Textbooks and JEE problems use two phrases that look like they describe different objects:

Students often wonder: at what moment does the first phrase get replaced with the second? Is there a conversion? Do you lose information?

The answer is a single line: the second phrase is the first phrase with B = A. Nothing is added, nothing is lost. What changes is the set of questions that become meaningful.

The one-line rule

"R is a relation on A" means exactly "R is a relation from A to A."

Both phrases say: R \subseteq A \times A. Every pair in R has both coordinates drawn from A. The second phrase is just shorter, and books reach for it whenever the source and target sets coincide.

Why: relations can pair elements of two different sets — students and subjects, say — or they can pair elements of one set with elements of the same set — integers with integers, for example. The second case is common enough that mathematicians invented the shorter name "relation on A" for it. The definition is identical, only the name is shorter.

When the phrase-shift happens

Watch for one of these signals:

  1. The problem says "R is a relation on \mathbb{R}" or "R is a relation on A = \{1, 2, 3\}." Both coordinates draw from the same set.
  2. The problem defines R by a rule that uses both coordinates from the same type. For instance, "aRb iff a^2 + b^2 = 1" with a, b \in \mathbb{R} — this is on \mathbb{R}.
  3. The problem asks whether R is reflexive, symmetric, transitive, antisymmetric, or an equivalence relation. These properties only make sense when source equals target — you need the phrase "on A."

If any of these hold, you are in the "on A" case. If the two sets are genuinely different — say A = students, B = subjects — you stay in the "from A to B" language, and those diagonal-style properties simply do not apply.

Why the properties demand "on A"

Take reflexivity. It requires (a, a) \in R for every a \in A. The pair (a, a) has both coordinates from the same set. If R \subseteq A \times B with A \neq B, the pair (a, a) is not even a valid element of A \times B — its second coordinate would need to be in B, but you wrote a which is in A. The question "is R reflexive?" is not just false, it is not well-posed unless B = A.

Same for symmetry: "(a, b) \in R \Rightarrow (b, a) \in R" asks (b, a) to live in the same product as (a, b). That is only possible when A = B, so both coordinates are drawn from the same set.

Transitive, antisymmetric — all of them are "on A" properties. The moment you see one of those words in a question, the relation is a relation on a single set, and you should read the problem statement to identify which set.

A picture: the two phrasings

Two arrow diagrams showing relation from A to B versus relation on ATwo side-by-side diagrams. The left shows two different ovals labelled A and B with arrows between them: a general relation from A to B. The right shows a single oval labelled A with arrows that start and end inside the same oval: a relation on A. relation from A to B relation on A (B = A) A B A pairs live in A × B pairs live in A × A; self-loops allowed
Left: a relation from $A$ to $B$ with $A \neq B$. Arrows cross between two different ovals. The notions reflexive, symmetric, transitive do not apply here. Right: a relation on $A$. Arrows start and end inside the same oval, and self-loops (dashed circle) become possible — this is what reflexivity asks about.

A JEE-style example to anchor the shift

From "on $\mathbb{Z}$" to the property check

Problem. Let R be the relation on \mathbb{Z} defined by aRb iff a - b is even. Determine whether R is reflexive, symmetric, transitive.

Reading the problem. The phrase "relation on \mathbb{Z}" tells you R \subseteq \mathbb{Z} \times \mathbb{Z} — both coordinates draw from \mathbb{Z}. The properties asked (reflexive, symmetric, transitive) are meaningful only because source equals target.

Reflexive? For every a \in \mathbb{Z}, is a - a even? Yes: a - a = 0, and 0 is even. Symmetric? If a - b is even, is b - a even? b - a = -(a - b), and negating an even number gives an even number. Yes. Transitive? If a - b and b - c are both even, is a - c even? a - c = (a - b) + (b - c), a sum of two evens, which is even. Yes.

R is an equivalence relation on \mathbb{Z} — but we only had the vocabulary to ask those questions because the problem told us source equals target.

Interactive: which phrase applies?

Slider switching between 'relation from A to B' and 'relation on A'A horizontal track with a draggable red point. On the left the readout describes a relation between two different sets and notes that reflexive, symmetric, transitive do not apply. On the right the readout describes a relation on a single set and lists which properties become meaningful. from A to B on A drag to switch phrasing
Drag across to see how the phrasing changes the available questions. The moment $B = A$, the list of meaningful properties expands to include reflexive, symmetric, transitive, antisymmetric, and equivalence — none of which can be asked when the source and target sets are different.

A short checklist

When you meet a relation in a problem, decide the phrasing in this order:

  1. Read the problem statement. Does it say "on A" or does it give two different sets? Use that phrase.
  2. Check the rule. Does the rule pair elements of the same type? If yes, the problem is on a single set.
  3. Check what is asked. If the question asks about reflexive, symmetric, transitive, you are on a single set. If the question asks only about domain, range, or how many relations exist, either phrasing is fine.

The grammatical switch from "from A to B" to "on A" has no mathematical depth — it is shorthand for B = A. But it signals which properties you are expected to check. Watch the phrase, and the next step of the problem writes itself.

Related: Relations · Relation On a Set vs Relation Between Two Sets · Ordered-Pair Plotter: Relation on the Grid · Equivalence Relations