Key Concepts and Formulas

• Equivalence Relation: A relation $R$ on a set $A$ is an equivalence relation if it satisfies three properties:
  1. Reflexivity: For all $a \in A$, $(a,a) \in R$.
  2. Symmetry: For all $a, b \in A$, if $(a,b) \in R$, then $(b,a) \in R$.
  3. Transitivity: For all $a, b, c \in A$, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.
• Set of Real Numbers: $R$ denotes the set of all real numbers.
• Set of Integers: $\mathbb{Z}$ denotes the set of all integers.

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Step-by-Step Solution

We are given two subsets of the plane $R \times R$, denoted by $S$ and $T$. We need to determine which of these is an equivalence relation on the set of real numbers $R$.

Analysis of Relation S

The relation $S$ is defined as: $S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$ We will check the three properties of an equivalence relation for $S$ on $R$.

• Step 1: Check for Reflexivity
  • Reasoning: For $S$ to be reflexive on $R$, every real number $x \in R$ must be related to itself, meaning $(x,x) \in S$.
  • Mathematical Verification: The condition for $(x,y) \in S$ is $y = x+1$ and $0 < x < 2$. For $(x,x)$ to be in $S$, we must have $x = x+1$.
  • Explanation: The equation $x = x+1$ simplifies to $0 = 1$, which is a false statement. This means that for any real number $x$, the pair $(x,x)$ does not satisfy the condition $y = x+1$. Consequently, $(x,x) \notin S$ for any $x \in R$.
  • Conclusion: Since $S$ is not reflexive, it cannot be an equivalence relation. We do not need to check symmetry or transitivity for $S$.

Analysis of Relation T

The relation $T$ is defined as: $T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$ We will check the three properties of an equivalence relation for $T$ on $R$.

• Step 2: Check for Reflexivity

  • Reasoning: For $T$ to be reflexive on $R$, every real number $x \in R$ must be related to itself, meaning $(x,x) \in T$.
  • Mathematical Verification: The condition for $(x,y) \in T$ is that $x-y$ is an integer. For $(x,x)$ to be in $T$, we must check if $x-x$ is an integer.
  • Explanation: For any real number $x$, $x-x = 0$. Since $0$ is an integer, the condition is satisfied. Thus, $(x,x) \in T$ for all $x \in R$.
  • Conclusion: $T$ is reflexive.

• Step 3: Check for Symmetry

  • Reasoning: For $T$ to be symmetric, if $(x,y) \in T$ for any $x,y \in R$, then it must also be true that $(y,x) \in T$.
  • Mathematical Verification: Assume $(x,y) \in T$. By definition, this means $x-y = k$ for some integer $k \in \mathbb{Z}$. We need to check if $(y,x) \in T$, which means checking if $y-x$ is an integer.
  • Explanation: From $x-y = k$, we can multiply by $-1$ to get $-(x-y) = -k$, which means $y-x = -k$. Since $k$ is an integer, $-k$ is also an integer. Therefore, $y-x$ is an integer, and $(y,x) \in T$.
  • Conclusion: $T$ is symmetric.

• Step 4: Check for Transitivity

  • Reasoning: For $T$ to be transitive, if $(x,y) \in T$ and $(y,z) \in T$ for any $x,y,z \in R$, then it must also be true that $(x,z) \in T$.
  • Mathematical Verification: Assume $(x,y) \in T$ and $(y,z) \in T$.
    • $(x,y) \in T$ implies $x-y = k_1$ for some integer $k_1 \in \mathbb{Z}$.
    • $(y,z) \in T$ implies $y-z = k_2$ for some integer $k_2 \in \mathbb{Z}$. We need to check if $(x,z) \in T$, which means checking if $x-z$ is an integer.
  • Explanation: We can add the two equations: $(x-y) + (y-z) = k_1 + k_2$. This simplifies to $x-z = k_1 + k_2$. Since $k_1$ and $k_2$ are integers, their sum $k_1 + k_2$ is also an integer. Therefore, $x-z$ is an integer, and $(x,z) \in T$.
  • Conclusion: $T$ is transitive.

• Step 5: Overall Conclusion for T

  • Reasoning: Since $T$ satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on $R$.

Step 6: Evaluate the Options

• $S$ is not an equivalence relation.
• $T$ is an equivalence relation.

Let's examine the given options: (A) Neither S nor T is an equivalence relation on R. (False, T is) (B) Both S and T are equivalence relation on R. (False, S is not) (C) S is an equivalence relation on R but T is not. (False, S is not and T is) (D) T is an equivalence relation on R but S is not. (True, as determined)

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Common Mistakes & Tips

• Scope of "on R": Remember that reflexivity, symmetry, and transitivity must hold for all elements of the set $R$, not just a subset. The condition $0 < x < 2$ in the definition of $S$ restricts the pairs $(x,y)$ that can be in $S$, but reflexivity requires checking $(x,x)$ for all $x \in R$.
• Single Counterexample: If you find even one instance where a property (reflexivity, symmetry, or transitivity) fails, the relation is not an equivalence relation.
• Integer Properties: Be comfortable with the closure properties of integers under addition, subtraction, and multiplication.

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Summary

We analyzed relation $S$ and found it to be not reflexive because $x = x+1$ is never true for any real number $x$. Therefore, $S$ is not an equivalence relation. We then analyzed relation $T$ and found it to be reflexive (since $x-x=0$ is an integer), symmetric (if $x-y$ is an integer, then $y-x$ is also an integer), and transitive (if $x-y$ and $y-z$ are integers, then $x-z$ is also an integer). Hence, $T$ is an equivalence relation on $R$. This means that $T$ is an equivalence relation on $R$ but $S$ is not.

The final answer is $\boxed{\text{(D)}}$.