Recipe

Prove a unique-existence claim

Use when: you must show $\exists! x, P(x)$ : exactly one object satisfies $P$ , not just at least one

Goal: both halves established together: one object satisfying $P$ , and a proof no second, different object could

Taught in Uniqueness proofs

Quick recipe

Prove existence first
Assume two objects both satisfy P
Chain justified steps to show they must be equal
Report both halves together

1
Exhibit at least one object satisfying $P$ , either by naming a specific one directly or by an indirect existence argument.

Example

To prove $2x-4=10$ has a unique real solution, first show existence: $x=7$ satisfies $2(7)-4=10$ .

Why: uniqueness alone never confirms anything exists at all, so this half cannot be skipped.
2
For the uniqueness half, suppose $x_1$ and $x_2$ are both fixed but arbitrary objects satisfying $P$ , exactly as a direct proof fixes an arbitrary object.

Example

Suppose $x_1$ and $x_2$ both satisfy $2x-4=10$ , so $2x_1-4=10$ and $2x_2-4=10$ .

Why: assuming two, rather than fixing one, is what lets the argument conclude they must actually be the same object.
3
Starting from the shared assumption, apply valid algebraic or logical steps until the two objects are shown equal.

Example

Since both equal $10$ , $2x_1-4=2x_2-4$ , so $2x_1=2x_2$ , and dividing by $2$ gives $x_1=x_2$ .

Why: this is the entire content of a uniqueness proof: any two objects satisfying $P$ collapse into one.
4
State existence and uniqueness as one combined conclusion; neither half alone establishes $\exists! x, P(x)$ .

Example

$x=7$ exists and satisfies the equation (Step 1), and any two solutions must be equal (Steps 2-3), so $2x-4=10$ has a unique real solution, $x=7$ .
- both halves were established — $\exists! x, P(x)$ holds.
- only uniqueness was established, with no existence check — the claim "at most one" holds, but not "exactly one."
Why: $\exists! x, P(x)$ is a conjunction of two separate facts, so both must be reported, not just whichever was easier to establish.

Common mistakes

Treating a uniqueness proof alone as sufficient, without ever checking existence.
Fixing a single object instead of assuming two arbitrary ones when starting the uniqueness half.
Concluding $\exists! x, P(x)$ from a vacuously true uniqueness argument, where no object satisfies $P$ at all.
Skipping straight to "the solution is unique" without showing the algebraic chain from $x_1$ and $x_2$ to $x_1=x_2$ .