Problem 2
Question
Define a relation \(\square\) on the integers by \(x \square y \quad \Longleftrightarrow x^{2}=y^{2}\). Show that \(\square\) is an equivalence relation. List the equivalence classes \(x / \square\) for \(0 \leq x \leq 5\)
Step-by-Step Solution
Verified Answer
The relation \( \backslash square \) is an equivalence relation. The equivalence classes for 0 ≤ x ≤ 5 are: \{0\}, \{-1, 1\}, \{-2, 2\}, \{-3, 3\}, \{-4, 4\}, \{-5, 5\}.
1Step 1: Check Reflexivity
A relation \(\backslash square\) on a set is reflexive if every element is related to itself. For any integer x, we need to verify if \(x \square x\):\[x \square x \Longleftrightarrow x^2 = x^2\]which is always true. Thus, \(x \square x \) which confirms reflexivity.
2Step 2: Check Symmetry
A relation \(\backslash square\) on a set is symmetric if for every pair (x, y), if x is related to y, then y is related to x. For any integers x and y, we need to verify if \((x \square y) \implies (y \square x)\):\[x \square y \Longleftrightarrow x^2 = y^2\]which also implies that:\[y^2 = x^2 \Longleftrightarrow y \square x\] Thus, symmetry holds.
3Step 3: Check Transitivity
A relation \(\backslash square\) is transitive if for any three integers x, y, and z, if x is related to y and y is related to z, then x is related to z. We need to verify if \((x \square y \land y \square z) \implies x \square z\):\[x \square y \Longrightarrow x^2 = y^2\]\[y \square z \Longrightarrow y^2 = z^2\]Then:\[x^2 = y^2 = z^2 \Longrightarrow x^2 = z^2 \Longrightarrow x \square z\] Thus, the relation is transitive.
4Step 4: List Equivalence Classes for 0 ≤ x ≤ 5
An equivalence class of x under \( \backslash square \) is the set of all elements y such that \( x \square y \) (i.e., \(x^2 = y^2\)).\[ x = 0 \Rightarrow \{0\}\] \[ x = 1 \Rightarrow \{-1, 1\}\] \[ x = 2 \Rightarrow \{-2, 2\}\] \[ x = 3 \Rightarrow \{-3, 3\}\] \[ x = 4 \Rightarrow \{-4, 4\}\] \[ x = 5 \Rightarrow \{-5, 5\}\]
Key Concepts
ReflexivitySymmetryTransitivityEquivalence Classes
Reflexivity
The property of reflexivity for a relation means that every element is related to itself. In simpler terms, given any element from the set, you should be able to pair it with itself.
For the given relation \(\backslash square\) defined by \(x \square y \Leftrightarrow x^2 = y^2\), to check if it is reflexive, we need to see if each integer is related to itself. Since \(x^2 = x^2\) is always true for any integer \(x\), we can conclude that \(x \square x\) holds for all \(x\).
This confirms that the relation is indeed reflexive. Reflexivity is crucial because it ensures that every element in our set is included in some way within the relation.
For the given relation \(\backslash square\) defined by \(x \square y \Leftrightarrow x^2 = y^2\), to check if it is reflexive, we need to see if each integer is related to itself. Since \(x^2 = x^2\) is always true for any integer \(x\), we can conclude that \(x \square x\) holds for all \(x\).
This confirms that the relation is indeed reflexive. Reflexivity is crucial because it ensures that every element in our set is included in some way within the relation.
Symmetry
A relation is symmetric if for any two elements \((x, y)\) in the set, whenever \(x\) is related to \(y\), \(y\) is also related to \(x\).
To check for symmetry in our relation \(\backslash square\), let's verify if \(x \square y\) implies \(y \square x\). Since our relation is defined as \(x^2 = y^2\), if \(x \square y\) holds, then \(x^2 = y^2\), which logically means that \(y^2 = x^2\), thereby \(y \square x\).
This confirms that the relation is symmetric. Symmetry ensures the bidirectional pairing of elements within the set.
To check for symmetry in our relation \(\backslash square\), let's verify if \(x \square y\) implies \(y \square x\). Since our relation is defined as \(x^2 = y^2\), if \(x \square y\) holds, then \(x^2 = y^2\), which logically means that \(y^2 = x^2\), thereby \(y \square x\).
This confirms that the relation is symmetric. Symmetry ensures the bidirectional pairing of elements within the set.
Transitivity
A relation is transitive if whenever an element \(x\) is related to \(y\) and \(y\) is related to \(z\), then \(x\) must also be related to \(z\).
To verify transitivity for our relation \(\backslash square\), we check whether \(x \square y\) and \(y \square z\) together imply \(x \square z\). If \(x \square y\) is true, then \(x^2 = y^2\). If \(y \square z\) is also true, then \(y^2 = z^2\).
So, if \(x^2 = y^2\) and \(y^2 = z^2\), it follows that \(x^2 = z^2\), thus \(x \square z\).
This confirms that the relation is transitive. Transitivity is key because it links chained relationships within the set.
To verify transitivity for our relation \(\backslash square\), we check whether \(x \square y\) and \(y \square z\) together imply \(x \square z\). If \(x \square y\) is true, then \(x^2 = y^2\). If \(y \square z\) is also true, then \(y^2 = z^2\).
So, if \(x^2 = y^2\) and \(y^2 = z^2\), it follows that \(x^2 = z^2\), thus \(x \square z\).
This confirms that the relation is transitive. Transitivity is key because it links chained relationships within the set.
Equivalence Classes
Equivalence classes are groups into which a set is partitioned by an equivalence relation. Essentially, they are collections of elements that are all related to each other under the given relation.
In our case, we need to find these groups for the relation \(\backslash square\) for the integers from 0 to 5.
\(\backslash text{Here are the equivalence classes:}\)
In our case, we need to find these groups for the relation \(\backslash square\) for the integers from 0 to 5.
\(\backslash text{Here are the equivalence classes:}\)
- For \(x = 0\): \{0\}
- For \(x = 1\): \{-1, 1\}
- For \(x = 2\): \{-2, 2\}
- For \(x = 3\): \{-3, 3\}
- For \(x = 4\): \{-4, 4\}
- For \(x = 5\): \{-5, 5\}
Other exercises in this chapter
Problem 2
The usual algebraic procedure for inverting \(T(x)=\left(x^{2}+x\right) / 2\) fails. Use your knowledge of the geometry of functions and their inverses to find
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Find a bijection from the set of odd squares, \(\\{1,9,25,49, \ldots\\},\) to the non-negative integers, \(\mathbb{Z}^{\text {acose }}=\\{0,1,2,3, \ldots\\} .\)
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Consider the relation A defined by \(\mathrm{A}=\\{(x, y) \mid x\) has the same astrological sign as \(y\\}\). Is A symmetric or anti-symmetric? Explain.
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The natural logarithm function \(\ln (x)\) is defined by a definite integral with the variable \(x\) in the upper limit. $$ \ln (x)=\int_{t=1}^{x} \frac{1}{t} \
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