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The following is a 'proof' that reflexivity is an unnecessary axiom for an equivalence relation. Because of symmetry \(X \sim Y\) implies \(Y \sim X .\) Then by transitivity \(X \sim Y\) and \(Y \sim X\) imply \(X \sim X .\) Thus symmetry and transitivity imply reflexivity, which therefore need not be separately required. Demonstrate the flaw in this proof using the set consisting of all real numbers plus the number \(i .\) Show by investigating the following specific cases that, whether or not reflexivity actually holds, it cannot be deduced from symmetry and transitivity alone. (a) \(X \sim Y\) if \(X+Y\) is real. (b) \(X \sim Y\) if \(X Y\) is real.

Short Answer

Expert verified
Examples show reflexivity is not deduced from symmetry and transitivity alone.

Step by step solution

01

- Understand Reflexivity, Symmetry, and Transitivity

First, recall the definitions: **Reflexivity** means for any element X, we have X ~ X. **Symmetry** means if X ~ Y, then Y ~ X. **Transitivity** means if X ~ Y and Y ~ Z, then X ~ Z. Reflexivity is not necessarily implied by symmetry and transitivity alone.
02

- Consider Set with Real Numbers and Imaginary Unit

Use the set of all real numbers plus the imaginary unit i. We will define two relationships and test if reflexivity can be deduced from symmetry and transitivity.
03

- Case (a): Investigate Relationship X ~ Y if X+Y is Real

For X ~ Y to hold if and only if X+Y is real, let's check the properties. Take real number 1 and i: 1 + i and 1 + (-i) are not real. Clearly, symmetry holds ((X+Y) real implies (Y+X) real), and transitivity holds too. However, reflexivity fails as X + X is not always real.
04

- Analysis of (a)

If X is i, then i + i = 2i is not real. Thus the relationship X ~ X does not hold for imaginary unit i, showing reflexivity doesn't follow from symmetry and transitivity.
05

- Case (b): Investigate Relationship X ~ Y if X * Y is Real

For X ~ Y to hold if and only if X * Y is real, let's check the properties. If X = 1 and Y = i: 1 * i is imaginary hence not real. Reflexivity fails as X * X is not always real.
06

- Analysis of (b)

If X is i, then i * i = -1 is real, showing reflexivity is valid. But for combinations like real and i, X ~ Y fails to hold true for real numbers with imaginary relationships.
07

Summary

These results demonstrate reflexivity cannot be deduced from symmetry and transitivity alone using this set and relations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reflexivity
Reflexivity is a crucial concept in the context of equivalence relations. It states that for any element X, X must be related to itself. Mathematically, this means that for every X in a set, we have X ∼ X. This property ensures that every element is self-related, offering a starting point for comparing other elements in the set.
Symmetry
Symmetry is another important property for equivalence relations. Symmetry means that if an element X is related to an element Y, then Y must also be related to X. In other words, if X ∼ Y, then Y ∼ X. This ensures that the relationship is mutual, providing an equal footing between pairs of elements.
Transitivity
Transitivity is fundamental to maintaining consistency in equivalence relations. A relation is transitive if, whenever X is related to Y and Y is related to Z, then X must also be related to Z. Formally, this is written as if X ∼ Y and Y ∼ Z, then X ∼ Z. This property guarantees that the relation can be extended through intermediary elements, ensuring coherence within the set.
Proof Flaw
The given proof incorrectly suggests that reflexivity can be derived from symmetry and transitivity. By using the set of all real numbers plus the imaginary unit i, we can show this is not true. Consider the relations defined in the problem:
  • Case (a) X ∼ Y if X + Y is real: This relation is symmetric and transitive, but not reflexive since X + X is not always real (e.g., i + i = 2i).
  • Case (b) X ∼ Y if XY is real: This relation, while also symmetric and transitive, is not reflexive as X * X is not always real (e.g., 1 * i = i).
Thus, whether or not reflexivity holds in specific cases, it cannot be deduced solely from symmetry and transitivity.
Real and Imaginary Numbers
Real numbers and imaginary numbers are fundamental in understanding the proofs provided. Real numbers are all numbers on the number line, including both positive and negative numbers, as well as zero. Imaginary numbers are multiples of the imaginary unit i, where i is defined as \(\sqrt{-1}\). Combining real and imaginary numbers forms complex numbers, which are used in advanced mathematics and engineering. Examining relationships with real and imaginary numbers helps illustrate why reflexivity cannot be assumed from just symmetry and transitivity.

