Solveeit Logo

Question

Question: Let A={1,2,3,....9} and R be relation in \(A\times A\) defined by (a, b) R (c, d) if a + d = b + c f...

Let A={1,2,3,....9} and R be relation in A×AA\times A defined by (a, b) R (c, d) if a + d = b + c for (a, b),(c, d) in A×AA\times A. Prove that R is an equivalence relation. Also obtain the equivalence class [(2,5)].

Explanation

Solution

Hint: For solving this problem, first we have to prove that the relation is reflexive, symmetric and transitive by using appropriate conditions. Once these three relations are established, then R becomes an equivalence relation on A. After that we can use the given relation to obtain the equivalence class.

Complete step-by-step answer:
The conditions for a set to be reflexive, transitive and symmetric are:

  1. For a relation to be reflexive, (a,a)R\left( a,a \right)\in R.
  2. For a relation to be symmetric, (a,b)R(b,a)R\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R.
  3. For a relation to be transitive, (a,b)R,(b,c)R(a,c)R\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R.
  4. For a relation to be equivalence, when it is reflexive, symmetric and transitive.
    Given: A = {1, 2, 3, ……9}
    We have to prove R belongs to the equivalence class.
    Now, (a, b) R (c, d) if (a, b) (c, d)A\left( a,\text{ }b \right)\text{ }\left( c,\text{ }d \right)\in A
    a+d=b+ca+d=b+c
    Consider, (a, b) R (a, b)
    a+b=b+a\therefore a+b=b+a
    Hence, R is reflexive.
    Consider (a, b) R (c, d) given by (a, b) (c, d)A×A\in A\times A
    Now, by using a + d = b + c => c + b = d + a, we get
    (c,d)R(a,b)=(a,b)R(c,d)\Rightarrow \left( c,d \right)R\left( a,b \right)=\left( a,b \right)R\left( c,d \right)
    Hence, R is symmetric.
    Let (a, b), (c, d), (e, f),A×A\in A\times A
    For (a, b) R (c, d) and (c, d) R (e, f), we have
    a + d = b + c and c + f = d + e
    ac=bd...(1) c+f=d+e...(2) \begin{aligned} & \Rightarrow a-c=b-d...\left( 1 \right) \\\ & \Rightarrow c+f=d+e...\left( 2 \right) \\\ \end{aligned}
    Adding equation 1 and 2 we get:
    a – c + c + f = b - d + d + e
    a + f = b + e
    (a, b) R (e, f)
    Hence, R is transitive.
    Since, R is reflexive, transitive and symmetric. Therefore, R is an equivalence relation.
    For, equivalence class of (2, 5): (2, 5) R (c, d)
    a and b such that 2 + d = 5 + c
    so, d = c + 3
    consider (1, 4) to make possible pairs of (c, d)
    (2, 5) R (1, 4) 2+4=5+1\Rightarrow 2+4=5+1
    [(2,5)=(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)]\left[ \left( 2,5 \right)=\left( 1,4 \right),\left( 2,5 \right),\left( 3,6 \right),\left( 4,7 \right),\left( 5,8 \right),\left( 6,9 \right) \right] is the equivalent class under relation R.

Note: The knowledge of equivalence of a relation is must for solving this problem. Students must remember all the necessary conditions for proving a set a reflexive, symmetric and transitive. All the possible pairs from the given number must be obtained for equivalence class.