If $R_1$ and $R_2$ are equivalence relations in set $A$, show that $R_1 \cap R_2$ is also an equivalence relation.

> [!TLDR] > Prove reflexivity, symmetry and transitivity of $R_1\cap R_2$ directly from the same properties holding for $R_1$ and $R_2$ individually. > [!FORMULA] > $(x,y)\in R_1\cap R_2 \iff (x,y)\in R_1 \text{ AND } (x,y)\in R_2$. $R_1,R_2$ equivalence $\Rightarrow$ each is reflexive, symmetric, transitive. 1. **Reflexive.** Let $a\in A$. Since $R_1$ is reflexive, $(a,a)\in R_1$. Since $R_2$ is reflexive, $(a,a)\in R_2$. Both hold, so $(a,a)\in R_1\cap R_2$. This is true for every $a\in A$, so $R_1\cap R_2$ is **reflexive**. 2. **Symmetric.** Let $(a,b)\in R_1\cap R_2$, so $(a,b)\in R_1$ and $(a,b)\in R_2$. - $R_1$ symmetric $\Rightarrow (b,a)\in R_1$. - $R_2$ symmetric $\Rightarrow (b,a)\in R_2$. Both hold, so $(b,a)\in R_1\cap R_2$. Hence $R_1\cap R_2$ is **symmetric**. 3. **Transitive.** Let $(a,b)\in R_1\cap R_2$ and $(b,c)\in R_1\cap R_2$. Then $(a,b),(b,c)\in R_1$ and $(a,b),(b,c)\in R_2$. - $R_1$ transitive $\Rightarrow (a,c)\in R_1$. …

If R₁ and R₂ are equivalence relations in set A, show that R₁ ∩ R₂ is also an equivalence relation.

Exercise 4.1 · Q5

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Concept understanding — Introduction

Introduction — What It Means and Why It Matters

You've been solving problems your whole life. When you see a new type of question, you first figure out what it's asking, then decide which tools to use, then carry out the steps. That first step — figuring out what's going on — is the heart of an introduction in mathematics and science.

An introduction is not a summary. It's a setup. It tells you the players, the rules, and the goal before the action starts.

The Intuition

Imagine you walk into a room and see a chessboard with pieces already arranged. Without knowing whose turn it is, what the last move was, or whether you're playing standard chess or a variant, you can't make sense of anything. An introduction is like being told: "White to move, standard rules, checkmate in two."

In a proof, an introduction declares the assumptions you're allowed to use. In a problem, it states the given data and the unknown you're after. In a definition, it names the new object and lists its essential properties.

Note

The word "introduction" here is about logical introduction — bringing a new idea or assumption into a discussion — not the first paragraph of an essay.

The Precise Statement

In formal logic and mathematics, an introduction rule tells you how to legitimately bring a new statement into a derivation. The most common example is the conditional introduction (also called "direct proof" or "→-introduction"):

If from assumption $P$ you can derive $Q$ , then you may conclude $P \implies Q$ .

That is, to prove "if $P$ then $Q$ ", you temporarily assume $P$ is true, work through the reasoning, and arrive at $Q$ . Then you discharge the assumption and write $P \implies Q$ as a proven statement.

Symbolically:

\frac{ \begin{array}{c} P \\ \vdots \\ Q \end{array} }{ P \implies Q }

The line means "from the top, infer the bottom." The dots represent a valid chain of reasoning.

Why This Matters for Exams

When you see a question that asks you to "prove that if $x > 2$ then $x^2 > 4$ ", you are being asked to perform a conditional introduction. You start by introducing the assumption $x > 2$ , work from there, and end with $x^2 > 4$ . Then you write the implication. …

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Prove reflexivity, symmetry and transitivity of $R_1\cap R_2$ directly from the same properties holding for $R_1$ and $R_2$ individually.

(x,y)\in R_1\cap R_2 \iff (x,y)\in R_1 \text{ AND } (x,y)\in R_2

Reflexive. Let $a\in A$ . Since $R_1$ is reflexive, $(a,a)\in R_1$ . Since $R_2$ is reflexive, $(a,a)\in R_2$ . Both hold, so $(a,a)\in R_1\cap R_2$ . This is true for every $a\in A$ , so $R_1\cap R_2$ is reflexive.
Symmetric. Let $(a,b)\in R_1\cap R_2$ $(a, b) \in R_{1} \cap R_{2}$ , so $(a,b)\in R_1$ $(a, b) \in R_{1}$ and $(a,b)\in R_2$ $(a, b) \in R_{2}$ .
- $R_1$ symmetric $\Rightarrow (b,a)\in R_1$ .
- $R_2$ symmetric $\Rightarrow (b,a)\in R_2$ . Both hold, so $(b,a)\in R_1\cap R_2$ . Hence $R_1\cap R_2$ is symmetric.
Transitive. Let $(a,b)\in R_1\cap R_2$ $(a, b) \in R_{1} \cap R_{2}$ and $(b,c)\in R_1\cap R_2$ $(b, c) \in R_{1} \cap R_{2}$ . Then $(a,b),(b,c)\in R_1$ $(a, b), (b, c) \in R_{1}$ and $(a,b),(b,c)\in R_2$ $(a, b), (b, c) \in R_{2}$ .
- $R_1$ transitive $\Rightarrow (a,c)\in R_1$ . …

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Q.If R1R_1R1​ and R2R_2R2​ are equivalence relations in set AAA, show that R1∩R2R_1 \cap R_2R1​∩R2​ is also an equivalence relation.

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Q.If $R_1$ and $R_2$ are equivalence relations in set $A$ , show that $R_1 \cap R_2$ is also an equivalence relation.

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