Animated Solution for Mathematics - Sets and Relations: The relation $R=\{(a,b):\gcd (a,b)=1,2a\ne b,a,b\in \mathbb{Z}\}$ is:

Welcome, fellow traveler, to the fascinating world of relations! Today, we are not just solving a problem; we are dissecting the very fabric of mathematical logic.

We are given a relation $R$ defined on the set of integers $\mathbb{Z}$, and our mission is to determine its properties: reflexivity, symmetry, and transitivity. The relation is defined as:

Think of this as a club with two strict bouncers at the door. The first bouncer, $\gcd (a,b)=1$, demands that you and your partner be co-prime. The second bouncer, $2a eq b$, demands that your partner cannot be exactly twice your value. To be in the club, you must pass both!

Reflexivity

Imagine standing in front of a mirror. A relation is reflexive if every element sees itself in the mirror and is accepted into the club.

Mathematically, this means for every $a\in \mathbb{Z}$, the pair $(a,a)$ must be in $R$. Let's test this. If we plug $(a,a)$ into our bouncers, the first one asks: Is $\gcd (a,a)=1$?

For any integer $a>1$, the $\gcd (a,a)$ is simply $a$. So, if we pick $a=2$, the $\gcd (2,2)=2$. The bouncer shakes his head—$2 eq 1$.

The condition fails immediately. Because we found at least one integer that fails the test, we can confidently declare that the relation $R$ is not reflexive.

Now, let's talk about symmetry. A relation is symmetric if, whenever you are related to your friend, your friend is also related to you.

If $(a,b)\in R$, then $(b,a)$ must also be in $R$. Let's test this with a pair that actually makes it into the club. Let $a=2$ and $b=1$.

First, $\gcd (2,1)=1$ (Pass). Second, $2(2) = 4 eq 1$ (Pass). So, $(2,1)\in R$.

Now, let's check the reverse: $(1,2)$. The first bouncer says $\gcd (1,2)=1$ (Pass). But the second bouncer, $2a eq b$, looks at $2(1)=2$.

Since $2=2$, the condition $2a eq b$ is violated! The pair $(1,2)$ is rejected. Because the reverse trip is not allowed, the relation is not symmetric.

Transitivity

Finally, we arrive at transitivity, the ultimate test of connection. Transitivity asks: If $a$ is related to $b$, and $b$ is related to $c$, does $a$ have to be related to $c$?

Let's build a chain. Let $a=2,b=3,c=4$. First, check $(2,3)$: $\gcd (2,3)=1$ and $2(2) = 4 eq 3$. It's in!

Now, check $(3,4)$: $\gcd (3,4)=1$ and $2(3) = 6 eq 4$. It's in too! Now, for the final link: $(2,4)$.

The first bouncer checks $\gcd (2,4)$. Wait, $\gcd (2,4)=2$. The bouncer stops us—$2 eq 1$. The pair $(2,4)$ is not in the relation. The chain is broken. Thus, the relation is not transitive.

We have walked through the landscape of this relation, testing its boundaries and finding its limits. We discovered it is neither reflexive, nor symmetric, nor transitive.

It is a beautiful example of how simple rules can create complex, non-trivial structures. Keep exploring, keep questioning, and remember: in mathematics, a single counter-example is all it takes to change the entire story. You have mastered this logic today!