What Is An Inverse Relation?
An inverse relation is the relation obtained by reversing the order of every ordered pair in a given relation. If $R$ is a relation, its inverse is written $R^{-1}$, and a pair $(a, b)$ belongs to $R$ exactly when $(b, a)$ belongs to $R^{-1}$.
$$R^{-1} = {(b, a) : (a, b) \in R}$$
If $R$ links set $A$ to set $B$ (so $R \subseteq A \times B$), then $R^{-1}$ links $B$ back to $A$ (so $R^{-1} \subseteq B \times A$). The swap is total: every pair flips, no pair is added or dropped.
Symbol | Meaning |
|---|---|
$R$ | the original relation (a set of ordered pairs) |
$R^{-1}$ | the inverse relation |
$(a, b)$ | an ordered pair in $R$ |
$A \times B$ | the set of all pairs with first element from $A$, second from $B$ |
How Do You Find the Inverse of a Relation?
For a relation listed as ordered pairs, swap each pair's coordinates. For a relation given by an equation, swap $x$ and $y$ and (where useful) solve for $y$.
A frequent reader question — "how do you find the inverse of $y = 3x + 2$?" — runs like this:
Swap the variables: $x = 3y + 2$.
Solve for $y$: $x - 2 = 3y$, so $y = \dfrac{x - 2}{3}$.
The original relation's domain becomes the inverse's range, and the original range becomes the inverse's domain. That swap is automatic — it falls straight out of reversing the pairs.
What Is the Inverse Relation Theorem?
The inverse relation theorem states that inverting twice returns the original relation:
$$(R^{-1})^{-1} = R$$
The proof is short. A pair $(a, b) \in R$ becomes $(b, a) \in R^{-1}$ by definition. Inverting again sends $(b, a)$ back to $(a, b)$, which lands in $(R^{-1})^{-1}$.
Since every pair returns to where it started, the two relations are identical. The flip is its own undo — which is exactly why the graph is a clean mirror image.
Examples of Inverse Relation
Six examples, building from a discrete list to algebraic equations and a graph reflection.
Example 1
Find the inverse of $R = {(1, 4), (2, 5), (3, 6)}$.
Swap each pair.
$R^{-1} = {(4, 1), (5, 2), (6, 3)}$
Final answer: $R^{-1} = {(4, 1), (5, 2), (6, 3)}$.
Example 2
State the domain and range of $R = {(a, 2), (b, 4), (c, 1)}$ and of $R^{-1}$.
Original domain $= {a, b, c}$; original range $= {1, 2, 4}$.
Inverse: $R^{-1} = {(2, a), (4, b), (1, c)}$.
Inverse domain $= {1, 2, 4}$; inverse range $= {a, b, c}$.
Final answer: the domain and range trade places — the inverse's domain is the original's range, and vice versa.
Example 3
A student inverts $R = {(1, 5), (2, 5), (3, 7)}$ and concludes the inverse is a function. Is it?
Wrong attempt. The student swaps to get $R^{-1} = {(5, 1), (5, 2), (7, 3)}$ and reasons, "the original was a function, so its inverse must be too."
Why it breaks. Look at the inverse's first coordinates: $5$ appears twice, paired with both $1$ and $2$. A function cannot send one input to two outputs. So the inverse fails the function test even though it is a perfectly valid inverse relation.
Correct. $R^{-1} = {(5, 1), (5, 2), (7, 3)}$ is an inverse relation but not an inverse function, because the original relation was not one-to-one.
Final answer: the inverse relation exists; it is a function only when the original is one-to-one.
Example 4
Find the inverse of the algebraic relation $y = x^2$.
Swap the variables.
$x = y^2$
Solve for $y$: $y = \pm\sqrt{x}$.
Final answer: $y = \pm\sqrt{x}$, a valid inverse relation, but the $\pm$ shows it is not a function (each positive $x$ gives two $y$-values).
Example 5
Find the inverse of $y = \dfrac{2x + 1}{5}$.
Swap the variables.
$x = \dfrac{2y + 1}{5}$
Multiply both sides by $5$: $5x = 2y + 1$.
Subtract $1$: $5x - 1 = 2y$.
Divide by $2$: $y = \dfrac{5x - 1}{2}$.
Final answer: $y = \dfrac{5x - 1}{2}$.
Example 6
A relation passes through $(0, 3)$, $(2, 7)$, and $(-1, 1)$. Where does its inverse pass through, and how does the graph relate?
