Existence and Uniqueness Proofs

Two Questions Inside “Exactly One”

The previous tutorials treated existence and uniqueness as distinct proof tasks. In Existence Proofs, we showed how to establish that at least one object satisfies a condition by producing a witness and checking it. In Uniqueness Proofs, we learned to compare two arbitrary candidates and show that they must be equal. A statement that exactly one object has a property requires both kinds of reasoning.

Let $D$ be a domain and $P(x)$ a condition on its elements. To prove that exactly one $x\in D$ satisfies $P(x)$, we must establish these two claims:

There is an element $c\in D$ such that $P(c)$.
Whenever $u,v\in D$ both satisfy $P$, we have $u=v$.

The first claim is existence: it answers whether any object works. The second is uniqueness: it rules out two different objects that work. Neither claim implies the other. A condition may have no witnesses, or it may have several. Even when we have found one successful candidate, other candidates may also satisfy the condition.

Keep the proof obligations separate. A witness proves existence only after its property is checked. A comparison of arbitrary candidates proves at most one only after their equality is derived. An exactly-one proof needs both conclusions.

This separation is useful when reading as well as writing proofs. If a proof names a candidate and verifies it, look for a later argument that excludes other candidates. If it begins with two arbitrary solutions and proves they are equal, look for a witness or another argument showing that a solution exists. If either part is absent, the proof may establish a weaker claim than the one stated.

A Combined Theorem for Linear Equations

A simple but useful setting for joining these methods is a linear equation with a nonzero coefficient. The nonzero condition matters: it allows us to divide by the coefficient when constructing a solution, and the same condition makes the solution unique. The resulting theorem packages existence and uniqueness into one conclusion.

Theorem. Let $a,b,c\in\mathbb R$, with $a\ne0$. There is exactly one real number $x$ satisfying $$ ax+b=c. $$

Proof. We first prove existence. Since $a\ne0$, the real number $$ x_0=\frac{c-b}{a} $$ is defined. Substituting this candidate into the left side gives $$ ax_0+b =a\left(\frac{c-b}{a}\right)+b =(c-b)+b =c. $$ Thus $x_0$ satisfies the equation, so a real solution exists.

Now we prove uniqueness. Suppose $u,v\in\mathbb R$ both satisfy the equation. Then $au+b=c$ and $av+b=c$. Subtracting $b$ from each equality gives $au=c-b$ and $av=c-b$. Therefore $au=av$, or $a(u-v)=0$. Because $a\ne0$, division by $a$ gives $u-v=0$, so $u=v$. Every two real solutions are equal. We have proved both existence and uniqueness, so exactly one real solution exists.

The proof’s two parts use different starting points. For existence, we chose a candidate designed to make the equation true and then verified it. For uniqueness, we did not assume that either candidate had a particular value; we assumed only that both satisfy the equation and showed they must agree. The nonzero hypothesis is used in both parts, but the logical jobs remain distinct.

Worked Example: Exactly One Solution to a Linear Equation

Claim. There is exactly one real number $x$ satisfying $4x-7=13$.

Existence. Choose $x_0=5$. Then $$ 4x_0-7=4(5)-7=20-7=13, $$ so $5$ is a real solution.

Uniqueness. Let $u$ and $v$ be arbitrary real solutions. Then $4u-7=13$ and $4v-7=13$, so $4u-7=4v-7$. Adding $7$ to both sides gives $4u=4v$. Since $4\ne0$, division by $4$ yields $u=v$. Thus at most one real number solves the equation. Together with the verified solution $5$, this proves that exactly one real number solves it.

The witness and the arbitrary candidates have different roles. The particular number $5$ establishes existence; it is not used as an assumption in the uniqueness argument. This keeps the proof from relying on the unproved claim that every possible solution must already equal the proposed one.

Choosing an Efficient Order

There is no requirement that every proof present its parts in the same order. Some arguments establish uniqueness first and construct a witness afterward. For elementary equations, however, existence first is often convenient: it gives a concrete candidate that can be named in the conclusion. For other problems, the uniqueness argument may be short enough to establish first. What matters is that both obligations are explicit and complete.

