Q.Let $R$ be a relation from $\mathbb{Q}$ to $\mathbb{Q}$ defined by $R = \{(a, b) : a, b \in \mathbb{Q}\ \text{and}\ a - b \in \mathbb{Z}\}$. Show that

Concept: Relation Arrow Diagram — the three properties shown are reflexivity, symmetry, and transitivity, which together make $R$ an equivalence relation on $\mathbb{Q}$.

Step 1 (Reflexivity): For any $a \in \mathbb{Q}$, $a - a = 0 \in \mathbb{Z}$. Hence $(a, a) \in R$ for all $a$.

Step 2 (Symmetry): If $(a, b) \in R$, then $a - b \in \mathbb{Z}$. Since $\mathbb{Z}$ is closed under negation, $b - a = -(a - b) \in \mathbb{Z}$. Thus $(b, a) \in R$.

Step 3 (Transitivity): If $(a, b) \in R$ and $(b, c) \in R$, then $a - b \in \mathbb{Z}$ and $b - c \in \mathbb{Z}$. Their sum $(a - b) + (b - c) = a - c \in \mathbb{Z}$. Hence $(a, c) \in R$.

The core idea here is that the condition $a - b \in \mathbb{Z}$ turns the difference of two rational numbers into a whole number. Think of rational numbers on a number line. Two numbers are related if the distance between them (in terms of difference) is an exact integer — no fractional part. So $2.5$ and $5.5$ are related because $2.5 - 5.5 = -3$, an integer. But $2.5$ and $3.2$ are not, because $2.5 - 3.2 = -0.7$, not an integer.

What does this look like as an arrow diagram? Imagine each rational number as a point. An arrow from $a$ to $b$ means $a - b \in \mathbb{Z}$. The three properties we need to check are exactly what make this relation behave nicely: every point has a loop (reflexive), every arrow can be reversed (symmetric), and if you can go from $a$ to $b$ and then $b$ to $c$, you can go directly from $a$ to $c$ (transitive). Let's prove each one.

1. Reflexive: $(a, a) \in R$ for all $a \in \mathbb{Q}$

   We need to show $a - a \in \mathbb{Z}$.

   $a - a = 0$, and $0$ is an integer.

   So the condition holds for every rational $a$.

   Note

   This is the simplest of the three — it only uses the fact that zero is an integer.

2. Symmetric: $(a, b) \in R$ implies $(b, a) \in R$

   If $(a, b) \in R$, then by definition $a - b \in \mathbb{Z}$.

   Consider $b - a$. We can write $b - a = -(a - b)$.

   Since $a - b$ is an integer, its negative is also an integer.

   Therefore $b - a \in \mathbb{Z}$, which means $(b, a) \in R$.

   Watch out

   A common mistake is to assume $b - a$ is automatically an integer just because $a - b$ is. You must explicitly note that the negative of an integer is an integer — it's a small step, but it's the logical link.

3. Transitive: $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$

   From $(a, b) \in R$, we have $a - b \in \mathbb{Z}$.

   From $(b, c) \in R$, we have $b - c \in \mathbb{Z}$.

   Add these two integers: $(a - b) + (b - c) = a - c$.

   The sum of two integers is an integer, so $a - c \in \mathbb{Z}$.

   Hence $(a, c) \in R$.

   Tip

   The clever trick here is that the $b$ cancels out when you add the two differences. This is why the condition $a - b \in \mathbb{Z}$ works so neatly — it's designed so that transitivity follows from the closure of integers under addition.

All three properties are satisfied. This means $R$ is an equivalence relation on $\mathbb{Q}$. In fact, the equivalence classes are sets of rational numbers that differ by an integer — for example, all numbers with the same fractional part (like $\dots, -1.3, 0.3, 1.3, 2.3, \dots$) form one class.