After working with nested quantifiers and their negations, we now refine existential quantification. Unique existence says not only that an element exists, but that exactly one element exists. This idea appears in inverses, identities, functions, coordinates, and solutions of equations. In this lesson, unique existence will be expressed using $\exists!$, expanded into ordinary quantifiers, and tested through examples involving quantifier order and universe dependence. :::definition[Unique Existence Quantifier] The symbol $\exists!$ is called the unique existence quantifier. The statement $$\begin{equation*} \exists!x\,p(x) \end{equation*}$$ is read as “there exists a unique $x$ such that $p(x)$.” It means that at least one element satisfies $p(x)$ and no two distinct elements satisfy $p(x)$. ::: :::definition[Formula for Unique Existence] The statement $$\begin{equation*} \exists!x\,p(x) \end{equation*}$$ is equivalent to $$\begin{equation*} (\exists x\,p(x))\land(\forall x\forall y[(p(x)\land p(y))\to(x=y)]). \end{equation*}$$ The first part says that at least one element satisfying $p(x)$ exists. The second part says that if two elements both satisfy $p$, then they must be equal. ::: Unique existence requires two checks: at least one value must work, and a second distinct value must not also work. Choose a finite universe and test either a one-variable predicate or a two-variable relation. The calculator counts successful values and reports whether the statement has no solution, exactly one solution, or more than one solution. This separates ordinary existence from uniqueness, which is the extra condition represented by $\exists!$. :::calculator[Unique Existence Tester] ::: :::solved-problem Use $\exists!$ to write: every nonzero real number has a unique multiplicative inverse. ::: :::solution Given that the universe is the set of all real numbers. The statement says that for every nonzero real number $x$, there exists a unique real number $y$ such that $xy=1$. Therefore, the symbolic form is $$\begin{equation*} \forall x[x\neq 0\to\exists!y(xy=1)]. \end{equation*}$$ If commutativity is also written explicitly, then the statement is $$\begin{equation*} \forall x[x\neq 0\to\exists!y(xy=yx=1)]. \end{equation*}$$ ::: :::solved-problem Use $\exists!$ to write: the sum of any two real numbers is unique. ::: :::solution Given that the universe is the set of all real numbers. The statement says that for every real number $x$ and every real number $y$, there exists a unique real number $z$ such that $z=x+y$. Therefore, the symbolic form is $$\begin{equation*} \forall x\forall y\exists!z(z=x+y). \end{equation*}$$ ::: :::solved-problem Use $\exists!$ to write: for each $x$-coordinate, the corresponding $y$-coordinate on the line $y=3x+7$ is unique. ::: :::solution Given that the line is $y=3x+7$. The statement says that for every real number $x$, there exists a unique real number $y$ such that $y=3x+7$. Therefore, the symbolic form is $$\begin{equation*} \forall x\exists!y(y=3x+7). \end{equation*}$$ ::: :::solved-problem Let $p(x,y):y=-2x$. The universe is the set of all integers. Determine the truth value of $\forall x\exists!y\,p(x,y)$. ::: :::solution Given that $p(x,y):y=-2x$ over the integers. The statement means that for every integer $x$, there exists a unique integer $y$ such that $y=-2x$. For any integer $x$, the value $y=-2x$ is also an integer. Moreover, one fixed integer $x$ determines exactly one integer $y$. Therefore $\forall x\exists!y\,p(x,y)$ is true. ::: :::solved-problem Let $p(x,y):y=-2x$. The universe is the set of all integers. Determine the truth value of $\exists!y\forall x\,p(x,y)$. ::: :::solution Given that $p(x,y):y=-2x$ over the integers. The statement means that there exists a unique integer $y$ such that for every integer $x$, $y=-2x$. This would require one fixed integer $y$ to equal $-2x$ for every integer $x$. For $x=0$, this gives $y=0$. For $x=1$, this gives $y=-2$. The same $y$ cannot be both $0$ and $-2$. Therefore, no such unique $y$ exists. Hence $\exists!y\forall x\,p(x,y)$ is false. ::: :::solved-problem Let $p(x,y):x+y$ is even. The universe is the set of all integers. Determine whether $\forall x\exists!y\,p(x,y)$ is true. ::: :::solution Given that $p(x,y):x+y$ is even over the integers. The statement means that for every integer $x$, there exists exactly one integer $y$ such that $x+y$ is even. This is false. For a fixed integer $x$, there are infinitely many integers $y$ having the same parity as $x$. If $x$ is even, every even integer $y$ makes $x+y$ even. If $x$ is odd, every odd integer $y$ makes $x+y$ even. Therefore $y$ is not unique. Hence $\forall x\exists!y\,p(x,y)$ is false. ::: :::solved-problem Consider $\exists!x(x>1)$. Find a universe where this statement is true and a universe where it is false. ::: :::solution Given that the statement is $\exists!x(x>1)$. This means that exactly one element $x$ in the universe satisfies $x>1$. If the universe is $\{0,2\}$, then exactly one element is greater than $1$, namely $2$. Hence the statement is true in this universe. If the universe is $\{0,2,3\}$, then both $2$ and $3$ are greater than $1$. Therefore more than one element satisfies $x>1$. Hence the statement is false in this universe. ::: :::exercise Use $\exists!$ to write each statement symbolically: every nonzero real number has a unique multiplicative inverse; the sum of any two real numbers is unique; for each $x$-coordinate, the corresponding $y$-coordinate on $y=3x+7$ is unique. Then determine the truth values of $[\forall x\exists!y\,p(x,y)]\to[\exists!y\forall x\,p(x,y)]$ and $[\exists!y\forall x\,p(x,y)]\to[\forall x\exists!y\,p(x,y)]$ for $p(x,y):y=-2x$ over the integers. ::: :::answer The symbolic forms are $\forall x[x\neq 0\to\exists!y(xy=1)]$, $\forall x\forall y\exists!z(z=x+y)$, and $\forall x\exists!y(y=3x+7)$. For $p(x,y):y=-2x$, the statement $\forall x\exists!y\,p(x,y)$ is true, while $\exists!y\forall x\,p(x,y)$ is false. Hence the first implication is false, and the second implication is true. ::: :::faq[Frequently Asked Questions] Q: What does $\exists!$ add to $\exists$? A: It adds uniqueness; exactly one element satisfies the predicate. Q: How can unique existence be expanded? A: Write existence first, then write that any two satisfying elements must be equal. Q: Does unique existence depend on the universe? A: Yes. A statement may be true in one universe and false in another. ::: :::call-to-action[Consolidate Predicate Logic] subtitle: Review quantifiers, negations, nested order, and unique existence before applying predicate logic in proofs. button: Review Section | /discrete-mathematics/mathematical-logic/predicates-and-quantifiers/open-statements-and-universe-of-discourse button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/predicates-and-quantifiers/negation-of-nested-quantified-statements :::