b = a + b + 1 \), the identity element is -1. The inverse element of -2 is the element \( x \) such that \( -2 \ast x = -1 \). Solving \( -2 + x + 1 = -1 \) gives \( x = 0 \).

The final answer is 0.

Was this answer helpful?

Important Questions from Groups

  1. Consider the following statements:

    S 1: If a group (G, *) is of order n, and a ∈ G is such that a m= e for some integer m ≤ n, then m must divide n.

    S 2: If a group (G, *) is of even order, then there must be an element a ∈ G such that a ≠ e and a * a = e

    Which of the statements is (are) correct
  2. If a group G is internal Direct product of its subgroups A, B, C, .... Z then G is isomorphic to ______.

  3. Every element of a group G when expressed as internal Direct product of a, b, c, ... z if and only of every element is uniquely expressed as ?

  4. The multiplicative group {1, -1, i, -i} is a cyclic group, its generators are

  5. Given:

    Statement A: All cyclic groups are an abelian group.

    Statement B: The order of the cyclic group is the same as the order of its generator.

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App