What Does “Unique” Mean?
In Existence Proofs, we established that an existential claim can be proved by giving a witness and verifying its required property. A statement of uniqueness asks for something more: no distinct object in the specified domain has that same property. The relevant question is not merely whether a candidate works, but whether two candidates that work must be equal.
Let \(D\) be a domain and let \(P(x)\) be a condition on its elements. The statement “at most one \(x\in D\) satisfies \(P(x)\)” means that any two elements of \(D\) satisfying \(P\) are equal: $$ \forall u,v\in D,\quad \bigl(P(u)\land P(v)\bigr)\Longrightarrow u=v. $$ The words “at most one” allow the possibility that no object satisfies \(P\). If an object does satisfy it, the condition says that every other object satisfying it is that same object.
By contrast, “exactly one \(x\in D\) satisfies \(P(x)\)” includes existence as well as uniqueness. In symbols, it means there is a witness and any two witnesses are equal: $$ \exists c\in D,\ P(c) \quad\text{and}\quad \forall u,v\in D,\ \bigl(P(u)\land P(v)\bigr)\Longrightarrow u=v. $$ The first part supplies at least one candidate. The second rules out two distinct candidates. Keeping the two parts separate prevents a common logical gap: a proof that no two different solutions exist has not necessarily shown that any solution exists.
The phrase “two arbitrary solutions” is important. If a claim permits any objects in \(D\), the proof must compare arbitrary objects from \(D\) that satisfy the condition. Checking one particular pair does not rule out other pairs. Likewise, showing that a familiar candidate works does not by itself establish uniqueness.
The General Comparison Test
The logical form of “at most one” gives a direct proof strategy. Suppose \(u\) and \(v\) are arbitrary members of the domain, and suppose both satisfy \(P\). Use those assumptions and established facts to derive \(u=v\). Since the choice of \(u\) and \(v\) was unrestricted except for the property \(P\), the result applies to every pair of candidates.
Proof. By definition, “there is at most one \(x\in D\) satisfying \(P(x)\)” says that any two elements of \(D\) satisfying \(P\) are equal. Written with variables, this is exactly the assertion that for all \(u,v\in D\), if \(P(u)\) and \(P(v)\), then \(u=v\). Thus each formulation implies the other: they express the same condition in words and in quantified form.
Although the proposition is direct, it provides a useful test when planning a proof. Translate the claim into the pairwise statement, name the two candidates, assume each meets the condition, and aim for equality. A proof should not assume from the outset that the candidates are equal; that is the conclusion to establish.
There is also a simple algebraic way this strategy frequently works. If two candidates satisfy the same equation, write down the two equations, combine them, and simplify until their equality follows. The simplification must be valid under the hypotheses. In particular, division by a quantity requires that it be nonzero.
Worked Example: At Most One Integer Solves a Linear Equation
Claim. There is at most one integer \(n\) satisfying \(5n+3=18\).
Proof. Let \(m,n\in\mathbb Z\) be arbitrary integers that both satisfy the equation. Then $$ 5m+3=18 \qquad\text{and}\qquad 5n+3=18. $$ The left sides are therefore equal, so \(5m+3=5n+3\). Subtracting \(3\) from both sides gives \(5m=5n\), and hence \(5(m-n)=0\). Since \(5\ne0\), this equality implies \(m-n=0\). Therefore \(m=n\).
We have shown that any two integer solutions must be equal, which proves “at most one.” The proof itself does not need to assume that a solution exists. In this particular example, \(n=3\) does satisfy the equation, since \(5(3)+3=18\); combining that check with the comparison proof would establish exactly one integer solution.
Proving Uniqueness by Cancellation
The preceding example illustrates a general uniqueness result for a linear equation. A nonzero coefficient can be cancelled from an equality, so two solutions cannot differ. The nonzero hypothesis is essential: if the coefficient is zero, the equation \(ax=b\) may have no solution when \(b\ne0\), or every real number may be a solution when \(b=0\).
Proof. Let \(u,v\in\mathbb R\) be arbitrary, and suppose both satisfy \(ax=b\). Then \(au=b\) and \(av=b\), so \(au=av\). Subtracting \(av\) from both sides gives $$ au-av=0, $$ or \(a(u-v)=0\). Since \(a\ne0\), division by \(a\) is valid and gives \(u-v=0\). Thus \(u=v\). Every pair of real solutions is equal, so there is at most one solution.
The proof uses the nonzero coefficient precisely at the cancellation step. If \(a=0\), that step is unavailable, and the conclusion is not true in general. For example, \(0x=0\) is satisfied by every real \(x\), so it does not have at most one solution. Stating and using the hypothesis \(a\ne0\) is part of a complete argument, not a technical detail that can be omitted.
Worked Example: Uniqueness for an Equation with a Parameter
Claim. For each fixed real \(r\), there is at most one real \(x\) satisfying \(7x-r=2x+10\).
Proof. Fix \(r\in\mathbb R\). Let \(u\) and \(v\) be arbitrary real solutions for this same fixed \(r\). Then $$ 7u-r=2u+10 \qquad\text{and}\qquad 7v-r=2v+10. $$ Subtract the second equation from the first. The terms involving \(r\) and the constant \(10\) cancel, giving $$ 7u-7v=2u-2v. $$ Subtracting \(2u-2v\) from both sides yields \(5u-5v=0\), so \(5(u-v)=0\). Since \(5\ne0\), it follows that \(u-v=0\), and therefore \(u=v\).
The parameter \(r\) is held fixed while the two possible solutions are compared. The proof establishes at most one solution for every real \(r\); it does not need to calculate a candidate solution or claim that a solution exists for every \(r\).