Swap each pair: the inverse passes through $(3, 0)$, $(7, 2)$, and $(1, -1)$.
Each original point and its inverse are mirror images across $y = x$.
Final answer: the inverse passes through $(3, 0)$, $(7, 2)$, $(1, -1)$, and its graph is the original reflected over the line $y = x$.
Where Inverse Relations Show Up
"Reversing a mapping is how we ask a question backwards."
The inverse relation exists because we constantly need to run a correspondence in reverse — and the swap formalises that.
Encoding and decoding. A cipher maps letters to symbols; reading a message back uses the inverse relation. The decode step is the inverse.
Unit conversion. Celsius-to-Fahrenheit and its reverse are inverse relations of each other. The reflection across $y = x$ is literally the "convert the other way" rule.
Databases and lookups. Any "who has this?" table can be flipped to "what does this person have?" — the same pairs, read backward.
The destination is the inverse function: once a relation is one-to-one, its inverse relation graduates into an inverse function, the form you will use across algebra, trigonometry, and calculus.
Common Errors With Inverse Relations
Mistake 1: Assuming the inverse of a function is always a function
Where it slips in: Any relation that is not one-to-one, as in Examples 3 and 4.
Don't do this: Conclude that because $R$ is a function, $R^{-1}$ must be too.
The correct way: Check whether the original is one-to-one. Every relation has an inverse relation; the inverse is a function only when the original passes the horizontal line test.
The first instinct is to treat "inverse" as a property that preserves functionhood — but the inverse of $y = x^2$ being $\pm\sqrt{x}$ is exactly where that assumption drops the second branch and goes wrong.
Mistake 2: Forgetting to swap domain and range
Where it slips in: Stating the domain and range of an inverse, as in Example 2.
Don't do this: Keep the original domain as the inverse's domain.
The correct way: The inverse's domain is the original's range, and the inverse's range is the original's domain. They always trade.
Mistake 3: Solving an equation before swapping the variables
Where it slips in: Finding the inverse of $y = mx + b$ style relations.
Don't do this: Solve the original for $x$ in terms of $y$ and call that the inverse.
The correct way: Swap $x$ and $y$ first, then solve for $y$. The rusher who reorders the steps gets an expression that looks similar but represents the wrong relation.
Practice Questions
Try these, then check the answers below.
Find the inverse of $R = {(2, 8), (3, 8), (5, 1)}$ and decide whether the inverse is a function.
Find the inverse of $y = 4x - 7$.
Reflect the points $(1, 2)$ and $(4, 0)$ across $y = x$.
State the domain and range of $R = {(p, 3), (q, 5)}$ and of $R^{-1}$.
Apply the inverse relation theorem: what is $(R^{-1})^{-1}$ for $R = {(1, 9)}$?
Answers
Answer to Question 1: $R^{-1} = {(8, 2), (8, 3), (1, 5)}$. It is not a function — the input $8$ maps to both $2$ and $3$, because the original relation was not one-to-one.
Answer to Question 2: Swap first: $x = 4y - 7$, so $4y = x + 7$ and $y = \dfrac{x + 7}{4}$.
Answer to Question 3: $(1, 2) \to (2, 1)$ and $(4, 0) \to (0, 4)$. Swap each pair's coordinates.
Answer to Question 4: Original domain $= {p, q}$, range $= {3, 5}$; inverse domain $= {3, 5}$, range $= {p, q}$. The two trade places.
Answer to Question 5: $(R^{-1})^{-1} = R = {(1, 9)}$. Inverting twice returns the original.
Conclusion
An inverse relation $R^{-1}$ is formed by swapping every ordered pair $(a, b)$ in $R$ into $(b, a)$.
The graph of an inverse relation is the original reflected across the line $y = x$, and the domain and range trade places.
The inverse relation theorem says $(R^{-1})^{-1} = R$, so inverting twice returns the original.
An inverse relation is always a relation, but it is a function only when the original is one-to-one.
Always swap the variables before solving for $y$ when inverting an equation.
Your Next Move
Work through the practice questions above to solidify your understanding. If you get stuck on the function-or-not question, return to Example 3 and check the inverse's first coordinates for repeats. At Bhanzu, trainers anchor the inverse relation on the $y = x$ mirror first, so the algebra never feels like a memorised swap. Want a live Bhanzu trainer to walk through more inverse relation problems? Book a free demo class.
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