$\frac{R^1}{R^1} = R^{1-1}= R^0= R^{-1}.R = I$

Matrix for $R^0$ will be = $\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0&0 &1 \end{bmatrix}$ for n=3 and it's always true for every non empty relation.

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+1 vote

Can someone help me in "part b" of this question- https://gateoverflow.in/1724/gate1998-10 .

I am still not able to understand why $R^0$ is considered here ?

and what is $R^0 $?

Is it Equality relation?

Do we have to consider it in every question of this type ?

+4 votes

Best answer

In relations $R^0$ is Always the Identity relation (Equality relation as you termed). We can prove it in 3 ways.

1. Just By-heart it and think of it as an Axiom of relations.

2. We know in Relation theory $R^n.R^0 = R^n$ where $.$ is Composition Operator (Composition/Composite of two relations)

We could even write $R.R^0 = R$, Now you need to think what should be $R^0$ in such a way that When any relation is Composed with it, results in the same relation. (Sounds like **Identity element** for Composition of relation)

3. Answer Why in General mathematics $n^0 = 1$ where $n \neq 0$. If you can answer that, You can answer why $R^0$ is always the Identity relation.

Try it.

+1

Thanks for point $2$ and $3$. Now it looks intuitive to me. :)

$\frac{R^1}{R^1} = R^{1-1}= R^0= R^{-1}.R = I$

Matrix for $R^0$ will be = $\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0&0 &1 \end{bmatrix}$ for n=3 and it's always true for every non empty relation.

$\frac{R^1}{R^1} = R^{1-1}= R^0= R^{-1}.R = I$

Matrix for $R^0$ will be = $\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0&0 &1 \end{bmatrix}$ for n=3 and it's always true for every non empty relation.

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