Before writing, translate “exactly one” into a plan. Identify what could serve as a witness, then separately ask how two possible witnesses could be compared. The witness may be found by solving an equation, using a definition, or applying a result proved earlier in the course. A uniqueness argument may use algebra, inequalities, or another established theorem. Do not assume that one part automatically supplies the other.

Specify the domain and property.
State which objects are allowed and exactly what condition they must satisfy.

Plan a witness.
Choose a proposed object from the domain and determine how its required property will be verified.

Prove existence.
Write down the candidate and check the defining condition explicitly.

Compare arbitrary candidates.
Let $u$ and $v$ be arbitrary objects in the domain, and suppose both satisfy the condition.

Derive equality and combine conclusions.
Show $u=v$, then state that the verified witness and the uniqueness argument together establish exactly one.

This order also helps reveal hidden assumptions. If finding the candidate requires dividing by a quantity, establish that the quantity is nonzero. If the uniqueness argument uses cancellation, check the same issue there. If a theorem from earlier in the course supplies the witness, verify that its hypotheses apply to the present domain and condition before invoking it.

Worked Example: Exactly One Number Equidistant from Two Endpoints

Claim. For any fixed real numbers $a$ and $b$, there is exactly one real number $c$ satisfying $$ c-a=b-c. $$

Existence. Define $c_0=(a+b)/2$, which is real because $a,b\in\mathbb R$. Then $$ c_0-a=\frac{a+b}{2}-a=\frac{b-a}{2} $$ and $$ b-c_0=b-\frac{a+b}{2}=\frac{b-a}{2}. $$ The two sides are equal, so $c_0$ satisfies the condition.

Uniqueness. Let $u$ and $v$ be arbitrary real numbers satisfying the condition. Then $u-a=b-u$ and $v-a=b-v$. Rearranging the first equality gives $2u=a+b$, and rearranging the second gives $2v=a+b$. Hence $2u=2v$. Dividing by $2$, which is nonzero, gives $u=v$. Every two real numbers satisfying the condition are equal.

The verified candidate and the comparison argument together show that there is exactly one such $c$, namely $(a+b)/2$. The proof is valid even when $a=b$: in that case the candidate is $a$, and the same equations still show that no other real number satisfies the condition.

Existence Does Not Imply Uniqueness

The theorem in Existence Proofs that every pair of real numbers $a<b$ has a real number strictly between them is an existence statement. It guarantees at least one such number, but it does not assert that only one exists. In fact, whenever $a<b$, there are at least two distinct real numbers between them. This makes the difference between finding a witness and ruling out other witnesses especially clear.

Theorem. If $a,b\in\mathbb R$ and $a<b$, then there are at least two distinct real numbers $d,c$ such that $$ a<d<c<b. $$

Proof. Since $a<b$, the result from Existence Proofs gives a real number $c$ such that $a<c<b$. Apply the same result to the pair $a,c$, for which $a<c$. There is a real number $d$ such that $a<d<c$. Combining these inequalities gives $$ a<d<c<b. $$ In particular, $d\ne c$, because $d<c$. Thus $d$ and $c$ are distinct real numbers, and each lies strictly between $a$ and $b$. This proves the claim.

The proof uses an existence result twice, with different endpoint pairs. The second application is legitimate because the first application produced $a<c$, exactly the hypothesis needed to apply the result again. This is a small but important proof-writing habit: when invoking an earlier theorem, make clear why its assumptions hold in each use.

Worked Example: An Interval Does Not Have a Unique Interior Point

Claim to test. There is exactly one real number $x$ such that $2<x<8$.

Analysis. The numbers $3$ and $6$ are distinct real numbers, and $$ 2<3<8 \qquad\text{and}\qquad 2<6<8. $$ Thus at least two real numbers satisfy the condition. The proposed uniqueness claim is false. The examples establish more than existence: they directly disprove the assertion that any two numbers satisfying the condition must be equal.

