If the group (z, ∗) of all integers, where a ∗ b = a + b + 1 for all a, b ∈ z, the inverse of -2 is
0
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.
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.
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.
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.
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
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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 ?
The multiplicative group {1, -1, i, -i} is a cyclic group, its generators are
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.