Show that the set of rational numbers (positive and negative fractions) together with multiplication is not a group. By removing one element, it can be made into a group. Which element? Explain.
Show that the set of rational numbers (positive and negative fractions) together with multiplication is not a group. By removing one element, it can be made into a group. Which element? Explain.
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Report on the concept of a group by sharing your understanding of this algebraic structure.

Transcribed Image Text:**Question 6: Group Theory and Rational Numbers**
Show that the set of rational numbers (positive and negative fractions) together with multiplication is not a group. By removing one element, it can be made into a group. Which element? Explain.
**Solution Explanation:**
To demonstrate why the set of rational numbers with multiplication is not a group, we need to consider the group axioms: closure, associativity, identity, and invertibility.
1. **Closure**: Multiplying any two rational numbers results in another rational number. Thus, closure is satisfied.
2. **Associativity**: Multiplication of rational numbers is associative. For any rational numbers \(a\), \(b\), and \(c\), \((a \cdot b) \cdot c = a \cdot (b \cdot c)\). Therefore, associativity holds.
3. **Identity**: The identity element under multiplication is 1, since for any rational number \(a\), \(a \cdot 1 = a = 1 \cdot a\). Hence, there is an identity element.
4. **Invertibility**: For a group, every element must have an inverse such that the element multiplied by its inverse gives the identity element (1). In the set of rational numbers, every number \(a\) (except 0) has an inverse \(1/a\). However, 0 does not have a multiplicative inverse, as there is no number \(b\) such that \(0 \cdot b = 1\).
Since 0 does not have a multiplicative inverse, the set of all rational numbers with multiplication is not a group.
**Modification to Form a Group:**
By removing the element 0, we make the set into a group. The set of non-zero rational numbers satisfies all group axioms with respect to multiplication:
- Closed under multiplication.
- Associative.
- Contains an identity element (1).
- Every element has a multiplicative inverse.
Therefore, the set of non-zero rational numbers under multiplication is indeed a group.
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