Finding a number between the endpoints would be enough to show that the interval contains an element. It would not be enough to show that the element is the only one. To disprove uniqueness, it suffices to produce two distinct witnesses; to prove uniqueness, by contrast, one must show that every pair of witnesses is equal.

Domain and Hypotheses Matter

An exactly-one statement is always relative to a specified domain. The same condition may have exactly one solution in one domain and several, or none, in another. Likewise, the assumptions attached to a theorem may be essential to both the construction of a witness and the argument that excludes other candidates.

For the linear equation $ax+b=c$, the condition $a\ne0$ guarantees that $(c-b)/a$ is defined and that a solution can be isolated by division. If $a=0$, the equation becomes $b=c$, independent of $x$. When $b=c$, every real $x$ satisfies it, so there is not a unique solution. When $b\ne c$, no real $x$ satisfies it. Thus omitting the nonzero hypothesis would make the exactly-one theorem false.

The domain can change the conclusion as well. For instance, the equation $x^2=1$ has two real solutions, $1$ and $-1$. Restricted to the nonnegative real numbers, however, $1$ is a solution and the nonnegative solutions are unique: if $u,v\geq0$ and $u^2=v^2=1$, then $$ (u-v)(u+v)=u^2-v^2=0. $$ Since $u,v\geq0$ and neither can be zero, $u+v>0$. Therefore $u+v\ne0$, so the first factor must be zero and $u=v$. Both the domain restriction and the existence check are necessary to make the exactly-one conclusion.

Proof establishes	Logical conclusion	What may still be missing?
A candidate satisfies the property.	At least one object exists.	Other candidates may also satisfy it.
Any two candidates satisfying the property are equal.	At most one object exists.	There may be no candidate at all.
A verified candidate exists, and any two candidates are equal.	Exactly one object exists.	Neither existence nor uniqueness is missing.
Two distinct candidates satisfy the property.	There is more than one object.	A claim of uniqueness is false.

A polished proof should state its conclusion at the same strength as the work supports. If the argument only verifies a candidate, conclude that a solution exists. If it only compares arbitrary candidates, conclude that there is at most one. Conclude exactly one only after both obligations have been met, and keep the domain and hypotheses visible throughout.

Worked Example: Exactly One Integer Solves an Equation

Claim. There is exactly one integer $n$ satisfying $3n+2=17$.

Existence. The integer $n_0=5$ is a witness, since $3(5)+2=15+2=17$.

Uniqueness. Let $m,n\in\mathbb Z$ be arbitrary integers that both satisfy the equation. Then $3m+2=17$ and $3n+2=17$, so $3m+2=3n+2$. Subtracting $2$ from both sides gives $3m=3n$. Since $3\ne0$, cancellation gives $m=n$. Therefore any two integer solutions are equal.

The equation has a verified integer solution, and every pair of integer solutions must be equal. Hence it has exactly one integer solution. The proof states the domain explicitly: the witness is an integer, and the comparison is between arbitrary integer solutions.

Final check. For an exactly-one claim, ask: Have I named an object in the correct domain? Have I verified its property? Have I compared arbitrary candidates and derived equality? Have I used every required hypothesis? A complete proof answers all four questions.

Check Your Understanding

For each question, identify the existence and uniqueness obligations separately. When a claim concerns exactly one object, make sure both parts are addressed.

In your own words, distinguish “at least one,” “at most one,” and “exactly one” object satisfying a condition $P(x)$.
Prove that there is exactly one real number $x$ satisfying $5x+4=19$. Identify the witness and show why any two real solutions must be equal.
Suppose a proof shows that any two elements of a domain satisfying $P$ are equal, but it does not exhibit an element satisfying $P$. What conclusion has been established, and what remains to be shown to obtain exactly one?
For real numbers $a<b$, use the result that there is a real number between any two distinct real numbers to show that there are at least two distinct real numbers strictly between $a$ and $b$.
Explain why the claim that exactly one real number $x$ satisfies $1<x<7$ is false. Give two distinct witnesses.
Why is the hypothesis $a\ne0$ needed in the theorem asserting that $ax+b=c$ has exactly one real solution?