A useful diagnostic is to ask what the proof has actually established. If it starts by supposing \(u\) and \(v\) both satisfy the property and ends by deriving \(u=v\), it establishes at most one. If the original claim says exactly one, a separate existence argument is still required unless the proof also supplies and verifies a witness.
Uniqueness Can Depend on the Domain
The domain is part of a uniqueness claim. A condition can have at most one solution in a restricted domain but more than one in a larger domain. For example, \(x^2=25\) does not have at most one real solution: both \(5\) and \(-5\) satisfy it. If the domain is restricted to nonnegative real numbers, however, at most one solution remains.
Worked Example: At Most One Nonnegative Solution to a Square Equation
Claim. There is at most one nonnegative real number \(x\) satisfying \(x^2=25\).
Proof. Let \(u,v\in\mathbb R\) be arbitrary nonnegative numbers satisfying \(u^2=25\) and \(v^2=25\). Then \(u^2=v^2\), so $$ (u-v)(u+v)=u^2-v^2=0. $$ Because \(u,v\geq0\), we have \(u+v\geq0\). Also \(u+v\ne0\): if \(u+v=0\) with both terms nonnegative, then \(u=v=0\), which contradicts \(u^2=25\). Thus \(u+v>0\), and in particular \(u+v\ne0\). Since the product \((u-v)(u+v)\) is zero while its second factor is nonzero, its first factor must be zero. Therefore \(u-v=0\), so \(u=v\).
This proves at most one nonnegative solution. The number \(5\) is a witness because \(5\geq0\) and \(5^2=25\). Thus, with that existence check included, there is exactly one nonnegative real solution. The domain restriction matters: on all of \(\mathbb R\), the two distinct numbers \(5\) and \(-5\) both solve the equation.
This example also shows why a factorization must be handled carefully. From a product equaling zero, one may conclude that at least one factor equals zero. To conclude specifically that \(u-v=0\), the proof first had to establish that \(u+v\ne0\). Without that check, the argument would leave open the possibility that the other factor is zero.
When a Uniqueness Claim Is False
A good uniqueness proof begins by checking whether the claim could be false. Two distinct examples satisfying the condition immediately disprove “at most one.” This is not a uniqueness proof; it is a counterexample to the proposed uniqueness statement. Testing simple cases can reveal whether a domain restriction is missing or whether the property is too broad.
Worked Example: Numbers Between Two Endpoints Are Not Unique
Claim to test. There is exactly one real number \(c\) such that \(0<c<3\).
Analysis. The numbers \(1\) and \(2\) are both real, and $$ 0<1<3 \qquad\text{and}\qquad 0<2<3. $$ They are distinct because \(1\ne2\). Hence there are at least two real numbers satisfying the stated condition. The proposed uniqueness claim is false.
The Existence Proofs tutorial established that when \(a<b\), the midpoint \((a+b)/2\) lies strictly between them. That result supplies one point in the interval, but it does not say that the midpoint is the only point there. Here, for example, \(1\) is the midpoint of \(0\) and \(2\), but the interval from \(0\) to \(3\) contains many other real numbers as well.
An existence result should not be read as a uniqueness result. Constructing one object with a property answers “does at least one exist?” It says nothing by itself about how many other objects might also have that property. Conversely, showing at most one does not answer whether any object works. A proof must match the exact quantifiers and conclusion of the claim.
| What has been proved? | What it establishes |
|---|---|
| A candidate has the property. | At least one object works; this is an existence conclusion. |
| Any two objects with the property are equal. | At most one object works; this is a uniqueness conclusion. |
| A candidate works, and any two candidates must be equal. | Exactly one object works. |
| Two distinct objects both have the property. | The claim of at most one is false. |
A Reliable Structure for a Uniqueness Proof
Before writing, identify the domain and the property precisely. Then introduce two arbitrary objects that satisfy that property. This opening is not claiming that two distinct solutions exist; it is a method for testing what follows if two solutions are given. The goal is to use their defining conditions to derive equality. When the target is exactly one, plan an existence argument as well and keep the two obligations visible.
State exactly which objects are being compared and what each must satisfy.
Let \(u\) and \(v\) be arbitrary elements of the domain that both satisfy the property.
Write down the equations, inequalities, or other facts that follow for \(u\) and for \(v\).
Combine the conditions, checking any restrictions such as a nonzero quantity needed for cancellation.
Conclude at most one from the comparison. For exactly one, also establish existence of a witness.
One common error is to name a proposed solution and verify that it works, then claim it is unique. That verification proves existence, not uniqueness. Another is to compare two specific candidates rather than arbitrary ones; that rules out only the pair checked. A third is to reach an equality after dividing by a quantity that might be zero. In each case, the repair is to return to the quantified claim, make the candidates and hypotheses explicit, and justify each step toward equality.
Check Your Understanding
For each question, identify whether the task concerns existence, at most one, or exactly one. In any uniqueness argument, begin with arbitrary candidates satisfying the property.
- State in quantified form what it means for at most one real number \(x\) to satisfy a condition \(P(x)\).
- Prove that there is at most one real number \(x\) satisfying \(6x+1=19\). Which algebraic step ensures that the two candidates are equal?
- Explain why proving that \(x=4\) satisfies \(x^2=16\) does not establish that \(4\) is the only real solution.
- Prove that there is at most one nonnegative real number \(x\) satisfying \(x^2=9\). Identify why the nonnegative domain matters.
- Give two distinct real numbers satisfying \(1<t<5\). What does this show about a claim that exactly one real number satisfies that condition?
- A proof establishes that any two real solutions of an equation are equal, but gives no solution. What has been proved, and what additional fact would be needed to conclude exactly one solution?