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Magazine Chapter 16: Problem 18
Let \(R\) be the real line. Consider the following subsets of the plane \(R
\times R\) :
\(S=\\{(x, y): y=x+1\) and \(0
Short Answer
Expert verified
(d): \(T\) is an equivalence relation on \(R\), but \(S\) is not.
Step by step solution
01
Define Equivalence Relation
For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive. We need to check if subsets \(S\) and \(T\) satisfy these properties.
02
Check Reflexive Property for S
For \(S\) to be reflexive, each \((x, x)\) must belong to \(S\) for every \(x\) in \((0,2)\). Since \(y = x + 1\), we would need \(x = x + 1\), which is not possible. Thus, \(S\) is not reflexive.
03
Check Symmetric Property for S
For \(S\) to be symmetric, if \((x, y)\) is in \(S\), then \((y, x)\) must also be in \(S\). For \((x, y) \in S\), this implies \(y = x + 1\); then \(x = y + 1\), but this cannot satisfy \(y = x + 1\). So \(S\) is not symmetric.
04
Check Transitive Property for S
For \(S\) to be transitive, if \((x, y)\) and \((y, z)\) are in \(S\), \((x, z)\) must be in \(S\). If \(y = x + 1\) and \(z = y + 1\), then \(z = x + 2\), which cannot belong to \(S\) since it violates \(y = x + 1\). Thus, \(S\) is not transitive.
05
Conclude about S
As \(S\) does not satisfy reflexive, symmetric, or transitive properties, \(S\) is not an equivalence relation on \(R\).
06
Check Reflexive Property for T
For \(T\) to be reflexive, \(x - x = 0\) is an integer for any real \(x\). Hence, \(T\) is reflexive.
07
Check Symmetric Property for T
For \(T\) to be symmetric, if \((x, y)\) is in \(T\), then \(x - y\) is an integer; thus \(y - x = -(x - y)\) is also an integer. Hence, \(T\) is symmetric.
08
Check Transitive Property for T
For \(T\) to be transitive, if \((x, y)\) and \((y, z)\) are in \(T\), then \(x - y\) and \(y - z\) are integers. So \(x - z = (x - y) + (y - z)\) is also an integer, making \(T\) transitive.
09
Conclude about T
Since \(T\) is reflexive, symmetric, and transitive, \(T\) is an equivalence relation on \(R\).
10
Final Conclusion
We have determined that \(S\) is not an equivalence relation, but \(T\) is. Therefore, the correct answer is (d): \(T\) is an equivalence relation on \(R\), but \(S\) is not.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Symmetry
Symmetry is a property that is crucial when dealing with equivalence relations. To say a relation is symmetric means this: if you switch the two elements in any pair and the new pair is still in the relation, then the relation is symmetric.
Think of it like a seesaw. If you sit on one end, and someone sits on the other, you can swap places without changing the balance. For a relation to be symmetric, if \( (x, y) \) is in the relation, then \( (y, x) \) must also be in the relation.
Now, let's see how this works with our sets, \( S \) and \( T. \)
Think of it like a seesaw. If you sit on one end, and someone sits on the other, you can swap places without changing the balance. For a relation to be symmetric, if \( (x, y) \) is in the relation, then \( (y, x) \) must also be in the relation.
Now, let's see how this works with our sets, \( S \) and \( T. \)
- For \( S, \) the relation requires \( y = x + 1, \) and swapping \( x \) and \( y \) so that \( x = y + 1, \) actually changes the relation. This is because \( y \) can never truly equal \( x + 1 \) in reverse, thus, \( S \) is not symmetric.
- In \( T, \) the difference \( x - y \) is an integer, which means \( y - x = -(x - y) \) is also an integer. This swap does not disrupt the property, so \( T \) maintains symmetry.
Reflexivity
Reflexivity is another important property for equivalence relations. It requires that every element be related to itself.
Imagine looking in a mirror; every time, you'll see yourself. In the world of equivalence relations, for a relation \( R \) to be reflexive, each element \( x \) in the set must satisfy \( (x, x) \) \in R.
Let's break it down with our exercise:
Imagine looking in a mirror; every time, you'll see yourself. In the world of equivalence relations, for a relation \( R \) to be reflexive, each element \( x \) in the set must satisfy \( (x, x) \) \in R.
Let's break it down with our exercise:
- For the subset \( S, \) reflexivity fails because there are no self-pairs \( (x, x) \) where \( x = x + 1 \) holds. It's mathematically impossible.
- However, for the subset \( T, \) reflexivity holds because \( x - x = 0 \) is an integer. Thus, every real \( x \) satisfies \( (x, x) \) \( \in T, \) making \( T \) reflexive.
Transitivity
Transitivity is the last property that defines equivalence relations and it involves a chain of relationships.
If one element is related to a second, and the second is related to a third, then the first must be related to the third. It's like passing a message along: if A tells B, and B tells C, then A effectively tells C.
In mathematical terms, for a set relation \( R \) to be transitive, whenever \( (x, y) \) and \( (y, z) \) are within \( R, \) it must follow that \( (x, z) \) is also in \( R. \)
If one element is related to a second, and the second is related to a third, then the first must be related to the third. It's like passing a message along: if A tells B, and B tells C, then A effectively tells C.
In mathematical terms, for a set relation \( R \) to be transitive, whenever \( (x, y) \) and \( (y, z) \) are within \( R, \) it must follow that \( (x, z) \) is also in \( R. \)
- Looking at \( S, \) the linkage breaks. Given \( y = x + 1 \) and \( z = y + 1, \) we get \( z = x + 2, \) which cannot round back to \( y = x + 1. \) So \( S \) is not transitive.
- Meanwhile, \( T \) sustains transitivity nicely. If \( x - y \) and \( y - z \) are integers, then adding them gives \( x - z = (x - y) + (y - z), \) also an integer, ensuring \( T \) is transitive.
