Animated Solution for Mathematics - Sets and Relations: Consider the following two binary relations on the set A={a,b,c}: $R_1=\{(c,a),(b,b),(a,c),(c,c),(b,c),(a,a)\}$ and $R_2=\{(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)\}$. Then

Welcome, fellow traveler, to the fascinating world of binary relations! Today, we are mapping connections between three mysterious islands: $a$, $b$, and $c$. Think of these islands as nodes in a network, where a relation is simply a set of one-way bridges between them.

Our mission is to analyze two specific networks, $R_1$ and $R_2$, and determine if they possess the elegant properties of symmetry and transitivity.

The Two-Way Street

Imagine you are standing on island $b$ and you see a bridge leading to island $c$. For a relation to be symmetric, it must be a "two-way street." Mathematically, we say $R$ is symmetric if:

Let us look at $R_1=\{(c,a),(b,b),(a,c),(c,c),(b,c),(a,a)\}$. We see a bridge from $b$ to $c$, represented by $(b,c)$. Is there a return bridge $(c,b)$?

Scanning our list, we find no such pair. Because this return path is missing, $R_1$ fails the symmetry test immediately. It is a one-way street in that specific section!

Now, consider $R_2=\{(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)\}$. We see $(a,b)$ and its partner $(b,a)$, as well as $(c,a)$ and its partner $(a,c)$.

The self-loops $(a,a)$, $(b,b)$, and $(c,c)$ are naturally symmetric. Since every bridge has a return path, $R_2$ is perfectly symmetric.

The Shortcut

Transitivity is the property of "shortcuts." If you can travel from $x$ to $y$, and then from $y$ to $z$, transitivity demands that there must be a direct shortcut from $x$ to $z$.

Formally, if $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)$ must also be in $R$.

Let us test $R_1$ again. We have $(b,c)\in R_1$ and $(c,a)\in R_1$. To be transitive, we need the shortcut $(b,a)$ to exist in $R_1$.

But alas, $(b,a)$ is missing! Therefore, $R_1$ is not transitive.

Finally, let us test $R_2$. We have $(b,a)\in R_2$ and $(a,c)\in R_2$. This sequence of bridges implies we should have a direct shortcut $(b,c)$ in $R_2$.

Let us search for it... it is not there! Because this crucial shortcut is missing, $R_2$ fails the transitivity test as well.

Through our journey, we have discovered the following:

$R_1$ is neither symmetric nor transitive. $R_2$ is symmetric but not transitive.

This exercise teaches us that properties of relations are independent; one does not guarantee the other. By visualizing these as paths, we avoid the confusion of abstract sets and see the logic clearly.