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Most popular questions from this chapter

Given that matrix \(M\) is a member of the multiplicative group \(G L(3, \mathbf{R})\), determine, for each of the following additional constraints on \(M\) (applied separately), whether the subset satisfying the constraint is a subgroup of \(G L(3, \mathbf{R})\) : (a) \(\mathrm{M}^{T}=\mathrm{M}\) (b) \(\mathrm{M}^{\bar{T}} \mathrm{M}=\mathrm{I}\); (c) \(|\mathrm{M}|=1\);

The group of all non-singular \(n \times n\) matrices is known as the general linear group \(G L(n)\) and that with only real elements as \(G L(n, \mathbf{R}) .\) If \(\mathbf{R}^{*}\) denotes the multiplicative group of non-zero real numbers, prove that the mapping \(\Phi\) : \(G L(n, \mathbf{R}) \rightarrow \mathbf{R}^{*}\), defined by \(\Phi(\mathrm{M})=\operatorname{det} \mathrm{M}\), is a homomorphism. Show that the kernel \(\mathcal{K}\) of \(\Phi\) is a subgroup of \(G L(n, \mathbf{R})\). Determine its cosets and show that they themselves form a group.

If \(\mathcal{A}\) and \(\mathcal{B}\) are two groups then their direct product, \(A \times \mathcal{B}\), is defined to be the set of ordered pairs \((X, Y)\), with \(X\) an element of \(\mathcal{A}, Y\) an element of \(\mathcal{B}\) and multiplication given by \((X, Y)\left(X^{\prime}, Y^{\prime}\right)=\left(X X^{\prime}, Y Y^{\prime}\right)\). Prove that \(\mathcal{A} \times \mathcal{B}\) is a group. Denote the cyclic group of order \(n\) by \(\mathcal{C}_{n}\) and the symmetry group of a regular \(n\)-sided figure (an \(n\)-gon) by \(\mathcal{D}_{n}-\) thus \(\mathcal{D}_{3}\) is the symmetry group of an equilateral triangle, as discussed in the text. (a) By considering the orders of each of their elements, show (i) that \(\mathcal{C}_{2} \times \mathcal{C}_{3}\) is isomorphic to \(\mathcal{C}_{6}\), and (ii) that \(\mathcal{C}_{2} \times \mathcal{D}_{3}\) is isomorphic to \(\mathcal{D}_{6}\). (b) Are any of \(\mathcal{D}_{4}, \mathcal{C}_{8}, \mathcal{C}_{2} \times \mathcal{C}_{4}, \mathcal{C}_{2} \times \mathcal{C}_{2} \times \mathcal{C}_{2}\) isomorphic?

If \(\mathcal{A}\) is a group in which every element other than the identity, \(I\), has order 2, prove that \(\mathcal{A}\) is Abelian. Hence show that if \(X\) and \(Y\) are distinct elements of \(\mathcal{A}\), neither being equal to the identity, then the set \(\\{I, X, Y, X Y\\}\) forms a subgroup of \(\mathcal{A}\). Deduce that if \(\mathcal{B}\) is a group of order \(2 p\), with \(p\) a prime greater than 2, then \(\mathcal{B}\) must contain an element of order \(p .\)

(a) Denote by \(A_{n}\) the subset of the permutation group \(S_{n}\) that contains all the even permutations. Show that \(A_{n}\) is a subgroup of \(S_{n}\). (b) List the elements of \(S_{3}\) in cycle notation and identify the subgroup \(A_{3} .\) (c) For each element \(X\) of \(S_{3}\), let \(p(X)=1\) if \(X\) belongs to \(A_{3}\) and \(p(X)=-1\) if it does not. Denote by \(\mathcal{C}_{2}\) the multiplicative cyclic group of order 2 . Determine the images of each of the elements of \(S_{3}\) for the following four mappings: $$ \begin{array}{ll} \Phi_{1}: S_{3} \rightarrow \mathcal{C}_{2} & X \rightarrow p(X) \\ \Phi_{2}: S_{3} \rightarrow \mathcal{C}_{2} & X \rightarrow-p(X) \\ \Phi_{3}: S_{3} \rightarrow A_{3} & X \rightarrow X^{2} \\ \Phi_{4}: S_{3} \rightarrow S_{3} & X \rightarrow X^{3} \end{array} $$ (d) For each mapping, determine whether the kernel \(K\) is a subgroup of \(S_{3}\) and, if so, whether the mapping is a homomorphism.

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