Question

If the relation \(R: A \rightarrow B\) where \(A=\{1,2,3,4\}\), \(B=\{1,3,5\}\) is defined by \(R=\{(x,y): x<y,\; x\in A,\; y\in B\}\), then \(R\circ R^{-1}=\)

• A. \(\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}\)
• B. \(\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\}\)
• C. \(\{(3,3),(3,5),(5,3),(5,5)\}\)
• D. \(\{(3,3),(3,4),(4,5)\}\)

Hint:

For composition \(R\circ R^{-1}\):
Domain and codomain are the same
Result is always a relation on the codomain set
Check existence of a common intermediate element

Answer 1

The Correct Option is C

Step 1: First write the relation \(R\). All ordered pairs \((x,y)\) such that
Step 2: Write the inverse relation \(R^{-1}\) by interchanging the coordinates: \[ R^{-1}=\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\} \] 
Step 3: By definition, \[ R\circ R^{-1}=\{(a,c)\mid \exists b \text{ such that } (a,b)\in R^{-1} \text{ and } (b,c)\in R\} \] 
Step 4: Since \(R^{-1}:B\rightarrow A\) and \(R:A\rightarrow B\), the composition \(R\circ R^{-1}\subseteq B\times B\). 
Step 5: Check elements of \(B=\{1,3,5\}\): 
For \(3\): there exist \(1,2\in A\) such that \(1<3,\,2<3\) \(\Rightarrow (3,3)\in R\circ R^{-1}\) 
For \(5\): there exist \(1,2,3,4\in A\) such that \(a<5\) \(\Rightarrow (5,5)\in R\circ R^{-1}\) 
Also, \[ (3,5)\in R\circ R^{-1}, \quad (5,3)\in R\circ R^{-1} \] 
Step 6: Hence, \[ R\circ R^{-1}=\{(3,3),(3,5),(5,3),(5,5)\} \]